Transcript for:
Introduction to Mathematical Thinking: Key Concepts and Summary

hello i'm keith devlin welcome to this online course on mathematical thinking the goal of the course is to help you develop a valuable mental ability a powerful way of thinking that people have developed over 3000 years what i want to do today is get you ready for the course and tell you a little bit about the way the course will work i'm doing this because for most of you this will be a very different perspective on what mathematics is apart from the final two lectures there's very little mathematical content in the course and you won't learn any new mathematical procedures but mathematical thinking is essential if you want to make the transition from high school math to university-level mathematics the quickest way to learn what mathematical thinking is is to take a course like this so by the time we're finished you should know what it is but for now let me give you an analogy if we compare mathematics with the automotive world school math corresponds to learning to drive in the automotive equivalent to college mathematics in contrast you learn how a car works how to maintain and repair it and if you pursue the subject far enough how to design and build your own car the only prerequisite for the course is completion or pending completion of high school mathematics that means many people could take the course and find it valuable in particular a key feature of mathematical thinking is thinking outside the box in contrast the key to success in high school math was to learn to think inside the box it's because thinking outside the box is such a valuable ability in today's world that this course could be valuable to many people but my primary student is someone in their final year of high school or their first year at college or university who's thinking of majoring in mathematics or a math dependent subject if that's you then you'll probably find the transition from high school mathematics to college level pure abstract mathematics difficult i certainly did and so did most mathematicians i know not because the mathematics gets harder once you've successfully made the transition i think you'll agree that college math is in many ways easier what causes the problem is the change in emphasis at high school the focus is primarily on mastering procedures to solve various kinds of problems that gives the subject very much the flavor of a cookbook full of mathematical recipes thinking inside boxes at university the focus is on learning to think a different way to think like a mathematician thinking outside the box that's not true of all college math courses those designed for science and engineering students are often very much in the same vein as the courses you had in high school it's the courses that form the bulk of the mathematics major that are different but some of those courses are usually required for more advanced work in science and engineering so if you are a student in those disciplines you may also find yourself faced with this different kind of mathematics if you did well at math and school you probably got good at recognizing different kinds of problems so you can apply different techniques you learned at university you have to learn how to approach a new problem one that doesn't quite fit any template you're familiar with it comes down to learning how to think about a problem in a certain way the first key step is learn to stop looking for a formula to apply of a procedure to follow approaching a new problem by looking for a template a worked example in a textbook or presented on youtube and then just changing the numbers often won't work sometimes it will so all that work you did at high school won't go to waste but it isn't enough for many of your college math courses if you can't solve a problem by looking for a template to follow or a formula to plug some numbers into or a procedure to apply what do you do the answer is you think about the problem a certain way not the form of the problem that's probably what you were taught to do at school and it served you well there rather you have to look at what the problem actually says that sounds as though it ought to be easy but most of us initially find it extremely hard and very frustrating it doesn't come quickly or easily you have to work at it you're going to have to accept going a lot slower than you're used to most of the time you won't feel as though you're making any progress your goal has to be understanding not doing you should definitely try all the exercises they're there to aid your understanding should also work with others few of us can master this crucial shift in thinking on our own that part's crucial a lot of what we'll be doing is not so much focused on right and wrong but on learning how to think about a problem yeah sure at the end of the day solutions are right or wrong but usually there are many different right answers or different ways to the right answer and many wrong ones when you're learning how to think mathematically it's how and why you've got something right or wrong that's important the only way to find that out to find out how well you're doing is for somebody else to look at your attempt and critique your work it's not possible to automate the grading process maybe one day artificial intelligence will have advanced far enough for a course like this to be automated although frankly i doubt it but right now you need feedback from other people for a regular class here at stanford to professor and the graduate student tas grade students work and provide feedback with an open online course like this where there are many thousands of students that's not possible so we have to go about things a different way i've designed the course so that the benefit comes primarily from doing the work and discussing it with other students getting it right is important in mathematics but in a massively open online course a mooc like this there's no way to guarantee that incidentally not being sure if we are right until others have seen our work is very familiar to we mathematicians even very famous mathematicians have had the experience of thinking they've solved the problem and writing it writing it up writing up their solution and sending it off for publication only for an anonymous referee to find an error in mathematics there is such a thing as right and wrong but deciding between them can be very difficult so even the professionals have to live with never being sure whether they're right or not part of this introductory session is a reading assignment it gives you a bit of a history that should explain why today's mathematics students need a course such as this in lecture one there'll be a short quiz on the reading let me say a few words about the quizzes if you were in one of my regular physical classes i'd talk with you to find out if you had understood the material sufficiently to progress but with a massively open online course a mooc that's not possible you have to monitor yourself the quizzes are one way to help you do that i'll say a little bit more about the quizzes later but before you start lecture one please read a file called background reading it's just over six pages long it's a pdf file so you can download it and read it offline well now you know what an intellectual quiz looks like the next thing i suggest you do is view the video finding your way around the course website it will point you to some particularly important things to check out before you start the course i'm going to start off with a question what is mathematics that might seem strange given you've probably spent several years been taught math but for all the times schools devote to the teaching of mathematics very little if any is spent trying to convey just what the subject is about instead the focus is on learning and applying various procedures to solve math problems well that's a bit like explaining soccer by saying it's a series of maneuvers you execute to get the ball into the goal both accurately describe various key features but they miss the whats and the why of the big picture if all you want to do is learn new mathematical techniques to apply in different circumstances then you can probably get by without knowing what math is really about but if that's the case then this isn't the course for you one thing you should realize is that a lot of school mathematics dates back to medieval times with pretty well all the rest coming from the 17th century at the very latest virtually nothing from the last 300 years has found its way into the classroom yet the world we live in has changed dramatically in the last 10 years let alone the last 300. most of the changes in mathematics over the centuries were just expansion but in the 19th century there was a major change in the nature of mathematics first it became much more abstract second the primary focus shifted from calculation and following procedures to one of analyzing relationships the change in emphasis wasn't arbitrary it came about through the increasing complexity of what became the world we are familiar with procedures and computation did not go away they're still important but in today's world they're not enough you need understanding in our education system the change in emphasis in mathematics usually comes when you transition from high school to university in the 1980s i was one of a number of mathematicians who advocated a new meme to capture what mathematics is today the science of patterns according to that description the mathematician identifies and analyzes abstract patterns can be numerical patterns patterns of shape patterns of motion patterns of behavior voting patterns in a population patterns of repeating chance events and so on there can be either real or imagined patterns visual or mental static or dynamic qualitative or quantitative utilitarian or recreational they can arise from the world around us from the pursuit of science or from the inner workings of the human mind different kinds of pattern give rise to different branches of mathematics for example arithmetic and number theory study the patterns of counting and number geometry studies the patterns of shape calculus allows us to handle patterns of motion logic studies patterns of reasoning probability theory deals with patterns of chance topology studies patterns of closeness and position fractal geometry studies the self-similarity found in the natural world and so on and so on and so on one major consequence of the increasing abstraction and complexity of the mathematics in the 19th century was that methods developed to solve important real-world problems had consequences that were counter-intuitive let me give you one example it's called the banach taski paradox it says you can in theory take a sphere and cut it up in such a way that you can reassemble it to form two identical spheres each the same size as the original one and that wasn't the only surprise mathematicians had to learn to trust the mathematics above our intuitions just as physicists did with the discovery of relativity theory and quantum mechanics of course if you're going to trust mathematics above intuition and common sense you'd better be sure the math is right this is why the mathematicians in the 19th and early 20th centuries developed the precise way of thinking i'm calling mathematical thinking what this course is about now for that quiz i promised you during my course introduction did you read that short document called background reading if not i suggest you pause the video right now and then come back after you've looked at it well the correct answer is money for the first question people certainly measured land um and they used various kinds of yardstick but they didn't use numbers and they certainly counted seasons but you can count without numbers you can count with notches in sticks and you can count with uh with pebbles and so forth our ancestors only invented abstract numbers in order to get money at least that's the best available evidence that we have and that we think happened about 10 000 years ago for the second question topology studies patterns of closeness if you thought it was geographical terrain you were confusing topology with topography how did you do well for this one the main focus in the 19th century became concepts and relationships that was a revolution in mathematics that took place in germany and for this question at least according to me and many of my colleagues i should point out the main mathematical ability today is being able to adapt to old methods and develop new ones yes use of technology is important yes you need to master your basic skills but the crucial ability in today's world is adapting old methods or developing new ones okay how did you do on that quiz yup i know it wasn't a math quiz it was really there just to get you used to the in lecture quiz format i gave the rationale for those quizzes on that very first quiz as i wrote there the intention is that you should find the answers to the quizzes immediately obvious if you do that's a sign that you are sufficiently engaged and not trying to move too fast if you find you have to spend time on a quiz question or go back and look at the lecture again then you will know that you are not engaging sufficiently closely and you need to slow down the secret to this entire course is reflection not completion doing all the in-lecture quizzes is required to get a certificate of completion for the course but does not contribute to your final grade as those of us giving and taking these first generation moocs have discovered in-lecture quizzes are valuable without the regular feedback from an instructor or a ta neither of which is possible with a course having many thousands of students the individual student has to monitor his her own progress the quizzes those simple have proved to be very useful in that regard oh by the way i know that some of you don't have constant broadband access and you have to download the videos and watch them locally so you can't do all the quizzes well they're all reproduced on the course website with minor changes so you can do them there when you go online they're grouped by lecture you should definitely try to do the quizzes as close in time to watching the lecture as that will help you decide if you are moving at a good pace which for this course means not too fast we professional mathematicians despair school systems that impose tight time limits on completing mathematics tests and encourage fast working real math takes time before i dive into the first topic let me say a couple of things about what to expect as we get into the course the main thing to realize is that a lot of what we do probably won't seem like doing math since the focus is on how to think mathematically not how to apply standard techniques to solve problems for most of you if you've had a fairly standard high school mathematics education this is a big shift second even when we do math that looks familiar to you you'll spend most of the time thinking rather than writing things down if at all possible you should work with somebody else or in a small group learning to think a different way is a lot harder than learning a new technique and few of us can do it alone if you find you need help or if you think there's a mistake in something i've said or written the first thing to do is discuss it with your study group or in the course discussion forum if after discussion your group thinks there's a mistake post it on the forum to see what others think that way issues that arouse a lot of interest will rapidly rise to the attention of me and my teaching assistants okay now we've gotten you orientated and disposed of the preliminaries let's get down to work in earnest the first topic is getting precise about how we use language the american melanoma foundation in its 2009 fact sheet states that one american dies of melanoma almost every hour like many mathematicians i can't help being amused by such claims not because we mathematicians lack sympathy for a tragic loss of life what we find amusing is that if you take the sentence literally it does not at all mean what the amf intended what that sentence actually claims is that there is one american let's call this person x who has the misfortune to say nothing of the remarkable ability of almost instant resurrection to die of melanoma every hour the sentence the amf writer should have written is this almost every hour an american dies of melanoma you see the difference misuses of language like this are fairly common so much so that they really aren't misuses everybody reads the first sentence as having the meaning captured accurately by the second such sentences have become figures of speech apart from mathematicians and others whose profession requires precision of language hardly anyone ever notices that the first sentence when read literally actually makes an absurd claim when people use language in everyday context to talk about everyday circumstances this share a common knowledge of the world and that common knowledge can be relied upon to determine the intended meaning but when mathematicians use language in their work there often is no shared common understanding moreover in mathematics the need for precision is paramount that means that when mathematicians use language in doing mathematics they rely upon the literal meaning they have to as a result mathematicians have to be aware of the literal meaning of the language they use this is why beginning students of mathematics in college are generally given a crash course in the precise use of language yeah that might sound like a huge undertaking given the enormous breadth of language but the use of language in mathematics is so constrained that the task actually turns out to be relatively small modern pure mathematics is primarily concerned with precise statements about mathematical objects mathematical objects are things like integers real numbers sets functions etcetera here are some mathematical statements there are infinitely many prime numbers for every real number a the equation x squared plus a equals 0 has a real root the square root of 2 is irrational if p n denotes the number of primes less than or equal to the natural number n then as n becomes very large p of n approaches n divided log e of n not only are mathematicians interested in statements like these they are above all interested in knowing which statements are true and which are false the truth or falsity in each case is demonstrated not by observation or measurements or experiments as in the natural sciences but by a proof in this course we look at some different ways of proving statements in the case of my four examples 1 3 and 4 are true but 2 is false let me show you a proof of the first statement it's due to the ancient greek mathematician euclid who lived around 350 bce we're sure that if we list the primes p1 p2 p3 etc the list continues forever well suppose we've reached some stage n so we've listed p1 p2 p3 up to pn can we find another prime to continue the list if we can always find another prime then the list goes on forever and we've shown that there are infinitely many primes well can we well here's a clever trick that euclid described in his famous book elements in 350 bc look at the number n defined as follows set n equal to p1 times p2 times p3 all the way up to pn multiply them all together and add one clearly n is bigger than pn so if n happens to be prime we found a prime number bigger than p n and we can continue the list if n is not prime then it's going to be divisible by a prime say p now p cannot be any of the primes p1 p2 p3 up to pn because if you divide any of those primes into n that prime will divide into this part and then there's a remainder of 1. so p cannot be any of those why because dividing them leaves a remainder of one so p is bigger than p n that means we've found a prime number bigger than p n either way if n is prime or if it's not prime we've shown that there's a bigger prime than p n which means the list can always be continued and that proves that there are infinitely many primes let's just take another look at what we've done we start with a list of all of the primes and we try to list all of the primes we have to show that we can do that and keep going okay so we start with a list of the primes we assume we've reached some stage n n could be a 10 a million a billion a trillion whatever we've reached some stage and we show that we can always find another prime bigger than the last one how do we do that well there's a clever idea we look at this number big n which means you multiply the first little n primes that's a little in there you multiply them together and you add one n is certainly bigger than the last one in in that sequence so if big n is prime then it's a prime bigger than p n now we're not saying that big n would be the next prime after pn in fact it almost certainly wouldn't be because big n is a lot bigger than these numbers so this number is a lot bigger than p n so this isn't going to be the next prime almost certainly but that doesn't matter we've shown that there is another prime and whatever the next prime is we'll put it onto the list the alternative was that n wasn't prime in which case it's divisible by a prime and we call it p now that prime p can't be any of these why well this is why we defined n the way we did if you define if you divide n by any of these primes you're left with a remainder of one so the prime that divides n can't be any of these it must be a different one if it's a different one it's bigger than p n so again we found a prime that's bigger than p n is this prime p the next prime after pn well it might be but there's no reason to assume it is and it doesn't matter the point is we found a prime bigger than p pn so once again the list can be continued either way the list can be continued this is the clever trick that makes it work defining in that way and we define in that way to make sure that if there's a prime dividing big n it won't be equal to any of those theorems proved there are infinitely many primes how about that so we've proved the first of our four examples there are infinitely many prime numbers that one's true what about the second one well it says that for every real number a the equation x squared plus a equals zero has a real root that turns out to be false to show that it's false all you need to do is find a single real number a for which the equation does not have a real root well why don't we just take a equals 1 then we know that the equation x squared plus one equals zero does not have a root because there's no number that you can square no real number that you can square such that when you add one to it you get zero the square of any real number is positive and you take a positive number and add one you end up with a positive number because there is a real number a for which the equation does not have a root that shows that the statement that for every real number there's a root is false what about number three well that one turns out to be true and we're going to prove that's true later in the course the fourth one is rather complicated looking statement about the distribution of the prime numbers that's a very famous result that was proved at the uh about just over 100 years ago at the end of the 19th century it's known as the prime number theorem so this one is true the prime number theorem and there we are well the correct answer is false the proof certainly involved looking at this number but it didn't require that this number be prime it was it was it was rather different so if you've got the if you thought the answer was true here i would strongly advise you to go back and look at that proof again because we most definitely did not assume that this number was prime clearly before we can prove whether a certain statement is true or false we must be able to understand precisely what the statement says above all mathematics is a very precise subject where exactness of expression is required this already creates a difficulty since words tend to be ambiguous and in real life our use of language is really precise in particular when we use language in an everyday setting we often rely on context to determine what our words convey for example an american can truthfully say july is a summer month but that would be false if spoken by an australian the word summer means the same in both statements namely the hottest three months of the year but it refers to one part of the year in america and another in australia in everyday life we use context on our general knowledge of the world and of our lives to fill in missing information in what is written or said and to eliminate the false interpretations that can result from ambiguities for example we would need to know something about the context in order to correctly understand the statement the man saw the woman with a telescope who had the telescope the man or the woman ambiguities in newspaper headlines which are generally written in great haste can sometimes result in unintended but amusing second readings among my favorites that have appeared over the years are sisters reunited after 10 years in checkout line at safeway large hall appears in high street city authorities are looking into it mayer says bus passengers should be belted okay to systematically make the english language precise so people can communicate effectively by defining exactly what each word is to mean would be an impossible task it would also be unnecessary since people generally do just fine by relying on context and background knowledge but in mathematics things are different precision is crucial and it cannot be assumed that all parties have the same contextual and background knowledge in order to remove ambiguities moreover since mathematical results are regularly used in science and engineering the cost of miscommunication through an ambiguity can be high possibly fatal at first it might seem like a herculean task to make the use of language in mathematics sufficiently precise but fortunately it turns out to be very doable though a bit tricky in places and what makes it possible is the special highly restricted nature of mathematical statements almost every key statement of mathematics the axioms conjectures hypotheses and theorems is a positive or negative version of one of four linguistic forms object a has property p every object of type t has property p there is an object of type t having property p if statement a then statement b or else the statement is a simple combination of sub statements of these forms using the connecting words which we call combinators and or and not for example 3 is a prime number or 10 is not a prime number every polynomial equation has a complex root or it's not the case that every polynomial equation has a real root there is a prime number between 20 and 25. there's no even number beyond two that is prime if p is a prime of the form four n plus one then p is a sum of two squares in their everyday work mathematicians often use more fluent variants of such statements such as not every polynomial equation has a real root or no even number is prime except for two but those are just that variance incidentally the final statement about the primes of the form four n plus one is a celebrated theorem of the early 19th century mathematician carl friedrich gauss the ancient greek mathematicians seem to have been the first to notice that all mathematical statements can be expressed using one of these simple forms and they made a systematic study of the key language terms involved namely and or not implies for all and their exists the greek mathematicians provided universally accepted meanings of these key terms and analyze their behavior when that study is carried out in a mathematically formal way it's known as formal logic or mathematical logic the study of mathematical logic is a well-established branch of mathematics studied and used to this day in university departments of mathematics computer science philosophy and linguistics it gets a lot more complicated than the original work carried out in ancient greece by aristotle and his followers and by the stoic logicians but that's well outside our present interest it's time for another quiz let me stress what i wrote to accompany the first in lecture quiz the quizzes are designed so that if you're progressing in a manner that will get you through the entire course you'll find the quiz is easy in most cases the answer will be obvious their role is so that if you find you're getting some quiz questions wrong or if you have to check back through the lecture before you answer then you'll know you need to put in more focus on mastering the material you have to do all the quizzes to get the course completion certificate but your course grade comes from the weekly problem sets and the final exam so here's the quiz the ancient greeks were the ones who began the formal study of language and reasoning that became the branch of mathematics known as formal logic and that brings us to the end of the first lecture your next task is to complete course assignment one you probably find it easier to print off the pdf file and work on it when you have time it's really hard to say how much time you should allow for the course assignments as people work at very different rates the first one is meant to be pretty light so i'm guessing that an hour or so should be enough for most of you but i really can't be sure if you're ever uncertain about anything in a lecture a reading or an assignment discuss it with your study team or go on to the course discussion forum in fact even if you don't feel uncertain you should discuss the course material regularly with other students many students do fine at high school working on their own in fact the best students usually work that way all the time i did when i was at high school but when i got to university i soon discovered that working alone was the worst possible strategy university mathematics is not focused on learning procedures to solve problems it's about thinking a certain way and mastering a new way of thinking is best learned by working in groups in a mooc where you don't have an instructor or a ta to regularly check on your progress working with a study group is pretty well essential the folks at core server are working on developing their platform to facilitate group work but that's not available to us now so you'll have to form and manage your study groups yourself of course the the course discussion forum provides a starting point but you should use whatever medium you prefer to keep in touch with other students forming a group on facebook google groups yahoo etc are all obvious ways to do it and you can share files and work together on google docs check out the course website for information recommendations suggestions and updates and keep in touch with the rest of the class by following activities on the forums good luck and i'll see you next time welcome to the second lecture i hope you're making progress with the first assignment i'll post my answers to some of the assignment questions later consult the course schedule on the website and check regularly for announcements of any changes you should not expect to solve all the problems in an assignment in a single session or even before the next lecture what you should do before the next lecture is attempt each question that's what i mean when i say complete the assignment remember the goal of this course is to acquire a certain way of thinking not to solve problems by a given deadline the only way to develop a new way of thinking is to keep trying to think in different ways without guidance that would be unlikely to get you anywhere of course but the point of a course like this is to provide that guidance and the assignments are designed to guide your thinking attempts in productive directions okay let's proceed as a first step in becoming more precise about our use of language in mathematical contexts we'll develop precise unambiguous definitions of the key connecting words and or and not the other terms we need to make precise implies equivalent for all and their exist are more tricky and will handle them later let's start with and we often want to combine two statements into a single statement using the word and so we need to analyze the way the word and works the standard abbreviation that mathematicians use for this is an inverted v known as the wedge sometimes you'll see the old familiar ampersand used but i'm going to stick to the common mathematical practice of using the wedge for example we might want to say pi is bigger than 3 and less than 3.2 we could do this as follows we could write pi is bigger than 3 and pi is less than 3.2 in fact for this example where we're just talking about the position of numbers on the real line there's an even simpler notation we would typically write 3 less than pi less than 3.2 but as an example illustrating the use of the word and this one is fine what does it mean well if we have two statements phi and psi phi and psi means that they're both true the official term for an expression like this is it's the conjunction of phi and psi relative to the conjunction the two constituents phi and psi are called the conjuncts of fine psi what are the circumstances under which a conjunction phi and size two well if phi and psi are individually true then the conjunction phi and psi will be true under what circumstances will phi and psi be false well if either phi is false or psi is false or they're both false this might seem very self-evident and trivial but already this definition leads to a rather surprising conclusion and here it is according to our definition phi and psi means the same as psi and phi they both mean that phi and psi are both true in mathematical pilots conjunction is commutative but that's not the case for the use of the word and in everyday english for example john took the free kick and the ball went into the net that doesn't mean the same as the sentence the ball went into the net and john took the free kick they're both conjunctions and the two conjuncts are the same one of them is john took the free kick the other one is the ball went into the net but anyone who's familiar with soccer realizes that these two sentences have very different meanings the fact is in everyday english the word and is not commutative sometimes it is but not always let's see what you make of this one let a be the sentence it rained on saturday and let b with a sentence it snowed on saturday question does the conjunction a and b accurately reflect the meaning of the sentence it rained and snowed on saturday well what do you think yes or no although i can think of situations in which the answer would be no in general i would be inclined to say the answer is yes a useful way to represent a definition like this is with a propositional truth table what we do is we list the component statements in this case will be phi and psi and they're going to go together to make the conjunction phi and psi and now we're going to draw a table that lists all the possible truth false combinations for phi psi and phi and psi so let me see phi could be true and psi could be true or phi could be true and psi could be false or phi could be false and psi could be true or phi could be false and psi could be false the next step is to list in the final column t or f according to our definition of what the conjunction means why don't you see if you can fill in trf in each of those four boxes to represent the definition of phi and psi that we've given according to the definition and psi is going to be true whenever phi is true and psi is true that's the first row so there's a t going to go here but that's the only condition under which phi and psi is true in all other circumstances it's false so the entries for these are all f so in one simple table we've captured the entire definition of phi and psi this emphasizes the fact that the truth of a conjunction depends only on the truth or falsity of the two conjuncts the definition was entirely in terms of truth and falsity what phi and psi meant was irrelevant it was only about truth and falsity that's going to be the case for all the definitions that we're going to give in order to make language precise they're going to depend upon truth or falsity not upon meanings or logical connections now let's look at the combinator o we want to be able to assert that statement a is true or statement b is true for instance we might want to say a is greater than 0 or the equation x squared plus a equals zero has a real root or maybe we want to say a b equals zero if a equals zero or b equals zero those are both statements that we get when we combine two substatements with the word or both statements are in fact true but there's a difference between them the meaning of or is not the same in the first sentence as it is in the second sentence in the first sentence there's no possibility of both parts being true at the same time either a is going to be positive or this equation will have a real root they can't both occur if a is positive then this equation does not actually have a rare root in the case of the second sentence they could both occur together to get a b equals zero it's enough if a0 it's enough if b is zero or they can both be zero so these two are different in the first case we have an exclusive or in the second case we have an inclusive or incidentally it doesn't matter if you try to enforce the the exclusivity by putting an either in front of it if you look at the way the word either operates if you say either this or that then what happens is that the either simply reinforces an exclusive or if one happens to be there in the case of the second one you could say a b equals zero if either a equals zero or b equals zero and in fact that doesn't enforce the exclusivity at all we just accept the fact that they could both be true in other words the word or in in everyday english is ambiguous and we rely upon the context to disambiguate in mathematics it's different we simply can't afford to have ambiguity floating around we have to make a choice between either the exclusive or or the inclusive or and for various reasons it turns out to be more convenient in mathematics to adopt the inclusive use the mathematical symbol we use to denote the inclusive or is a v symbol it's known as the disjunctive symbol so given two sentences phi and psi phi v psi means phi or psi or both this sentence phi or psi is called a disjunction of phi and psi and relative to the disjunction the constituents phi and psi are called the disjuncts remember phi or psi in mathematics means at least one of those two is true they could both be true for example the following rather silly statement is true three is less than five or one equals zero i can't imagine a mathematician actually writing that down except as an example as i'm doing now silly examples like this are actually quite useful in mathematics because they help us understand exactly what a definition means this thing is true even though one of the disjuncts is apparently false so this emphasizes the fact that for a disjunction to be true all you need to do is find one of the disjuncts which is true it doesn't matter if one or more of the other disjuncts is is patently false okay let's see how well we do understand that here's a quick quiz let a be the sentence it will rain tomorrow and let b be the sentence it will be dry tomorrow here's the question does the disjunction a or b accurately reflect the meaning of the sentence tomorrow it will rain or it will be dry all day well what do you think the answer is clearly no if that comes as a surprise to you uh you need to think about the definition of oh a little bit longer and see what's going on here i'll leave you to that one to wrap up this discussion of disjunction let's see if we can complete the truth table for phi or psi okay if you got this one right your truth table should look like this true true true false the disjunction is true if both are true if one is true or if the other is true the only time when a disjunction is false is when both disjuncts are false okay so now we've sorted out the meaning of the word or the next word i want to look at is not if psi is a sentence then we want to be able to say that psi is false so given psi we want to create the sentence not psi the standard abbreviation mathematicians use today is this symbol which is like a negation symbol with a little vertical hook older textbooks will you'll find with the we'll use the tilde that's not the one i'm going to use i'm going to stick with the modern notation this sort of negative sign with the hook and we call this the negation of psi if psi is true then the negation of psi is false and if psi is false the negation of psi is true we often use special notations in particular circumstances for example we would typically write x not equal to y instead of not x equals y but you have to be a little bit careful for example i would write not the case a less than x less than or equal to b you might be tempted to write something like a not less than x not less than or equal to b i would i would advise against that that one is better than this one this one is completely unambiguous it means that it's not the case that x is between a and b in that fashion this one well it could mean i mean you could agree that it means that but you've it's really ambiguous as to exactly what's going on here so i would say avoid things like that use something like this we should always go for clarity in the case of mathematics remember the whole point of this precision that we're trying to introduce is to avoid ambiguities to avoid confusions because in more advanced situations all we're going to have to rely upon is the language and and then we need to make sure that we're using language in a non-ambiguous and reliable way negation might seem pretty straightforward and in many ways it is but it's not trivial if we took something like not the case that pi is less than three then that's pretty straightforward that means pi is greater than or equal to three okay that's easy no problems there let me give you one that's not quite so obvious look at this sentence all foreign cars are badly made what's the negation of this sentence let me give you four possibilities possibility one or possibility a all phone cards are well made possibility b all phone cards are not badly made possibility c at least one foreign car is well made possibility d at least one foreign car is not badly made well i'm not giving you this as a quiz but i would like you to think for a minute as to which one of these you think is the negation of that original sentence or maybe you think it's something else maybe you think it's something to do with domestic cars domestic is after all the negation of fallen so what do you think it is well let's look at them a is actually a very common one for beginners to to pick i've been teaching this material for many years now it's one of the examples i've always gave it's in in the textbook i've written for this course and previous textbook i've written and this one is a common answer that i that i often get but if you think about the what the sentence really means it's obviously not this one this is not the negation why is the original sentence true no of course it's not there are many good cards that the phone made okay so it's not the case that all foreign cars are badly made so this sentence is in fact a false sentence we know that just by our knowledge of the world if that sentence is false then it's negation is going to be true no this isn't true it's not the case that all foreign cars are well made that's false so that can't be the negation what about b same reasoning that can't be the negation because it's simply not the case that all fallen cars that all foreign cars are not badly made okay those are false statements these are false so they can't be the negation of a false sentence the negation of a false sentence is going to have to be true so whatever the negation of this original sentence is that negation will have to be something that's true and we know what's true and false in terms of cars being well made well is this one true yeah that's true is this one true well these are both true so these are both possibilities for the negation of that and this is still not a quiz but i'm going to leave you for a little while to think about this one which one of these do you think actually is the the negation of this we'll come back to this i'm going to introduce some some formal notation from sort of algebraic notation and eventually we'll be able to reason precisely to say which one of these two here or maybe a different thing is the actual negation of this but let me stress a point i made a minute ago and didn't write anything down look at the following sentence all domestic cars are well made i've actually had students over the years who have thought that that was in fact the negation of this i know why they're saying that because they're saying this says something about all foreign cars and this says something about all cars that are not foreign so there is a sort of negation going on between these two but it's not the negation of the original sentence how do i know it's not the negation of the original sentence because the original sentence is false therefore whatever the negation is is going to have to be true but this isn't true this is also false and because this is false it can't possibly have been a negation of the original sentence and in fact this one uh really falls a long way from being the negation of that for the following reason the original sentence is about foreign cars that's what it's talking about it has nothing to do with domestic cars it's purely talking about foreign cars so the negation can only possibly talk about foreign cars these were good candidates for the negotiation because they talked about foreign cars this one isn't even in the ballpark for billigation because it's not talking about foreign cars it's talking about domestic cars negating a word in a sentence is not at all the same as negating the sentence so this one is a really bad choice let me finish with a very simple quiz let me ask you to fill in the truth table for negation this one's a much simpler table because there's only one statement involved phi and then we're going to negate it so very simple truth table what do you think the values are yup this one was an easy one if phi is true the negation is false if phi is false the negation is true and with that you should be in a position to complete assignment two that last example about the negation of the sentence all foreign cars are badly made should i think illustrate why we're devoting time to making simple bits of language precise to figure out what the correct negation is we relied on our knowledge of the everyday world that's fine for statements about the everyday world we're familiar with but in a lot of mathematics we're dealing with an unfamiliar world and we can't fall back on what we already know we have to rely purely on the language we use to describe that world when we've taken our study of language far enough we'll be able to look at that foreign car statement again and use rigorous mathematical reasoning to determine exactly what its negation is well that brings us to the end of the first week how are you getting on for most of you this will seem like a very strange course and certainly won't look much like mathematics that's because you've only been exposed to school math this course is about the transition to university level mathematics which in some ways is very different there isn't much material and as a result the lectures are short i'm not providing you with new methods or procedures i'm trying to help you learn to think a different way doing that is mostly up to you it has to be if you're at all like me and pretty well every other mathematician i know you're going to find it hard and frustrating and it's going to take some time you definitely need to connect to other students and start working together you're able to scan pages of work into pdf or just use your smartphone to take good clear photos of your work i advise you to start showing your work to other students to get their feedback send images as email attachments put them on google docs or upload them to whatever networking site you choose you should definitely attempt all the assignments that i give out after each lecture doing those assignments both on your own and in collaboration with others is really the heart of this course yeah sure you can watch the lecture several times but you'll find that it almost never tells you the answer or even how to get the answer in the way that you're familiar with from high school it's like learning to ride a bike someone can ride up and down in front of you for hours telling you how they do it but you won't learn to ride from watching them or having them explain it to you you have to keep trying it for yourself and failing until it eventually clicks this is a very different way of learning than you're used to at least in mathematics as well as the assignments there's also a weekly problem set the problem sets comprise assignment questions that count directly towards your grade each set as a submission deadline because this course is designed for many thousands of students it's impossible for me or my tas to look at everyone's work and provide feedback so we have to rely on automated grading this means that the questions are posed in multiple choice formats but these are not at all like the in lecture quizzes those are supposed to be answerable on the spot the problem set questions will require considerable time this is not ideal for the material in this course whether you get particular questions right or wrong is pretty insignificant it's your thinking process that's important but we can't check that automatically asking you to answer multiple choice questions is like checking your health by taking your temperature it tells us something and can alert you and us that something is wrong but it's pretty limited still checking temperatures is better than nothing and the same is true for the problem set grading what i'd like you to do is to try to grade your own work and that of others in whatever study group you form and you should definitely try to get into one there's a mechanism for doing this it's called calibrated peer review it's been tried a number of times and does appear to offer learning benefits it's the method we're going to use to grade the final exam so anyone who wants to earn a distinction in this course is going to have to learn and use it eventually the advantage of using it now is that it does yield positive learning outcomes check out the description on the course website and then give it a shot when assignment one most of the questions are just there for you to think about and uh to discuss with your fellow students and i'm going to leave you to to do that and uh and sort them out on your own uh discuss them on the on the forums or with their with whatever group you put together and please do put together a group what i will do is say a little bit about question eight well question eight is related to euclid's proof that there were infinitely many primes another crucial point in that proof we looked at this number where p1 through pn uh enumerates the first n primes we multiply them together and we add one and i mentioned during the proof that this number is not always prime and question 8 asks how do you how would you prove that it's not always prime and the answer is you have to find a number n such that when you do multiply the first in primes and add one it's not prime well how would you go about doing that well the most obvious way is to just start searching so you would start by looking at the first two primes say multiply them together add one you get seven well that is prime try another one two times three times five add one that's 31. that's also prime let's try another one two times three times five times seven plus one that's two hundred and eleven that's also prime well at this point you're probably starting to lose heart but if you keep on just two more primes two three five seven eleven and thirteen multiply them together and add one you end up with thirty thousand and 31 which is not prime because it's the multiple of 59 and 509 okay so here we've got an example of a number of this form which is not prime and in order to show that not every number of that form is prime all you have to do is find a single single example of a number of that form which isn't prime and we've done it so that's the answer to question eight that's how you prove that those numbers are not always prime okay well that's really all i want to say about assignment one um assignment two has a little bit more meat to it in the sense that there are more things that i'll need to show you okay let's see what we've got well number one has been done for you uh number two the simple word of like that is to say seven less than or equal to p less than twelve the next one we would write as five less than x less than seven the next one well if x is let me see if x is less than four then it's automatically less than six so the second conjunct here is super airflows we could just write that as x less than four what about the next one well y squared less than nine means uh let me see negative three is less than y is less than three that's what the the first what the that's what the second conjunct means okay but if y satisfies this condition then certainly y is less than four so the first conjunct is superfluous and the only one that counts as a second one so we could simplify that as y squared less than 9. okay that was just our thinking on the way what about this one well x is greater than or equal to zero and it's less than or equal to zero there's only one possibility and that's x equals zero well for number three in order to show that the conjunction is true what you would do is show that all phi 1 phi 2 etc up to phi n the true and for number four to show that the conjunction is false is you show that one of phi 1 phi 2 phi n is false so you have to find one of these at least one of these which is false and then it falls as follows that the conjunct is false okay well that's that one okay well with part a if pi is bigger than 10 then it's automatically bigger than three so in terms of the disjunct pi bigger than 3 dominates that's that says more if we draw a picture we've got 0 we've got 3 we've got 10 and the first one says that x is to the right of this point and the second disjunct says that x is to the right of that point and the disjunction will be true if at least one of them is true and at least one of them will be two if we start at three first of all only the first disjunct is correct and then when we get beyond this point both these chunks are correct so it's the first one that works this says that either x is less than zero or x is greater than zero so the simplest way to write that is to say that x is not equal to zero it's either less than zero it's negative or it's bigger than zero it's positive this says at x equals zero or x is greater than zero the standard abbreviation for that is x greater than or equal to zero and once you've got x greater than or equal to zero that's going to dominate over the first disjunct infer in terms of a disjunction so you're going to have x greater than or equal to zero that makes a more general claim of the two and for this one um let me make note that x squared greater than 9 this part means that either x is greater than 3 or x is less than negative 3. well one of these two disjuncts is the one that's here so this one is super airflows so that's the one that counts so x squared greater than 9 is a simple way of saying it and again as i was as with the previous case as with numbers one and two as i was going through them i actually articulated what these things are and these were really just to sort of prompt you to thinking about how you would express them in english okay well that's that one well in number seven to show that a disjunction is true you have to show that one of them is true at least one of them and for number eight to show that it's false you have to show that all phi 1 phi 2 etc to phi n are false okay well that's number seven and eight taken care of i'm turning to number nine now to say that it's not the case that pi is greater than 3.2 is to say that pi is less than or equal to 3.2 when you negate a strict greater than you get less than or equal to say that it's not the case that x is negative is to say that x is greater than or equal to zero to say that it's not well this these are these are real numbers uh from the context of the expression uh we can assume that these are talking about real numbers and for a real number every real number has a square which is strictly positive with one exception and that exception is zero so the only real number for which it's not the case that x squared is strictly positive is x equals zero the standard deviation for not x equals one is x not equal to 1 and when you negate a negation it takes you back to the original statement and again as i went through them i essentially answered number 10. we're turning to question 11 now here are the answers that i get dollar and yarn both strong okay dollar strong and yan strong i think that one's fairly straightforward what about this one well there's these words despite and but by my mind those are just nuanced forms of conjunction they both mean and so we've got the yarn week and there's a trade agreement and the dollar strong so the despite and the but they sort of get at the whether these things are contrary to our expectations or consistent with our expectations but they still say that all of these three things hold together okay this one i think is is fairly straightforward they can't both be strong at the same time so it's not the case that the dollar's strong and the young strong for part d well if that doesn't prevent that it means that happens and both of those fail so the trade agreement signed but the dollar falls and the yan falls the trade agreement holding doesn't prevent the fall in the dollar and the yen that means the dollar and the undo fall okay and then for the last one u.s china trade agreement fails but both currencies remain strong again the but i think uh is is and its conjunction so we've got the trade agreement failing the currencies both remain strong the dollar strong and the young strong okay those are my answers and i i would hope you get the more or less the same you know you might end up writing slightly differently but even though we had to think a little bit about what these words mean to my mind there's no real dispute as to whether these are correct or not okay well how did you do on that one we're judging from the discussions on the on the course forum in problem set one the questions that caused people the most difficulty were numbers six and seven those are those questions about alice and uh let's just see uh and see what's going on here notice that in each of these we've got the same phrase works in a bank works in a bank works in a bank works in a bank works in a bank but for four of these there's an additional requirement not only does she work in a bank but something else works in a bank something else works in a bank something else works in a bank something else each of these additional requirements each time you've got one it makes it less likely to happen because you've got something else to satisfy they all have to satisfy the fact that alice works in a bank but for the first four alice also has to satisfy something else so each of these makes it less likely it makes it less likely to be true so this one is the one most likely to be true now it's possible that in a particular instance another one has the same likelihood we're not saying that there's a unique most likely one but we are seeing is the most likely one incidentally it doesn't matter whether you express in terms of the word likely or the word probable or probability you can do it numerically you can do it however you want the issue is not what the word is the issue is the information you have and in each of these cases you have an extra restriction and whenever you've got an extra restriction with a conjunction and it's critical at this conjunction here when you've got an extra restriction with a conjunction it makes it less likely because instead of just having to satisfy one thing you have to satisfy two things and when you satisfy two things it makes it less likely to happen so regardless of whether you use the word likely probable whatever you want to do the most likely one is this one because this is the easiest one to satisfy and this actually i think is a great example of mathematical thinking i mean this is a superb example of the kind of thing i'm talking about in this course because we're arguing purely in terms of mathematics in fact it's the mathematics of the word and the reason this is in this problem set is because this is all about the way and works so if you really understand how and works this one just drops right out uh that's why it's there it's it's to to to make you really reflect on on conjunction and the power of conjunction okay so it's uh there you go it's all to do with the amount of information you get incidentally all the probability theory does well i say all i mean probability theory is a powerful theory but what probability theory does is adds numbers to the amount of information we have okay and in this case the extra information is more restrictive and and when you assign probabilistic probabilities to that uh that gets reflected okay well that was number six let's take a look at number seven well number seven is slightly different because we don't have a common expression in in all of the all of the five in the case of works in a bank uh in this one we have works in a bank and something else this one we've got is a rock star uh this one we've got works in a bank and something else this one we've got works in the bank there we've got works in the back okay the other thing here is being quiet being honest and so forth but if you look through these this one has works in a bank and something else so this is less likely than working in a bank this has been a rock star this is honest and so this is this one is also less likely to work in the bank so if we're looking for the most likely thing these make them less likely so let's just sort of ignore that part that makes it less likely it's already more likely that she works in the bank this one there's nothing to be more this one she works in the back so we've got either working in a bank you know this is more likely than it was originally so i'm making things more likely here i've got alice been a rockstar in this one either she's a rock star or she works in a bank so this is the most general one this is more likely because there's two chances here either working in a bank or being a rock star here there's just one of them here's just one of them here's one of them and here's one of them okay so the most likely one is this one because of disjunction there were two ways that she could satisfy this but i've been a rock star and working in a bank in each of these there's one of them and i've ignored the thing that makes it even less likely so having modified that i've made b more likely having modified d i've made it more likely so now i've just got works in the bank works in a bank rockstar works in a bank but there's only one of them in each of that in this one there are two of them so again purely by using mathematical thinking i've arrived at the one that's most likely and as in number six it doesn't matter whether you take expression in terms of probabilities or relative probabilities however you want incidentally probabilities are many different types of probability there's frequencies probability there's bayesian probability there's subjective probability there's epistemic probability assigning numbers to the information you have and calling them probabilities is a pretty tricky thing there are many different ways of doing it uh and it's uh it's just a very complicated issue so those of you who actually tried to use probability theory that's fine you can do it but it actually gets more complicated not least because you end up having to talk about whether things are independent events and so forth and the point is you don't need any of that you know remember at the beginning i said this course is not about rushing in and applying techniques or procedures it's about thinking basically about the issues and you don't need the machinery here if you think about the issues it comes out remember my my exhortations look at a problem ask yourself what it means what is it saying and try to reason in terms of the problem itself don't look for techniques to apply you often need to do that but in this case um that's not what the course is about and many of the questions i'm giving in this course should be solved not by applying a technique but by uh by just thinking about the problem and let me finish this this tutorial on the problem section and these are the only two problems i want to look at let me finish that this tutorial session with a different problem of the same kind where you can solve it just by thinking about the problem not applying a technique and here is a problem okay you've arrived late at an airport you're rushing to catch your plane unfortunately your gate is at the far end of the terminal it always seems to be the found of the terminal isn't it the fastest you can walk is a constant speed of four miles per hour for parts of the way there's a moving walkway moving at two miles per hour you decide to take the walkway and continue to walk so you're gonna walk all the way but part of when there's a walkway you're gonna walk on the walkway so you can go faster right just as you're about to step onto the walkway you notice your shoelace is loose and the last thing you want to do is get your shoelace caught in the mic i mean the mechanism of the walkway um so you've got a choice you either stop and fasten your shoelace just before you step on the walkway or you step on the walkway and then stop to fasten your shoelace whilst the walkway is moving you along and there's no other options those are the two options you have which option will get you to the gate fastest now there are two ways you could try to go about this you could try to argue in terms of speeds and adding the speeds together and so forth and if you do that you're going to run into a problem there's going to be some information that you need that you don't have for example i haven't told you how long the walkways and how long you had to walk in general so there's this there's a problem if you try to use standard methods however you do have enough information to solve the problem but if you try to apply relative speed the kind of methods you were taught to use in in the high school for solving problems about speed relative speeds and so forth you're going to run into problems and you don't need that if you think about the problem you should be able to solve it well i'm just going to leave you with that one um it's it's an it's an interesting problem i think it's a rather neat problem but there are many problems like this that you really can just solve by mathematical thinking and the point is this again is all about this important notion mathematical thinking it's powerful if you can master mathematical thinking many problems can be resolved without going into the complexity of applying mathematical techniques mathematical techniques are what you need when mathematical thinking alone doesn't work but often you can get by with just mathematical thinking so good luck on that one and uh and enjoy it it's a lot of fun welcome to the third lecture i hope you made good progress with the assignment from the last session our next step in becoming more precise about our use of language for use in mathematics is to take a close look at the meaning of the word implies this turns out to be pretty tricky brace yourself for several days of confusion until the ideas sort themselves out in your mind just like learning to ride a bike where all the actual learning occurs before you finally get the hang of it so too most of the benefit from understanding the way language is used in mathematics comes from trying to figure it out the benefit in this case is helping to develop your mathematical thinking ability and it's the process of trying to understand the issues that does that for you yeah sure once you've sorted this language stuff out you'll be able to use language more accurately but by that stage you're just using it the part that helps you learn how to think mathematically is largely over we need other tasks to develop your mathematical thinking ability further so as you get into this bear in mind that the payoff for struggling with the issues is significant in terms of being able to think like a mathematician here we go in mathematics we frequently encounter expressions of the form phi implies psi indeed implication is the means by which we prove results in mathematics starting with observations or axioms so we'd better understand how the word implies behaves in particular how does the truth or falsity of a statement phi implies psi depend upon the truth or falsity of phi and of psi well the obvious answer is to say that phi implies psi if the truth of psi follows from the truth of phi but is that what we want well let me give you an example let 5 of the statement root 2 is irrational and let's cy by the statement 0 is less than 1 and let's ask ourselves is the statement phi implies psi true well phi is true as i've mentioned once already we're going to prove that later in the course and we all know that psi is true zero is certainly less than one so we have the truth of this we have the truth of that does that mean the phi implies psi obviously not there's no relationship between phi and psi this takes some effort to prove as you'll see this we all know this one so yes this is true and that's true but the truth of this doesn't follow from the truth of that and now we realize there's a complexity with implication that we didn't meet before when we were dealing with and with or and with not and the complexity is that implication involves causality and causality is an issue of great complexity that philosophers have been discussing for generations now we're facing a problem it didn't arise before because when we were dealing with conjunction and disjunction it didn't matter whether there was a any kind of relationship between the two conjuncts or the two disjoint for example let's look at the sentence julius caesar is dead and let's conjoin it with the sentence one plus one equals three the mathematical sentence and let's do the same thing with disjunction forming a conjunction and disjunction didn't require any kind of relationship between these two these are actually clearly they're independent one's a statement about a long dead individual and the other one is a mathematical statement incidentally while we've got these in front of us uh let me just give you a quick quiz is the first one true or is it false is the second one true or is it false what do you think well remember what the definition was a conjunction is true if both conjuncts are true this one's true but this one's false so the conjunction's false this one's true this one's false all you need for a disjunction to be true is at least one of the disjuncts to be true so that one's true and the fact that there's no meaningful relationship between the two conjuncts in this case or the two disjuncts in that case played no role in determining what the truth value was it was purely in terms of truth and falsity but that's not the same with implication because implication involves causality so let me just express that explicitly implication has a truth part and a causation part what we're going to do is ignore that part we're going to leave that to the philosophers if you like and we're just going to focus on the truth part now that might sound to be a very rash thing to do throwing away causation now we're going to be left with anything useful it turns out we are it might seem a dangerous thing to do to slow away this important constituent of implication but it turns out that when we focus on the truth part we're left with enough to serve our needs in mathematics so much so that we're going to give this a name we're going to call it the conditional or sometimes the material conditional that's the part we're going to focus on so we're going to split implication into two parts the conditional and causation the first part the conditional we're going to define entirely in terms of truth values and the second part we're going to leave to the philosophers the symbol that we use normally for conditional at least the symbol i'm going to use is a double arrow lots of textbooks use a single arrow but that's got um that has many uses in mathematics uh we're certainly going to use that for something else later on in the course so to avoid confusion i'm going to stick to this notation for the conditional so i'm going to write conditional expressions like this that's the truth part of phi implies psi when we have a conditional we call phi the antecedent and we call psi the consequent and we're going to formally define the truth of phi conditional psi in terms of the truth of phi and the truth of psy now you might worry that by throwing away a causation we're going to be left with a notion that's really of no use whatsoever that actually is not the case even though we're throwing away something of great significance hanging on to the truth part leaves us something very useful and the reason is whenever we have a genuine implication which are actually the only circumstances in which we're ultimately going to be interested whenever we have a genuine implication the truth behavior of the conditional is the correct one it really does capture what happens with truth and falsity when we have a genuine implication that probably seems a bit mysterious at this stage but when we start to look at some examples i think it should become clear what i mean the advantage is that the conditional is always defined for real implication you've got that issue of causation the the you know the square root of two and the zero less than one example the truth of falsity wasn't the issue it was whether there was a relationship between those two statements that's a complicated issue but because we're going to define the conditional purely in terms of the truth value of the two constituents the antecedent and the consequent it turns out that the conditional will always be defined when we do have a genuine implication the definition of the conditional will agree with the way implication behaves and when we don't have genuine implication the conditional will still be defined and so we can proceed again this probably seems very mysterious when i describe it in this way but as we develop some examples i hope you'll be able to understand what i'm trying to get at okay i think we need to catch our breath let me do that by giving you a quiz the truth of the conditional from phi to psi is defined in terms of one the truth of phi and psi or two whether phi causes psi or both which is it it's number one we define the truth of a conditional in terms of the truth and falsity of the antecedent and the consequence and because we define the truth of the conditional in terms of truth and falsity in that way it has a truth table so let's see if we can figure out what it is well i'm not giving you this as a quiz by the way i'm going to work this this one through this is tricky we're not out of the woods yet by any means so i'm going to lead you through this one this part we've already looked at we defined the conditional as the truth part of implication and implication has a property that a true implication leads to a true conclusion from a true assumption so because we take the conditional from real implication we have to have truth all the way through the top level now the fun begins we have to fill in these c values and when i say fun i mean fun let's look at that first row of the truth table suppose phi is the statement n is greater than 7 and suppose psi is a statement n squared is bigger than 40. if phi is true in other words if n is bigger than 7 then n squared is certainly bigger than 40. in fact it's bigger than 49. so certainly in this case phi does imply psi so nothing is true if n is bigger than 7 then n squared is certainly bigger than 40. so that's true we have true with everywhere so so this is consistent with the truth table now let's look at a different example 5 of the statements our old friend julius caesar is dead and that side with the statement pi is bigger than three phi is true psi is true according to the truth table it follows that phi conditional psi is true in other words julius caesar is dead conditional pi is bigger than three now if you read this as julius caesar is dead implies pi bigger than three then you're in a nonsensical situation but remember this isn't implication this is just the truth part of implication and in terms of the truth part there's no problem that's true that's true the conditional is true in the first example there is a meaningful relationship between phi and psi when we know that n is bigger than 7 then we can conclude that n squared is certainly bigger than 40. there's a connection between the two and in that case the behavior of the conditional is certainly consistent with what's really going on in the second case there's no connection between the two the conditional is true but it's got nothing to do with one thing following from the other the value of doing this is even though this has no meaning in terms of implication its truth value is defined in both cases we have a well-defined truth value in the first case it's a meaningful truth value that does follow from that in the second case it's purely a defined truth value but that's not going to cause us any problems because we're never going to encounter this kind of thing in mathematics we encounter this kind of thing all the time but we're not going to encounter this kind of thing so all we've done is we've extended the notion to be defined under all circumstances and we've done it in a way that's consistent with the behavior we want when something meaningful is going on this is actually quite common in mathematics to extend the domain of definition of something so that it's always defined so long as it has the correct behavior the correct definition for the meaningful cases and providing we do the definition correctly it really doesn't cause any problems in fact it solves a lot of problems and eliminates a lot of difficulties if we extend the definition so that it it covers all cases is it just something we do in mathematics all the time may seem strange when you first meet it but it is a part of modern advanced mathematics incidentally if you think this is just playing games let me mention that the computer system that controls that aircraft that you'll be flying in next time depends upon the fact that expressions like this are always well defined that software control system doesn't depend upon knowing whether julius caesar is dead or things like that it doesn't depend on those kind of facts of the world computer systems by and large don't depend upon understanding causation which is just as well because they don't what computer systems depend upon is that things are always accurately and precisely defined and this expression phi conditional psi occurs all over the place in software systems so quite literally your life depends upon the fact that this is always well defined it doesn't depend upon the fact that the computer doesn't know whether julius caesar is dead or not okay time to look at the second row of the truth table what goes in here t or f well we've only got two to choose from i guess we could just make a guess but we're not going to do that we're going to figure it out and get the right answer but what would happen if we put a t here let me put it in very lightly because i'm not sure it's going to be a t if this were true and we think about it in terms of genuine implication because we are trying to capture the truth behavior of genuine implication remember so if it was the case that phi really did imply psi if that statement was true when we interpret it as real implication then the truth of psy would follow from the truth of phi that's how we began remember that's what real implication means real implication means the truth of this would follow from the truth of that so if that were a t and it was real implication then when we have a t here we would have to have a t there but we don't we've got an f so we can't have a t here because if we put a t here the conditional is contrary it contradicts real implication and we're trying to extend implication to be defined in all cases when there's no causation so this has to be an f if we put a t there we're in trouble our notion doesn't agree with real implication in order that the conditional agrees with real implication that has to be an f if it's a if it's a true if it's a truth then we would have a true antecedent and a false consequence from a true implication so we've argued backwards to conclude that this has to be f let me write that down just to make sure everyone's following what i'm trying to say if there were a genuine implication phi implies psi and if that implication were true then psi would have to be true if phi were true so we cannot have phi true and psi false if phi implies psi is true that means that in the case where phi is true and psi is false we have to have a false here let me write that down are you confused this is tricky there's no getting around the fact that sorting out uh implication and extracting the conditional from implication is tricky you're probably going to have to replay the video several times and look at what's been written and think about it in order to sort this out i remember when i first met this it took me quite a while to to to get on top of it it really isn't an easy thing to do so you certainly have my sympathy that you're going to have to struggle with this thing it's probably one of the hardest things in the course to really come to terms with the way we define the meaning of the conditional okay final straight let's see if we can fill in the last two entries in the truth table for the conditional when we've done that we'll all go off in search of some aspirin now if you're like me you have no intuitions as to what to put here and the reason you have no intuition is that even though you're used to dealing with implication you've never dealt with an implication where the antecedent was false you were only ever interested in drawing conclusions from true assumptions so you never dealt with a true one of these guys when this guy was false you don't have any intuitions you do however have intuitions about this guy and the reason that's going to help us out is that negation swaps around f and t so corresponding to the f's here when we look at this guy we'll have truths we'll have t's for this so you are used to having to deal with this when this is true and that will be equivalent to dealing with this when that's false because negation swaps truth and falsity so the trick or at least the idea by which we're going to figure out what goes here is to stop looking at implication and look at not implication well phi does not imply psi if even though phi is true psi is nevertheless false just think about that for a minute phi does not imply psi if even though phi is true psi is nevertheless false that's how you know that this guy holds you know that phi doesn't imply psi if you can check that phi is true but psi is nevertheless false that's the only circumstance under which you can conclude this thing is true in all other circumstances this guy will be false so let me write that down okay this guy is true if you have phi true and psi false in all other circumstances this guy will be false you still with me because we're only one line away from the conclusion now let me just rewrite that as follows because negation swaps f and t this guy will be false when this guy is true so if we have t and f which we have here then negation is false which means this guy is true in all other circumstances this guy is true so those are truth and we're done phew that was difficult in fact to my mind this is the most difficult part of the entire course this is not easy stuff it's going to take you some time before you sort it out in your mind you're going to have to keep working at this look through the lectures several times and think about it it will take effort but it's it's you'll get there we all get there in the end but it's just not easy in the meantime you've got this far we should start thinking about opening a bottle of champagne but before you crack open the bottle let me leave you with one more little quiz which of the following are true phi conditional psi is true whenever 1 phi and psi are both true two phi is false and psi is true three phi and psi are both false and four phi is true and psi is false and i want you to check all that are true well which of these four other case which of these four conditions tell you when five psi is true the answer is one two and three phi conditional psi is true either when one is true or when two is true or when three is true the only time when five conditional psi is false is when four holds so the correct answer is one two and three and five is not okay now you can open that champagne let me sum up what we've done we've defined a notion the conditional that captures only part of what implies means to avoid difficulties we base our definition solely on the notion of truth and falsity our definition agrees with our intuition concerning implication in all meaningful cases the definition for a true antecedent is based on an analysis of the truth values of genuine implication the definition for a false antecedent is based on a truth-value analysis of the notion does not imply in defining the conditional the way we do we do not end up with a notion that contradicts the notion of genuine implication rather we obtain a notion that extends genuine implication to cover those cases where a claim of implication is irrelevant because the antecedent is false or meaningless when there's no real connection between the antecedent and the consequence in the meaningful case where there is a relationship between phi and psi and in addition where phi is true namely the case is covered by the first two rows of the truth table the truth value of the conditional will be the same as the truth value of the actual implication remember it's the fact that the conditional always has a well-defined truth value that makes this notion important in mathematics since in mathematics we can't afford to have statements with undefined truth values floating around let's finish this lecture with a short quiz when you've done that you should complete assignment 3. i've kept assignment 3 fairly short since i expect you'll need most of your time simply understanding our analysis of implication and the definition of the conditional remember to discuss these assignments with other students don't struggle for too long on your own on this kind of material it's much more effective to work with others here's the quiz well the answer to the first one is that it's true the antecedent is true and the consequent is true so the conditional is true in fact there's a deeper result going on here providing you take a positive number instead of pi any positive number then if the square of that positive number is bigger than 2 that number must be bigger than 1.2 because the square root of 2 is 1.4 etc etc etc so for positive numbers it doesn't have to be pi it can be anything any positive number whose square is bigger than 2 must be bigger than 1.2 and that would be a case of genuine causation genuine implication but in terms of the conditional it's enough that the antecedent is true and the consequence is true for this one it's also true now the consequent is false pi certainly does not equal three but the antecedent is false and if you have a false antecedent a conditional is always true pi squared is most certainly not less than zero so you've got false false that makes the conditional true number three that one's false the antecedent is true and the consequent is false and you cannot obtain a false conclusion from a true assumption what about the next one well that one's true the antecedent is true and the consequent is true what about this one do triangles have four sides no the squares have five sides no but anything with a false antecedent is true so that's true you've got false conditional false and that's always true what about this one well we don't know when euclid's birthday was at least i don't know when euclid's birthday was i suspect you don't either rectangles certainly have four sides however either this is true or it's false either way since the consequence is true the thing is true we've got an unknown yielding a true consequence so either we have true true in which case it's true or we have false true in which case it's true okay how did you get on with that how did you get on with assignment three i kept it short since i wanted to give you a lot of time to come to terms with the tricky notion of implication this lecture is going to be fairly short as well though the assignment that goes with it will be the longest of the entire course i'm not expecting you to complete it all by the next lecture though the next topic i want to look at is logical equivalence equivalence is closely related to implication two statements are said to be equivalent or more fully logically equivalent if each implies the other equivalence is a central notion in mathematics many mathematical results are proofs that two statements are equivalent in fact equivalence is to logic as equations are to arithmetic and algebra and you already know that equations play a central role in mathematics just as we had to introduce a formal version of implication that avoids the complex issue of causation namely the conditional we have to introduce an analogous version of equivalence it's called a biconditional fortunately we did all the difficult work with implication now we can reap the benefits of those efforts two statements phi and psi are said to be logically equivalent or just equivalent if each implies the other since we have a formal version of implication namely the conditional it follows that there's a formal version of equivalence we call it the biconditional the biconditional of phi and psi is denoted by means of a double-headed arrow like this and formally the biconditional is an abbreviation of phi conditional psi and psi conditional phi since the conditional is defined in terms of truth values it follows that the biconditional is defined in terms of truth values if you work out the truth table for phi conditional psi and for psi conditional phi and then you work out the conjunction you'll get the truth table for phi by conditional psi and if you do that what you will find is that phi by conditional psi is true if phi and psi are both true or if they're both false one way to show that two statements phi and psi are equivalent is to show they have the same truth tables actually to avoid confusion let me change those to capital phi and capital psi because i want to use the lowercase phi and psi for something else because here's the example i'm going to look at i want to show that phi coinjoined with psi or not phi is equivalent to phi conditional psi so this is actually going to be my capital phi and this is going to be my capital psi yeah if i was teaching this material at a high school level i'd be very careful to choose different letters to denote everything but we're looking at college university level mathematics now and uh university mathematicians professional mathematicians frequently use open lowercase symbols uh in in the same context and part of being able to master university-level mathematics is actually getting used to disambiguate the notations okay what we have to do is work out the truth table for this the truth table for that and show that they're the same so we're going to have to start with phi with psi i'm going to need to work out phi and psi i'm going to need to work out not phi then i'm going to need to work out phi and psi disjoined with not phi and then i'm going to have to compare that with phi conditional psi okay as usual we start with the four combinations t t t f f t f f by the way it doesn't matter which order you write these in so long as you get all four possible combinations the way i've written it is the way that most mathematicians write it conjunction conjunction is true if the two conjuncts are true t t gives t here there's an f so you're going to get an f here there's an f so you're going to get an f we have two f so you're going to get an f negation simply flips t and f so we have t t f f becomes f f t t now i'm going to disjoin this with that and this junction picks out at least one t where there's a t there so i'm going to get a t here there's no t so i'm going to get an f there's a t there which gives me a t and there's a t there which gives me a t i know the truth values for phi conditional psi they were t f t t if you look back at the earlier work you'll see that that's what we worked out before and now i just compare that column with that column t f t t is the same t f t t because these two columns are the same i can conclude that that is equivalent to that i should mention that proving equivalence by means of truth tables is very unusual it's only a very special case of equivalence in general proving equivalence is really quite hard you have to look at what the two statements mean and develop a proof based on their meaning equivalence itself is not too difficult a notion to deal with what is problematic is mastering the various nomenclatures that are associated with implication we start out with the notion phi implies psi and i'm really going to be thinking of genuine implication here but you can interpret everything i say in terms of the conditional and i will look at that i'll talk about that a little bit later if mathematicians always describe this situation in this way then life would be very simple but we don't there are many different expressions we use to describe phi implies psi some of them are intuitively obvious and some of them are actually counter intuitive when you first meet them the following all mean phi implies psy one if phi then say okay that's very obvious two phi is sufficient for psi in order to conclude that psi it suffices to nullify well on the face of it that's fairly straightforward but in uh in complicated situations uh people can sometimes get misled by the use of the word sufficient the use of the word sufficient is by no means the most problematic in the in the context of implication this one causes people a lot of problems at first phi only if psi that's not the same as if psi then phi the reason i'm emphasizing that is that the if here goes with the psi just as it does here let's put quotes on that one the if goes with the side the if goes with the sigh but when it's expressed this way with and only if it flips around the order of the implication for example some of you may know that i'm a very keen cyclist on the other hand i've never qualified nor have i ever ridden in the tour de france however i could i could honestly say a person can ride in this a person can ride in the tour de france only if they have a bicycle well that's true because if you don't have a bicycle you can't imagine the tour de france so a person can ride in the tour de france only if they have a bicycle now that's true for me but it doesn't mean the same as if you have a bicycle then you can ride in the tour de france i actually do have a bicycle i have several but i can't ride in the soda france so uh it's true to say keith could ride in the tour de france only if you had a bicycle because that's true for everybody you can only ride in the tour de france if you have a bicycle otherwise you can't ride in this but it's not the same as saying it this way around that if you have a bicycle then you can ride in the tour de france so i've got to be a little bit careful these are often confused which is why i'm stressing the distinction okay let's look at this one because this one that i'm going to write next is one that i've already really encountered in my discussions about the tour de france psi if phi notice that we've flipped the order now phi is the antecedent psi is a consequent let me write those down just to remind us phi is the antecedent and psi is a consequent and those are the rules going to be played by phi and psi in all of these so phi is the antecedent psi is phi the consequent antecedent psi is the consequence phi is the antecedent psi is the consequent they're flipped fire is still the antecedent psi is still the consequence this still means that phi implies psi but now the consequent comes first the antecedent comes second psy if phi if like me you're a cyclist you can think about that in terms of owning a bike and riding in the tour de france or use your own favorite example think of something that that you do that requires something that requires some equipment or some preparation or something and just interpret these things in terms of something that's meaningful in your life and that should help you to to sort these out okay there's a couple more i want to talk about psy whenever phi again the order is flipped from the order and implication that's still the consequence that's the antecedent but they've been written the opposite way around psy whenever phi and finally psi is necessary for phi again the order is flipped round that's still a consequent that's still the antecedent psi is necessary for phi in order to ride in the tour de france it's necessary that you have a bicycle it's not sufficient that you have a bicycle dividing the tour de france it's necessary that you have a bicycle but it's not sufficient that you have a bicycle that's the example that i would use for all of these but you can pick your own favorite example to to understand these things it's important to really master this terminology because it's used all the time not just in mathematics but in science and in analytic reasoning and thoughts in general this is not just mathematical language this is the language that people use in in legal documents uh in in logical arguments in analytic arguments and in discussions and so forth so understanding language as it's used is is very important in many walks of life and having introduced terminology commonly associated with implication we have an associated terminology for equivalence phi is equivalent to psi is itself equivalent to by the way this already shows how ubiquitous equivalences because the obvious word to describe this let me put quotes around that just to disambiguate the obvious way to describe that this is equivalent to something is to use the word equivalent as well i mean equivalence is just a very basic concept in mathematics so this sentence this statement this claim is itself equivalent to well phi is necessary and sufficient for psi this combination necessary insufficient is very common in mathematics and b phi if and only if psi that's also very common in mathematics if and only if notice that we're combining necessary and sufficient with necessary we have psi before phi with sufficient we have phi before psi and that gets us the implication in both directions and equivalence means implication in both directions so the fact that it's in both directions is captured by the fact that here we have the side before the phi and here we have the five before the psi similarly with b if and only if combines only if where phi comes before psi with if where psi comes before phi so in both of these cases we have an implication from phi to sine from psi to phi final remark this expression is often abbreviated iff iff is a standard mathematician's abbreviation for if and only if so if and only if or if if means the two things are equivalent okay once you've mastered this terminology you should be able to read and make sense of pretty well any mathematics that you come across that doesn't mean to say you understand the mathematics itself but at least you should be able to understand what it's talking about and that's the first step towards understanding the mathematics itself and that's all of us to say about this the rest is really up to you to spend some time mastering the concepts and the associated terminology time for a quiz this quiz comes in four parts so how did you do well let's see what's going on here which of the following conditions is necessary for the natural number n to be a multiple of 10. so the question we have to ask ourselves is does n being a multiple of 10 imply the statement to be necessary n being a multiple of 10 has to imply the statements well let's see does n being a multiple of 10 imply that it's a multiple of 5 yes it does so that one's necessary does n being a multiple of 10 imply that it's a multiple of 20 no i mean 10 is itself a multiple of 10 but 10 is certainly not a multiple of 20. does n being a multiple of 10 imply that n is even and that is a multiple of 5 yes does n being a multiple of 10 imply that it's a multiple of 100 no those n being a multiple of 10 implied that n squared is a multiple of 100 yes so the three conditions that are necessary for the number n to be a multiple of ten are condition one condition three condition five okay let's move on to part two well this time we have to ask the question does the statement imply n is a multiple of 10 okay well does this statement imply that n is a multiple of ten no it doesn't five is itself a multiple of five but five is not a multiple of ten look at number two is it the case that if n is a multiple of 20 then it has to be a multiple of 10. does this imply that the answer is yes does n being even and the multiple of five imply that it's a multiple of ten yes if n is a hundred does that imply that it's a multiple of ten yes if n squared is a multiple of a hundred does that imply it's a multiple of ten yes so in this case two three four and five are the correct answers they're all sufficient for n being a multiple of ten okay let's move on to part three for this one we have to compare our answers for the two previous questions the first question was necessity and the second question was sufficiency for necessity we had one we had three and we had five for sufficiency we had two three four and five this asks for necessity and sufficiency here we've just got necessity there we've just got sufficiency there we have both of them there we just have sufficiency there we have both of them so the ones that are necessary and sufficient this one and that one okay let's move on to part four for this one the question we have to ask ourselves is what does the implying in number one this does the implying so that's the antecedent in question two of statement two well even though it's written the opposite way around it's essentially the same statement it's the alarm ringing that does the implying what about number three keith cycles only if the sun shines what's doing the implying keeps cycling if you see me cycling you can conclude that the sun's shining because i cycle only if the sun shines incidentally i was brought up in england so that's not the case i'm quite happy to when i'm not happy to ride in the rain but i do ride in the rain but it's a good example number four what does the implying well amy arrives to the implying so far i've distinguished between genuine implications and equivalence and their formal counterparts the conditional and the biconditional in their daily work however mathematicians are rarely that particular for instance we often use the arrow symbol as an abbreviation for implies and the double-headed ammo there's an abbreviation for is equivalent to although this is invariably confusing to beginners it's simply the way a mathematical practice has evolved and there's no getting around it in fact once you get used to the notions it's not at all as confusing as it might seem at first and here's why the conditional and biconditional only differ from implication and equivalence in situations that do not arise in the course of normal mathematical practice in any real mathematical context the conditional effectively is implication and the biconditional effectively is equivalence so having made note of where the formal notions differ from the everyday ones mathematicians simply move on and turn their attention to other things the very act of formulating formal definitions creates an understanding of implication and equivalence that allows us to use the everyday notions safely of course computer programmers and people who develop aircraft control systems don't have such freedom they have to make sure all the notions in their programs are defined and give answers in all circumstances okay that's the end of lecture four as i said at the start it's been a fairly short lecture my reason for keeping the lecture short is that the upcoming assignment is much longer than the others it has to be implication and equivalence are at the heart of mathematics mastery of those concepts and of the terminology associated with them is fundamental to mathematical thinking you simply have to master implication and equivalence before you can go much further and there's only one way to achieve mastery right remember the story of the elderly lady who approached a new york city policeman and asked officer how do i get to carnegie hall the officer smiled and said lady there's only one way practice practice practice so i suggest you carve out some time grab some food and drink and head off somewhere quiet to complete as much of assignment for as you possibly can well question one picks up the same theme as question 11 in the previous assignment so let's see uh see how it works it's actually much more difficult i think than than question 11 of assignment two some of these really are tricky okay so let's take them one by one new trade agreement will lead to strong currencies in both countries so there's a trade agreement that leads to that results in so i think that's implication the fact that both currencies are stronger dollar strong and the young strong okay that one's fairly fairly straightforward strong dollar means that weak yarn or strong dollar implies that the yan is weak well again i think that one's fairly straightforward the dollar strong it follows the yen is weak trade agreement fails on news of weak dollar so we hear that there's a weak dollar and that implies that the trade agreement fails again i i think that one's a i'm fairly confident in writing that one down i think that's i can't think of another one that would really capture part c okay part d if a trade agreement is sad well this one is signaling very loudly implications for the trade agreement signed then the account both remain strong so it's not the case that the dollar strong and the yan strong okay now with part e i think we're into a an example where you know your mileage might be different from mine okay the question really is um well there's a butt and i think the book means and following however i think is it could mean one of two things it could mean the following the dollar's weak and the yarn is strong and there's a trade agreement in other words there's a trade agreement and after there's that trade agreement the dollar is weak and the yan's strong in other words it's all conjunction so you could interpret following is just meaning that happens first and then those two happen however you could also say well it was the new trade agreement that led to these currency strengths and weaknesses so you could i think quite legitimately say the following there's a trade agreement and as a result the dollar's weak the yarn is strong i think that's that's perfectly legitimate i think there's two interpretations depending on whether you mean following just one thing follows after another in time or it follows because of causality okay what about number five well this one is again signaling implication so if the trade agreement is signed then the rise in the yen will result in the fall in the dollar you could actually write that a different way you could say that there's really two assumptions here two things that tend to happen if there's a trade agreement signed and then if the young rises it follows that the dollar falls now those are actually equivalent uh as you'll be able to demonstrate using truth terms if you want um but they're both ways of interpreting that you could say that you've got an implication of an implication or you could say you basically got an implication with two assumptions i think this is like this is more of a literal interpretation and this one is more sort of semantic interpretation where you you look at what these things mean okay this one um the way i would be i think i would write this one was to say if there's a trade agreement signed then these are linked which is basically equivalent dollar equivalent to the yen okay or if you want to spell it out you would say trade agreement implies that the dollar rise leads to the young rise and the young rise reached the dollar rise okay they're both the same thing not actually is an abbreviation for that the biconditional abbreviates the two conditionals let's try and clean that one up a little bit okay okay new treated agreement will be good for one side but no one knows which okay this is what we're really doing here is he's talking about a an exclusive or so that was something that caused lots of you some grief uh in the uh in an area of example in an early assignment okay new trade agreement will be good for one side so either the dollar strong or the yan strong but and but his conjunction we've already noted that it's not the case that they both hold okay this really forces the fact that it's an exclusive or one of them holes but not both of them well that was question one and uh as i mentioned uh some of these uh you may you may come up with different answers uh i'm pretty confident about these that when i look at these sentences these are what these things mean to me but we are taking something in everyday language in natural language in fact we're taking stereotypical newspaper headlines which are abbreviated natural language and we're trying to interpret them in the formal language of logic so we're taking something that's imprecise and depends upon context and knowledge of culture and we're trying to express it in a logical formalism and that's uh that's that's they have interpretation differences your mileage may differ from mine on one or two of these things as i said i could defend each of these and i tried to as i went through but you might come up with an answer that you could defend and that's fine okay let's move on well for number two i've put my answers in bold uh if that doesn't show up on your screen these are the ones that have been bold okay i started with phi and i negated it so 2 becomes false 2 becomes false false becomes 2 false becomes true i have psi we've already worked out the truth table for uh for the conditional and we've seen it's t f t t now we can take phi but now we can take not phi and psi we can combine these two columns with a disjunction to get this guy so this will be true whenever one of these is true which means one of them's true here neither is true here so we get a false they're both two here we get a true one of them is true here we get a two so this column comes from combining these two with disjunction and now we simply compare these two and this is really getting to number three um we can compare these two and observe that every entry is the same and since the truth values are the same the conclusion we can draw is that phi e psi is equivalent to not phi or psi and that takes care of two and a three we're turning to number four now we've got four columns to fill in and again i've written them in boldface but you may not be able to see the ball face on on your screen but these are the four entries that i put i began with the the two columns were given the standard range of possibilities for t and f for phi and psi i negated the psi values to give me not psi so t became an f and f became t t to f after t then i wrote down the values for far yield psi which we know we've already worked that out uh it's t f t t then i negated those to give me phi does not yield psi so t becomes an f f becomes t t to f t to f and then i combined the first column with the third column using conjunction to give me this one and with conjunction you have to have both of them true to get a true well we've got a t and an f and that f means we're going to have an f there here we've got a t and a t so i get a t there here i've got two f's gives me an f and here i've got one f so i've also got an f well that fills in all of the columns and then to answer question three we just notice that these guys are all the same and hence we may conclude that phi does not yield psi is equivalent to phi and not psi and in fact what i'd like to do now is this recall the the discussion we had to obtain the truth table for fire yield psi if you remember to do that we had to look at phi does not yield psi in order to find the truth values in the case where phi was false we ended up having to look at phi does not yield psi and this truth table here sort of shines a light on what was going on it illustrates what it explains to us why looking at that enabled us to work out the the two problematic truth values uh for the truth table for yields for implication when fire was false because it's this equivalence that we were capitalizing on that phi does not yield psi if phi is true and nevertheless psi is false okey dokey well that's uh questions four and five for you how did you do okay let me finish this tutorial with this little puzzle i'll leave you with a woman was driving in her car along a black road she did not have her car lights on there was no moon and no light from the stars a black dog was asleep in the middle of the road as the woman approached the dog she swerved to avoid it and the animal slept on how did the woman know to swerve around the dog and let me just give you one clue remember the focus in this part of the course is on being precise about the use of language and being very careful about the information that language conveys and with that clue i'll leave you to puzzle this one out on your own bye bye for now well assignment 4 was pretty long but i would hope that by now uh working with other students in the class you were able to get through most of the ones that you attempted i'm just going to focus on three of them i think i'll look at number six 10 and 12. most of the other ones were just straightforward truth table calculations and i'm pretty sure that by now you should have done those and uh and if not then do go back and talk with friends and and other students and try and sort it out um and you may not have had too much difficulty with six ten and twelve as well but uh but let me just play it safe uh and work through these there's at least one here that's a little bit tricky it's not the first one um incidentally the first one and it's just it's a true statement this number is a prime number but that's not what the question about is about the question is what's the negation or the denial and then the simplest way to just say well 34 159 is not a prime number or else you might say 34 159 is a composite number either way i think that's fairly straightforward this one is straightforward too but there's i can't think of any nice way of writing the answer that sounds that sounds like good english you would have to say roses are not red or violets are not blue which is the kind of sentence that you will only see or hear in a in a class on logic as we're looking at now you know mathematical uses of language precision in language so it's correct but it just is not an elegant sentence you you wouldn't normally say something like that this one is tricky actually and it's tricky because there's a negation floating around in here remember when you've got a a conditional far yield psi when you deny it what you get is you get the antecedent conjoined with the negation of the consequence well in this case the antecedent is there there are hamburgers i'm sorry in this case the antecedent is that there are no hamburgers that way i almost sucked into that mistake myself couldn't be very careful with this the antecedent is there are no hamburgers so let's just write that down there are no hamburgers conjoined with the fact that i won't have a hot dog well i think in this case it's more natural to write the conjunction using butt rather than and and say but i won't have a hot dog okay so we've got the antecedent that there are no hamburgers conjoined in this case written with a butt because i think it's more natural to say that there's a butt but i won't have a hot dog okay um well think about that one for a bit in my experience students have often have trouble with this because of that negation in there it throws you off so you have to be a little bit careful um fred will go um but he will not play again the butt is really just the same as an and and so when we negate that we'll go we'll get red won't go the and becomes an o we got all he'll play okay in fact i think in english we're probably more likely to write it differently we're probably more likely to say fed will play or he won't go i think we do that because we would we'd read into this some causality some connection between the two um now that's going beyond just doing the straight negation but i think that's typically how we would understand it so to me that seems more natural way of saying it um just just to cause with the kind of thing people people normally do but in terms of the negation we just took the negation of the two parts um together with the negation of the book which becomes a disjunction um which is an or so got fed won't go or he will play or more naturally fred will play or he won't go this one is is actually messy if we do it in english but it's much easier to do it symbolically so let's do it symbolically first the original statement is that x is negative or x is greater than 10. now if we take that statement and negate it then the x less than zero becomes an x greater than or equal to zero the or becomes an and x less than x greater than 10 becomes x less than or equal to 10 so that just becomes 0 less than or equal to x less than or equal to 10. if you do it in english you're going to have to say something like the number x is non-negative you can't say positive because the negation of negative is not positive because of zero you've got to say non-negative the o becomes an and and the negation of greater than ten is not less than ten it's less than or equal to ten so you have to say less than or equal to 10. and i don't see any way of doing it in english other than with a with an awkward sounding sentence like this symbols provide a much more efficient way of expressing this idea and indeed for dealing with the negation okay but then we've found an answer to the question um the last one part f we'll win the first game or the second the negation is that we'll lose the first game and we'll lose the second game and the simplest way to write that is to say we lose the first two games remember the question asked for a denial and that's just not the strict negation it's it's a natural not necessarily natural but it's a an equivalent version of the uh of the negation and for some of these were able to find pretty good answers um in others it was it was kind of messy but there were a couple of tricky ones floated around in there so i think you need to be a little bit careful about these kind of things uh dealing with language precisely is actually not easy because our mind jumps ahead uh how many of you jumped from this one to something like if there are hamburgers i won't have a hot dog students often do that one you've got to be careful about the precision of language human beings are very smart with using language in everyday terms but the cost is that we we we drop precision we think in terms of meanings rather than what the we think in terms of the way we understand the meanings rather than what the literal meanings are the whole point of this this analysis of language we're doing now is to be very very precise um and then examples like this remind us that precision uh is not easy you have to work at it okay well uh that was number six and what was the next one i was gonna i was gonna do number 10 now so let's look at number 10. well for number 10 we have to refer back to number nine i'm not going to do number nine i would hope that by now you're able to do a truth table verification that these two are equivalent wake up the truth table of that work out the truth table of that and observe that they're the same so how do we go about uh about proving the the same equivalence but by means of a logical argument well there's two directions to prove i'm going to begin by proving left to right okay i'm going to assume this and conclude that okay so i'm going to assume phi yields psi and theta okay that means we can deduce sine theta that formula from phi but we know that we can deduce psi from sine theta and we know that we can deduce theta from sine theta because from any conjunction we can conclude either of the two conjuncts if you know sine theta then you know psi and you know theta okay hence by churning two things together by churning the deduction of this from that and then those two from that it means we can deduce psi from phi and we can deduce theta from phi okay you with me this means we know if we can deduce zeta if we can deduce psi from phi that means we know phi yields psi and we also know if i use theta hence we know if we know that and we know that we know their conjunction so assuming this which is the left-hand side we have concluded that which is the right-hand side okay let's go the other direction so let's assume phi u psi and phi use theta then we may conclude the first conjunct phi heal psi and we may conclude phi u theta okay that is we can deduce psi from phi and we can deduce whoops what did i do here that was meant to be a theta i hope you spotted that okay one of the problems with doing this over a video is that you can't yell out when i do it but at least i spotted it that time if i use theta sorry about that miss that's a theta okay we can deduce psi from phi and theta from phi oh i'm really making a mess of this aren't i we can deduce theta from phi okay let's take this from the top okay we can assume we're going to assume this that means we can assume the first conjunct then we can conclude the second conjunct right if we're assuming that then we know the first conjunct and we know the second conjunct if you know a conjunction you know the two conjuncts so that's the first conjunct and the second one the first one means we can deduce psi from phi the second one means we can deduce theta from phi well if we can deduce psi from phi and we can just use theta from phi that means we can deduce psi and theta from phi if you can deduce psi from it and you can deduce theta from it then you can deduce the conjunct from it that means we know phi yields psi and theta so now assuming the right hand side i've concluded the left-hand side now this is going to look a bit circular right um to figure out what's going on to follow what's going on you have to sort of separate out specific formulas from what they mean so this is a single formula and what we're doing is we're pulling it apart into what it tells us and then we're arguing with what it tells us and then at the end we put it put it back into a formula again so we start with a formula we end with a formula in the middle we're arguing with what the formula tells us and ditto here now since this really corresponds in all real cases to to implication and since that is conjunction we end up using words like candid use it follows from and we use the the word and a lot so there is a sort of forced circularity in that this actually does say this that's the point so if this seems as though there's a sort of a shallowness to it indeed there is in that this is specific these formulas are specifically set up to correspond to these things but if you really look into it this isn't totally trivial we have gone from formulas to what the formulas tell us okay well enough of that let's move on to the next one okay and that next one i want to look at is question 12 and that uh is talking about the the contrapositive and these are the four statements um that we have to write down the contrapositives of so if two rectangles are congruent they have the same area so the contrapositive would be if two rectangles do not have the same area they're not congruent number two contrapositive would be if in a triangle with sides a b c with c being the largest it is the case that a squared plus b squared is not equal to c squared then the triangle is not right angled so if in a triangle with sides a b c c the largest a plus b squared is not equal to c squared then the triangle is not right angle contrapositive to this one if n is not prime and 2 to the n minus 1 is not prime or if you like if n is composite then 2 to the n minus 1 is composite and then for part d if the dollar does not fall then the yarn won't rise well actually writing these down was fairly straightforward it's just a case of looking what the contrapositive does it flips the order of the things and puts negations in front of them so actually writing them down was not what the challenge was the point of this exercise was to give you an example to sort of help cement the fact that the contrapositive is actually logically equivalent because is this one true yes congruent rectangles have the same area if rectangles don't have the same area they're not congruent those are both true if your triangles have right angles so we've got pythagoras theorem if a triangle is added each one of these is is clear that one is equivalent to the other if this thing happened to be true then that thing would happen to be true if that thing was true then that would be true so i think these four examples will illustrate the fact that contrapositives are logically equivalent to the original statement okay i said i was just going to do 6 10 and 12 but as i was working through this one i thought it would be a good idea to look at question 14 which is very similar so let me do question 14 as well okay and number 14 is about converses uh we're again as with con as with a contrapositive you you flip the odd around but in this case you don't inject any negation signs it's simply the implication in the opposite direction okay um which means actually these four are going to be very straightforward because we're just going to flip the order so this one would be if two triangles have the same area then they are congruent now already we've shown by an example now that the converse is not necessarily logically equivalent to the uh to the original statement because the original statement is true but the converse is false now that's not always the case let's look at this one the converse of this one is the following if in a triangle which sides a b and c we see the largest it's the case that a squared plus b squared equals c squared then the triangle is right angled now this is true it's pythagoras's theorem and the converse of pythagoras theorem is true so these are both true but what's going on is that we've actually started out with something that's not just an implication but it actually is an equivalence and when you've got an equivalence and you and you take the convex you've just got this you just got another part of it okay the the statement and the uh and the convex are both the two halves of an equivalent so um you can get the same truth values but only when there's an equivalence already there in the case of this one the converse would be if n is prime then 2 to the n minus 1 is prime and the converse of this one is if the dollar falls the yarn will rise and in this case there's no reason to assume that one necessarily follows from the other unless in some way though the two currencies are linked okay well i i hope you managed to get the uh uh the four converses written down correctly but as i indicated a moment ago that wasn't really the point of this the point was to sort of give us examples of the fact that converses don't necessarily yield equivalent statements when you when you take a converse you don't get something that's equivalent uh whereas with the contra positive that we looked at in in the previous example in example 12 when you take the contrapositive you do get something something that's equivalent okie dokie well that should i think take care of assignment uh number four okay let's see what we've got with the problem set two um by the way i this part of the course we've been looking at taking uh expressions in everyday language or problems that might arise from the real world in a certain sense and and try to make them precise uh and that of course means that for people whose language was native language is not english this is even more complicated because we have to deal with the complexes of the english language and try to eliminate the ambiguity from that and make things precise um when it comes to these issues about necessary insufficiency i actually don't think it makes a lot of difference if you're if you're a native speaker in english this is tricky when you first meet it uh and it's it's easy to make a slip so this is one of those occasions where i don't think there's an advantage to uh be a native english speaker we native english speakers find this difficult too okay in the case of necessity a condition isn't it is necessary if it follows from uh from this so what we're asking for is is it the case that if six divides n then the condition x holds that's what we're asking does the condition x follow from uh in been divisible by six well let's see um is it the case for example that if six divides n does it follow that 3 divides n well the answer is yes if 6 divides it then 3 divides it okay so that one's necessary what about this one well notice that n equals 6 itself satisfies this condition okay if n equals 6 then then 6 divides n but it's 6 divisible by 9 and answer no again n equals 6 satisfies the condition have been divisible by 6 but does it then follow that n divisible by 12 in other words just 6 divisible by 12 no it's not what about n equals 24 well again n equals 6 gives us an example of a number that's divisible by 6. but that number's not necessarily 24 right because it's six so it's not that one okay n squared divisible by six n squared divisible by three well let's just say six divides n certainly implies three divides n if six divides it then three divides it and if three divides n then three divides n squared so that one's okay again if 6 divides n then of course 2 certainly divides n because 6 does and if 6 divides n then 3 divides n so 2 divides n and 3 divides n in other words n is even and divisible by 3 so it's that one so we've got parts a e and f and this is the condition this is this is the way it cashes out in terms of implication and in the case of the counter examples that we used to to prove some of these false the simplest counter example was n equals six itself okay well that takes care of number one let's move on to number two where in the case of sufficiency we have to look to see if the if the condition implies that n is divisible by 6. okay sufficiency means implies n is divisible by 6. so for each of these statements we have to ask ourselves does it imply that n is divisible by 6 well if n is divisible by 3 in order to if we think the answer is it's not sufficient then we have to find an example we have to find a counter example of a number that's divisible by three that's not divisible by a six well while it would check n equals three three is divisible by 3 but 3 is not divisible by 6. so that's a counter example so that one's not sufficient for being divisible by 6. that doesn't imply x is divisible by 6 because in particular 3 is divisible by 3 but not divisible by 6. what about part b let's take n equals 9. that's a counter example n is divisible by 9 but n is not but but if we take n equals 9 then the n is divisible by 9 but 9 is not divisible by 6. so that's a counter example n is divisible by 12. is it the case that if 12 divides into n does it follow that 60 divides into n well the answer is yes so that one's okay what about n equals 24 is it the case that if n equals 24 does it follow the six dimension in other words does six divided 24. again the answer is yes so that one's true n squared is divisible by three does it automatically follow that then is divisible by six well why don't we take the same counter example we did in part a let's take n equals three then certainly three divides n squared okay but the 65 n the 6 divided 3. the answer is no 6. um does not divide in so 3 divides n squared 6 does not divide n so it can't be that one uh finally n is even indivisible by 3. um i could write that as saying n is even is to say that 2 divides n and it's divisible by 3. so i can just rewrite this statement to say that 2 divides n and 3 divides n but if 2 divides n and 3 divides n then it's the case that 6 divides n okay so that one's true and when we come to question 3 in a minute it's really going to be a matter of combining questions one and two so we've now got all the information we need to to answer the next one well question three ask us for necessary and sufficient so let's just see what we did in question one we dealt with necessity in question two we dealt with sufficiency uh let's remind ourselves what we had let me see now we got um necessity that was true in a wasn't it it was true in [Music] e it was true in f remember correctly yeah okay actually i'm not remembering i'm working these out as i go through because i don't do the one in front of me anymore sufficiency let's see that was c it was d and it was f okay um by the way if i'm making this look fluent it's because i've been doing this for many years um although you can still catch me out with a well with a puzzling with a question it's described in a puzzling way um even the experts can be caught out with this kind of thing okay so we just have to look for the ones which have an x in both rows in both columns and the only one that does is this one in all of the other ones you've either got necessity or sufficiency or neither in the case of part b okay so uh you've got a one necessity you've got a neither you've got a sufficiency you've got a sufficiency you've got a necessity the only one that's both is part f and that takes care of number three well we're moving along rapidly now let's go ahead well question four starts out simply enough because it's a very straightforward if then statement and when you've got an if then statement um identifying the antecedent is pretty straightforward because it's the part that goes with the if okay so it's this one if the apples are red then they're ready to eat okay so this was an easy start but these things are going to get a little bit more tricky as we move forward um let's have a look at the part b okay well we're talking about sufficiency and sufficiency is the thing that does the implying okay so what does the implying here well f being differentiable that's the one that does the implying because the differentiability of a function implies that it's continuous okay sufficiency does he implying so sufficiency is the thing we're looking for so the antecedent is a sufficiency which is differentiability well i said these become a little bit tricky as soon as you get into them even though in one sense this is straightforward my experience is that many students including myself when i was a student and occasionally today if i don't really put my mind to it i have to think a little bit to just flesh these out all righty let's move on to part c okay remember we're still looking for the antecedent the thing that does the uh does the implying um this case again it's uh this actually is fairly straightforward because it doesn't matter whether you put the if clause first or second it's the thing that goes with the if so long as it's an if or not and only if um the thing that just goes with with a naked if is the thing that's the antecedent so in this case uh it's this guy it's if he's integral that's the thing that does the implying and whenever actually it's sort of the same as if it's another way of using a way of saying if this this tells us the condition under which something happens this is bounded under the conditions and that is it's convergent whenever s is convergent or if s is convergent so it's that guy okay so it's this is the antecedent this is the antecedent let's move on to paddy when the case of necessity necessity is the thing that follows so unnecessary condition is the thing that follows okay so we have to also ask ourselves what is it that's following in this case okay and the answer is this thing okay so this is the antecedent because n being prime is necessary so n being prime is the thing that's the consequence of the antecedent so it's this that implies that so this is the antecedent all ready let's uh move on to uh to part f how did you do with that one well this is where we start combining words like if and when with an only and because it says only when carl is playing this guy is the consequence the team wins only when carl's playing so if you know that the team wins you can conclude that kyle is playing because he only win when carl is playing so that's just another way of saying that card is playing is a consequence of the team winning okay so this is the antecedent and as i mentioned at the beginning of this discussion even if you're a native english speaker these typically cause people a lot of trouble it's just the way the human mind works it's not to do with the native languages to do with the the way the mind works well these two are a little bit more straightforward than the last one or two because the the when is really almost the same as if uh that just uh tells you the condition on which something is so the thing that just goes with a when without and only compared with it is uh is the antecedent okay so the antecedent in this case is that kyle is playing because it's when he's playing that the team wins if you know that uh if you know that carl is playing then you can conclude that the team's going to win on the basis of this statement so that's the antecedent there and it doesn't matter whether the the when clause comes first or second and as was the case with an if clause uh it can come first or it can come second that's still the antecedent okay being the antecedent uh is not directly related to whether you're the first clause or the second clause in a sentence it gets to go with which word you're combined with if it's an if or it's a if it's a when um then that's uh combines with the antecedents if it's an only if on only when then then that flips it around and then you're dealing with the consequence okay well well that's uh that's takes care of those kind of examples and the the last two parts or the last three parts of this question were this problem set were were a little bit different okay well in this case it's certainly true that if m and n are even then m n is even we know that so the question is is it the case that if m n is even then m and n are even so that's what this boils down to now this this this implication okay we know what there's there are two implications here it's an if and only if so that means the equivalence it means the the implication holds in both directions and one equivalent and one implication is certainly true if m and n are even then m n is even and so it boils down to the question if the product is even then are the two numbers necessarily even and once you get it down to that stage all you need to do is just observe there's a counter there are many counter examples um we could take m equals 2 n equals 3 then m n equals 6. so here we've got a product that's even but it's not the case that both numbers are even so this is a counter example so the answer to the question is no this is a statement about any pairs of integers and if we find one pair of integers that makes it fail then the whole statement fails so it's a counter example that we need to find and we found one m equals two n equals three the product is even it is not the case that both numbers are even okay now let's move on to number six so number six asks us is it the case that mn is odd if and only if m and n are odd well i'm sure you all realize that the the answer is yes and you almost certainly realize the reason you know the reason why because there were two facts we know that odd times odd i better say is not equals is odd and we know this even times anything any number odd or even is even so if you take two odd numbers multiply them together you get an odd number if you take a number a pair of numbers any one of which is even and multiply them together you get even and when you combine those two this falls out of it and if you don't see that i mean i'm going to leave it to you actually if you want to sort of give a little bit more detail and express it as an implication in both directions that's fine uh you could discuss this and i'm sure you could discuss this endlessly on the forums and that would be a good idea if you want to but i'm just going to leave it with the observation that it's really just these two facts that give you this result okay this tells you how the parity even a nod works and once you know that you know that but uh you know have fun with this one and and discuss it amongst yourselves and settle to your own satisfaction what constitutes a rigorous proof of this thing remember the there's actually no sort of gold standard of what is or is not a rigorous proof it depends on the on the experience of the the audience a proof in many ways uh involves audience design you've got to cast that proof at the level of detail and precision that matches the audience um you know typically a mathematician a professional mathematician would simply say and you see this actually in books and in papers a mathematician might very well say this is trivial okay um and that would be the proof in advance works on mathematics you often see remarks like the proof is trivial um it has to be said that a beginner might take several days to see why something's trivial and when mathematicians use that kind of expression they're doing it with a particular audience in mind namely other professional mathematicians so you know if you read that and it doesn't seem trivial to you it doesn't mean you're stupid it just means you haven't spent many years working as a professional mathematician it's just the way we we classify things um and proofs involve a lot of audience design when you write and formulate proofs okay well one way to do all of these is by by truth tables and if you work out the truth tables you find that a is true that b is true c is not true d is true e is true and f is true so simply by using truth tables you could answer this um this one we've already seen in in in the lectures and discussions we've we've sort of looked at these um this equivalence um this one and where is it that one those are examples of what's known as de morgan's laws after a mathematician augustus de morgan that if you take a negation with a disjunction you end up with a conjunction of the two negations and if you take an addition a negation of a conjunction you end up with the disjunction of the two negations okay so that was a basic fact about implication that a conditional is true um if either the antecedent is false or the consequence is true so that was that was the truth table that we worked out for for the conditional um those are de morgan's laws that one's not that one's not for that was not the case anyway um this one um you could probably reason this one out uh in terms of implication uh just think in terms of when implica under which circumstances can you start with an assumption p and deduce two conclusions and then see when that doesn't happen okay and you should end up with this so it's possible to reason this one out just in terms of implication and the same is true for this one you could reason it out uh this one i think is easier to reason out than that one because of fewer symbols to deal with this says that um if you have a con if you have an assumption and then you have another assumption you can make a conclusion from it well do you do it in two steps do you assume p and then on the basis of that show that if q is true then r is true or do you simply assume that both p and q is true true and deduce r those two things would be equivalent um you know this is sort of doing it in two steps and this is combining the two assumptions they both really tell us that there are two assumptions there's one assumption then there's another assumption and in this case we've explicitly said there are two assumptions and and in both cases it's the r that's following from them okay so the question is how and when do you get from p and q to a conclusion are so this you could reason it out and if you feel uneasy about that you could just work out the the truth table okay um but since we spent a lot of time on truth tables in the assignments my recommendation would be that you would go through these and actually try to reason them out in terms of what they mean you know truth tables are good if you're a computer but people are not computers they're much more interesting creatures than that and i think we have the power of reasoning and so i would suggest you go through these and try to to reason them out in terms of what these things mean because that after all is really what this entire course is about when question 8 we make use of the course evaluation rubric this is the first of many questions that you'll you'll be asked to do making use of this rubric in order to evaluate purported mathematical proofs what is the first one the claim is that for any two propositions p and q not p conjoined with not q is equivalent to not p and q uh the argument is fairly short in fact all of the ones i'm going to be using as examples of short that's why they're good as examples in particular i want to be able to give the solutions on a single slide as i'm going to do here so in a way these are not typical in fact if you do test flights at the end of the course you will almost certainly see proofs presented by other students which are much more complicated and much longer and perhaps have all sorts of mistakes in them the arguments i'm giving you they are they all began as arguments that were produced by students over the years when i've taught this material but what i've done is i've picked particular aspects of proofs where students typically go wrong and so the examples will have one or two common mistakes embedded in them just so you get used to looking at proofs from the different perspectives as captured by the rubric and seeing how they work and don't work as i mentioned in the in the description of the use of the rubric on the course website the way it will work is that because i'm using short examples in each particular example some of these these these features won't really apply in which case you have to sort of default and give four or zero depending on how the student handles it typically you would end up giving full marks because it just doesn't really apply okay well let's uh let's take a look at how the uh how this one was done and so we'll i'll pretend that it was done by a single student even though this is a composite of the kind of things that students have done over the years okay well it's an equivalence so we have to prove it in two directions we have to prove that this implies that and that implies that so there were two equivalences to prove here so let's just uh check the left right implication so suppose that not p and not q is true then a conjunction is true if and only if the two conjuncts are true so if the conjunction is true then both not p and not q are true okay that's correct that's what conjunction means if not p is true that means p is false that's what negation means if not q is true that means q is false so the truth of these two negations means the falsity of p and q well if p and q are both false then again because of the way conjunction works p and q is false hence not p and q is true again because of the way negation works so that's absolutely okay okay left to right was proved great what about right to left well in this case the student uh makes an attempt to be fairly sophisticated by saying the other argument about the argument in the other direction works the other way if it does work the other way it's absolutely okay to simply say that there's no requirement that you would have to repeat something that's that's obviously the case but we'd better check that it is obviously the case because we're actually evaluating whether this is the case okay so let's um let's spell out what this person didn't do in other words we'll try to do the same as here going to the direction so we're going to assume it's not the case the p and q okay let me just say we'll assume that's true and that's sort of redundant but i want to talk about truth and falsity so we'll assume that not p and q is true that means that p and q is false because of the negation if the negation is true it means the p and q is false well what does that mean that means at least one of p and q is false okay because you only need one of them false to make the conjunction false ah but the other one whichever it is could be true it doesn't have to be but there's nothing to rule out the fact that the other one's true so one f p and q is true that means one of not p and not q could be false okay at least one of them is false so the other one could be true that means one of those could actually be false one of these guys not being q could be false that means if one of them could be false it means not p and not q could be false that's not ruled out in other words that implication does not work just because not p and q is true it does not necessarily follow that not p and not q is true because it could be false we're not saying it is we're saying it could be so there isn't an implication there is no implication from right to left because you could have that without having that that thing could in fact be false in other words the original claim is a false claim it's simply not true so this statement though it was a it was a nice attempt to be somewhat sophisticated uh it didn't work because in fact the argument does not work the other way i mean it is essentially the same kind of argument and if it had been correct that would have been fine you wouldn't need to give this but it's not correct so now we're going to have to sort of put some numbers in here that capture what we've just said okay well what am i going to say there are formats available for logical correctness the left to right part was absolutely correct this is logically correct so i'm just going to give half marks with that i'm going to say 2 captures the fact that left to right was correct but remember logical correctness is just one aspect of proof there are other factors too and that's what the the rubric items are doing so let's look at what we've got clarity this is absolutely clear even here it was clear it was wrong but it was clear so in terms of clarity this is very clear so i'm going to give full marks for clarity is there an opening insofar as there's any reason to give an opening here yes it begins by stating that we're going to go from left to right and so i'm going to give four for that now you could say maybe the person should have begun by saying to prove an equivalence you have to prove implications from left to right and right to left well you know um it's a judgment call i would say that for someone at this stage where they're producing arguments like this and this is a nice argument this is perfectly correct for someone that's producing this um i think it's okay to say yeah it's obvious by the way they're doing it that they're doing that you know but these are remember even though the rubric does break a difficult task into smaller pieces they're still not easy so these are going to be judgment calls um i would say that there's enough demonstration of of knowing the person knowing what he or she is doing to just say i'm not going to insist that they say proving an equivalence is enough to prove left to right and right to left i think it's clear from the way they're doing it but they do start out by saying assume one and prove the other one okay what about stating the conclusion well absolutely it was stated we have implication in both directions up there well this just justifies what i've just said so the person has now said we have implication in both directions so just emphasizing the fact that he or she knew exactly what they were doing um remember also that we're really trying to do uh formative assessment rather than summative the process is is one of seeing what the person has done well and rewarding what they did well it's not about trying to take marks off when we do this kind of thing we should always be looking for reasons to to give marks to to to acknowledge what's been done correct and then to point out things to improve so we begin by adopting a positive giving max attitude that's the best way to do this kind of thing so i'm always going to be looking for reasons to give marks on the other hand if there's any you can't give marks if they're not justified okay what about reasons uh i mean reasons were given but it was all it was really the same as in the first case the the reasons for the other part that should have been given weren't because the person thought it was the same and it wasn't so some of the reasons are here they were given as a person was going along so i would have to give two for that one because it was only half a proof um i mean the proof was listed the result's false so this is absolutely false um so the most i can give here is two because even though reasons were given that's not a reason um that would have been that would have actually been a reason if it was correct if this kind of argument was correct that would have been okay but it's not um and then overall um again i'm just gonna have to give two because it's basically just half a result okay that's really what it's coming down to the person has laid it out well so so he or she is going to get a lot of marks for laying it out correctly but they're going to be losing some marks because they're only half of the thing i'm just saying there's only half available and so i'm giving half the max so the total is going to be 18. it may be a little bit generous actually um after all this thing is false um in fact if this was a mathematics course and and i was sort of grading people for doing mathematics work i think i'd have been much more harsh i'd probably come down at 10 12 maybe even less but this is about much more than mathematics it's about mathematical thinking it's about communication it's about understanding proofs um in reality the mark i've given the person is two right i've given them uh two out of the four so essentially i've said this is worth half marks in terms of the overall evaluation because half the proof is there okay but because we are also rewarding or looking to reward proof structure communication all of those other aspects of proofs which are important um i'm gonna give credit for the fact that the person has got the general idea okay so i've got 18. as i said maybe regarded as a little bit generous i'm inclined to think it's a little bit generous myself on the other hand um this is a 75 okay um which means 25 percent has been docked if you like or not given um and let's look at what happened this was essentially just one mistake the person looked at it and and didn't look close enough but it comes down to just one error it's one error in logic the person who could produce that argument almost certainly could produce the correct argument in either direction to show that it was false so it really comes down to one slip and frankly it seems to me that if you if you docked more than 25 or maybe 30 percent for just one error um that's kind of harsh um we are after all trying to turn turn people into better better mathematicians to make them better mathematical thinkers so let's look at what they do right and then give due credit for that and this person has done a lot of things right there's just one mistake and it was a mistake that almost certainly this person shouldn't have made they had the ability not to make the mistake but people do make mistakes so we're not going to give maxwell for free but um having hesitated a little bit on the 18 i've actually now talked myself back into saying 18 is actually a pretty good grade already well that's the end of that problem set how did you get on with that massive assignment for to give you a chance to complete it this lecture is also fairly short and this time the assignment is down to a single page so keep working on that last assignment it really is crucial to master those ideas and that terminology knowing when and why one statement implies another and being able to distinguish between necessity and sufficiency are crucial abilities in today's world there are two more language constructions that are fundamental to expressing and proving mathematical facts and which mathematicians therefore have to be precise about the two quantifiers there exists and for all when we've made these terms precise and are sure we know how to use them properly our analysis of language will be over the word quantifier is used in a very idiosyncratic fashion here in normal use it means specifying the number or amount of something in mathematics it's used to refer to the two extremes there is at least one and for all these are all we need to look at because of the special nature of mathematical truths when mathematics is viewed as a subject in its own right as opposed to being a set of tools used in other disciplines and walks of life the core of the subject is the theorem and the majority of mathematical theorems will have one of two forms there is an object x having property p for all objects x property p holes i'll take these one at a time we're looking at statements of the form there's an object x having property p for example the equation x squared plus two x plus one equals zero has a real root we can emphasize that this is an existence statement by writing it in the following form there is a real number x such that x squared plus two x plus one equals zero or we could write it this way there exists a real number x such that x squared plus two x plus one equals zero doesn't matter whether we write it with an is or exists it's an existence statement the symbolic abbreviation that mathematicians use for exists is a back to front e e for exists of course so a mathematician would write this as follows there exists an x such that x squared plus two x plus one equals zero this symbol is called the existential quantifier the simplest way to prove an existential statement of this form is to find an actual x that satisfies the property and with this example it's easy take x equals negative one then x squared plus two x plus one equals negative 1 squared which is 1 plus 2 times negative 1 which is negative 2 then we've got the plus 1 1 minus 2 plus 1 equals 0 which solves it well that was a simple example sometimes we can prove an existential statement without finding an actual object that satisfies the property let me give you an example of that i'm going to prove this statement there exists an x such as x cubed plus 3x plus 1 equals 0. i'm not going to find an actual x that solves this cubic equation but i am going to show that there is a solution i'm going to start by looking at the function y equals x cubed plus three x plus one this is a continuous function continuity of functions is actually a fairly deep topic uh we'll touch on it briefly at the end of the course but for my present purpose all i need to know is that a continuous function is one whose graph is a smooth curve with no breaks and jumps and you should be sufficiently familiar with cubic equations by now to know that the cubic equation has a nice smooth curve that looks something like that in many cases up and down two curves okay the point is it doesn't have any jumps if x equals negative one this curve has value or this function has a value negative one cubed which is negative one plus three times negative one which is negative three plus one which is negative three if x equals plus one the curve or the function has value y equals one plus three plus one which equals five so the curve lies below the x-axis for x equals negative one and above the x-axis for x equals plus one so it looks something like this put the orange in here we're gonna look at negative one and we're gonna look at plus one it's gonna lie somewhere down here at negative one it's gonna last somewhere up here at plus one and somewhere it's gonna cross the axis i mean it looks like that that who knows the point is it's going to have to cross the axis somewhere or rather to the left of the origin or to the right of the origin doesn't matter where the important point is somewhere between negative one and plus one this curve crosses the axis and when it crosses the axis the y value is zero so we haven't said which x solves the equation we haven't even said whether the x is negative or positive but we do know there is such an x we've shown that then x exists that satisfies the equation without finding such an x this is an example of an indirect proof we haven't proved it directly by finding an x that solves the equation we're simply shown that there is an x that solves the equation a lot of mathematical proofs are of this nature we show that there is a solution to some equation or that there is an object that satisfies some property over there without finding such an object pretty cool huh time for a quiz a simple modification of that last argument yields a more specific result which of the following is it is it one there is an x less than zero such that x cubed plus three x plus one equals zero or is it two there's an x greater than zero such that x cubed plus three x plus one equals zero okay which one do you think it is well the answer is one and here's how we can get that in the previous argument i looked at the cave at x equals negative one and x equals plus one and i'm sure it looks something like this at negative one the curve was below the x-axis at plus one the curve was above the x-axis and so it has to cross the x-axis somewhere between negative one and plus one but instead of going between negative one and plus one i could go between negative one and the origin itself well note that if x equals zero then y equals x cubed plus three x plus one is one so at zero it's somewhere up here which means it's negative at negative one it's positive zero so it's somewhere between negative one and zero that it crosses the axis since the curve lies below the x-axis at x equals negative one and above the x-axis at x equals zero it must cross the x-axis between negative one and zero or i could prove it a different way i could simply observe that if x is greater than or equal to zero then x cubed plus three x plus one is greater than zero and in particular is not equal to zero so two is false and that just leaves one we know that one of these two has to be true as a result of the previous argument and so if we can eliminate two as i've just done here then i can conclude one so either way either by refining the previous arguments or by deducing it from the previous results by this observation i end up showing that the equation has a root somewhere between negative one and zero i still haven't found that root but i have sure that there is a root it's an indirect proof incidentally in giving that last proof i was careful to distinguish between the terms negative one and minus one chances are your middle or high school teacher made a big deal of that like many professional mathematicians maybe even most of us i'm generally far less careful i tend to use negative and minus interchangeably it's not that we don't know the distinction it's just that at university level we focus on other issues and since this is a university course i'll assume you can use your knowledge of arithmetic to determine the intended meaning just as we do with everyday language in this course a similar issue arises with implication in the conditional and since that distinction is likely new to you i'm going to try to always use the correct term but when i'm talking with my professional colleagues we just talk about implication to cover both cases and rely on our shared knowledge of the concepts to disambiguate in both cases the distinctions are important which is why we emphasize them in our teaching but once we're sure everyone has fully grasped the issues we adopt a more relaxed attitude confident that our understanding will keep us out of trouble the same kind of argument i just used to show that a certain cubic equation has a real root can be used to prove the wobbly table theorem suppose you're sitting in a restaurant at a perfectly square table with four identical legs one at each corner because the floor is uneven the table wobbles one solution is to fold a small piece of paper and insert it under one leg until the table is stable but there's another solution simply by rotating the table you'll be able to position it so that it doesn't wobble you might enjoy trying to prove this the solution's simple but it can take a lot of effort before you find it it's very much a thinking outside the box question it would be an unfair question on a timed exam and i'm not giving it as a course assignment but it's a great puzzle to keep thinking about until you hit up on the right idea i'll leave you with it even if you can't think of a mathematical solution you could go off in search of a square wobbly table and confirm the theorem experimentally meanwhile let's get back to our main theme sometimes it's not immediately obvious that a statement is an existence assertion in fact many mathematical statements that do not look like an existence statement on the surface turn out to be precisely that when you work out what they mean for example the statement that a number x is rational is an existence statement and here's why let's take the specific statement square root of two is rational as it happens this is a false statement but uh we're just using it as an example of a statement so on the face of it this doesn't look like an existence statement but it actually is because we can write it as there exists natural numbers p and q such that square root of two equals p divided by q or using our symbolic abbreviation exist p exists q such that square root of 2 is p over q incidentally when we do come to prove that the square root of 2 is not rational we'll do it by assuring that there are no values of p and q that satisfy this equation so we'll prove a non-existent statement well this expression is fine so long as we know in advance that p and q denote natural numbers i mean here i said it explicitly here i didn't so how do you know that p and q do not natural numbers i mean maybe they should denote real numbers or complex numbers the answer is you make it more specific and you do that by writing it in the following way there exists a p in n and there exists a q in n such that root two equals p over q where n denotes a set of natural numbers incidentally another font that's often used to denote the set of natural numbers is something like this an n with a double line for the diagonal in this course i'm assuming that you're familiar with set there with basic set theory and you know what the the the membership symbol means uh if you're not then there is a course supplement and i'm just going to leave it to you to to read that supplement and of course there's of course textbook if you if you acquire the textbook that's my own book introduction to mathematical thinking in any case i'm going to assume that you are familiar with set theory and i'm not going to talk about it in the course itself you sometimes see mathematicians write it in an even more abbreviated fashion there exists a pq in n square root of 2 is p over q that's fine if you're confident about the material and you're familiar with the notations but we're going to be looking at examples where there are many different types of quantifiers coming in together and then if you start putting them together in this fashion there's a possibility of getting confused and going off course so i would say at the early stages keep things distinct as they are here and try to avoid that for now as i say professional mathematicians write this kind of thing all the time but just as when we're beginning to learn mathematics it's important that we distinguish things like negative and minus and then when we master them we tend to forget the distinctions and we don't forget them but we don't make them explicit so here when we're beginning to look at quantifiers i think it's important to be very explicit and then once you once you really understand it then you can do what we do in the business we just write things in the in the simplest fashion confident that we know what's going on and that the the way we write it won't mislead us incidentally see if you can prove that the square root of two is not rational we'll do that in class later but see if you can do it now what that amounts to is proving that there do not exist p and q in the natural numbers such that the square root of 2 equals p over q or the 2 equals p squared over q squared this is the statement you actually prove in order to show that the square root of 2 is irrational it's not a difficult argument but it's rather clever and ingenious it's fairly short and the chances are that you're not going to come up with it but it's worth giving it a shot you don't spend a half an hour an hour or so thinking about it to see if you can if you can prove that statement good luck on that one but we will come back to that in class how are you doing do try to show that the square root of two is irrational not so much to find the proof but to get used to what it's like to do university level mathematics one feature you need to get used to in mastering college mathematics or more generally what i'm calling mathematical thinking is the length of time you may need to spend puzzling about a problem or even one particular detail for the most part without seeming to be making progress high school mathematics courses particularly in the us are generally put together so that most problems can be done in a few minutes with the goal of covering an extensive curriculum at college there's far less material to cover but the aim is to cover it in more depth that means you have to adjust to the slower pace with a lot more thinking and less doing at first this comes hard since thinking without seeming to be making progress is initially frustrating but it's much like learning to ride a bike for a long time you keep falling or relying on training wheels and it seems you'll never get it then suddenly one day you find you can do it and you can't understand why it took so long to get there but that long period of repeated falling was essential to your body learning how to do it training your mind to think mathematically about various kinds of problems is very much like that okay sermon over the one remaining piece of language we need to examine and make sure we fully comprehend is the universal quantifier which asserts that something holds for all x this notation means for all x it's the case that something of other the symbol here is an upside down letter a means for all for example if i wanted to say the square of any real number is greater than or equal to zero i could write it like this for all x x squared greater than or equal to zero so the short way of actually saying that is for all x x squared greater than or equal to zero just as we would write things like there exists an x such that x squared equals zero notice that i didn't use the word all in this sentence i use the word any the same thing happened with exists there are different words we can use to express an existence assertion and there are different words we can use to express a a for all assertion this is the universal quantifier how do i know what the x means well i have to specify what it means if i want to be explicit i would have to write something like for all x in the set of real numbers x squared greater than or equal to zero okay let's take a look at combinations of quantifiers most statements in mathematics involve two or more quantifiers combined for example if i want to say there's no largest natural number what i would write is this for all m in the set of natural numbers there is an n in the set of natural numbers such that n is bigger than m for all natural numbers m there is a natural number n such that n is bigger than m that clearly says there's no largest natural number note that the order of the quantifiers is important if i swap those quantifiers round and write exists and then in n such that for all m in n n is bigger than m the result says there is a natural number n which has the property that for all natural numbers m n is bigger than m in other words this says there is a natural number bigger than all natural numbers which is false all i've done is swap the quantifiers around but in so doing i've turned a true sentence into a false sentence remember that example from the american melanoma foundation in their fact sheet they wrote one american dies of melanoma almost every hour using our quantifiers we could write that like this there exists an american such that for every hour he dies in our age there is an american such that every hour a dies in our h poor guy i mean quite amazing what the writer obviously meant was the following for every hour there is an american such as a dies in our age it will be different americans for different hours now as i mentioned earlier in the case of everyday english these are almost never a problem everyone understands the context we know what's meant uh we know that this one is not meant because this is this is clearly wrong we know that that one's the one that's meant and with a simple mathematical example like this one arguably it's not important either because everybody knows what's meant we know that there is a there is no largest natural number we know it must be this one and not this one so in these examples there's probably no problem but we want to use mathematical notation in complicated situations and to talk about things we don't yet understand that means we can't disambiguate it's really important that we say things in the right order okay i think it's time for a quiz so how did you get on well the literal meaning of this statement is captured by number one that's the literal meaning it says there is a license for which there are two distinct states that that license comes from so this is the literal meaning but it's false licences are issued by states you can't get a license that's issued by two separate states so the correct answer to the question is number one but it's not what the person who wrote that sentence in the in the form in the driver's license application meant to say let's look at number two that says there are two different licenses they're different there are two licenses which are issued by one state well i guess it's possible that you could own two licenses that are both issued by the same state if you had two identities or something like that so this is a possibly true statement but it's not what the sentence means let's look at number three that says there are two licenses and two states different states different licenses there are two licenses in two states one license comes from one state the other license comes from the other state this is the sentence that the driver's license application meant to say so the literal meaning is number one which is false number two may be true but it's it's not what the what the sentence is about number three is is clearly the intended meaning it is a meaning that we would all understand from this because we know about licenses and about states and so forth so this is this is the one we understand but it's not the literal meaning doesn't matter when you're applying for a driver's license unless i guess if you get into trouble you get your lawyer might try to get you off by arguing on the basis of the mathematical meaning good luck if you want to try that but in mathematics we can't allow this kind of thing to happen we get all sorts of problems and in mathematics we can't hire a lawyer to get us out of difficulty okay let's look at the next question in this quiz well the correct answer is number three let's see why what does the first one say it says there is a license and there are two different states such that that license is valid in one state and valid in the other state so it says there's a license that's valid in two states well okay it's true because in america when you have a license valid in one state it's valid in any state you can use it anywhere so it doesn't say anything particularly interesting what about number two that says for every license and for every pair of states which could even be the same state the license is valid in one state or it's valid in another state well this is actually false as a statement because there can be invalid licenses okay if you've got an invalid license that means the whole thing is wrong this one is the one that captures that that sentence let's just read it it says for every license any license if there is a state in which that license is valid then that license is valid in all states so three is the one that captures this sentence one is actually a true statement but of uh very little relevance and two is actually a false statement it's false remember because you can have invalid licenses so it's not the case that for all licenses and for all states the license is valid in one state or another state okay how did you do on that you might have had difficulty reading the formulas in that last quiz if so check out the supplementary video tutorial called how to read mathematical formulas you'll find it in the same place you access all the regular tutorial videos in this video i'll explain how you read mathematical formulas just as you're probably familiar with in arithmetic and algebra you need to know the conventions regarding the order in which logical operators apply well let me begin by just quickly summarizing the precedence order for applying the logical connectives the ones that bind the most tightly the ones that are sort of the strongest if you want but that holds something close together the quantifiers for all and exist and a quantifier applies to whatever comes adjacent to it okay now typically what comes adjacent to it uh involves various other things like hands and nose and knots so you would put them in in parentheses or brackets square brackets or whatever so very often in fact i almost always make a habit of putting whatever i want next to it in parentheses because for all then applies to everything that comes there okay it binds tightly to this applies to everything the same with existing here now if there's only something very simple coming next for example supposing i wanted to say all the balls are red i could say for all balls red b if red is a predicate that applies to balls so i could say for all b red b and that would apply to the red balls and if there was something else here you know if it was and da da da then the fall would not apply to that if i wanted the for all to apply to that i'd have to put parentheses around there i didn't put parentheses here i could do again especially when i'm giving introductory level courses i usually put parentheses around quantifiers but if you look at some of my research work and advanced courses you'll find i often don't do that that's very consistent among among instructors um you know the golden rule is if there's any doubt whatsoever and if you're beginning on this material there certainly would be doubts if there's any doubts put parentheses in um you know you can't have too many well that's not quite true if you have too many parentheses it gets hard to read it um so you have to strike a balance but always if there's going to be any ambiguity put the parentheses in and mix parentheses you know along square brackets or even spaces i'll try to remember to give an example in a minute with a space because you can sometimes use spaces to disambiguate but the golden rule must be you want to avoid someone being left unclear as to what the meaning is okay negation is about the same strength as well it is the same strength as the quantifiers so uh the negation applies to whatever's immediately next to it and since we usually want a whole bunch of things to be negated then the negation is followed by parentheses and then it applies to everything between there um let me give you the following example suppose we wanted to say not the case that 3 is greater than 0 and 3 is less than 0. okay well is 3 bigger than 0 yes is three less than zero or no so here i've got a conjunction of something that's true and something that's false so this conjunction is false so it's negation is true so this guy is true but supposing i wrote it this way not the case three greater than zero and three less than zero supposing the parentheses didn't include both of those they just included one of them in that case what have i got this guy is false this guy so he is greater than zero is true so that guy's fault so here i've got a conjunction of false things so i've got something that's false so these clearly aren't the same because this one's true and that one's false here the negation applies to everything in between which makes it true here the negation only applies to the thing next to it now i could have gone back here and put parentheses here and i'll mention that in a moment actually comes up in the next the next priority that wouldn't have changed things that wouldn't have changed things because the negation would apply to what was in the next parenthesis negation applies to whatever comes next and what comes next is the whole thing because of parentheses includes the whole thing so simply putting parentheses inside doesn't change anything it makes it maybe a little bit clearer although this is one of those cases where adding parentheses arguably makes things a little bit less clear but in terms of the logic the issue between these two wasn't whether there were parentheses around the three greater than or less than zero the issue was whether the parenthesis governed everything that was next or just the one thing that was next okay so these are not the same okay um the next one is conjunction let me now just pick up that thing i mentioned before when i did that the first time i wrote this i said three is bigger than zero and 3 is less than 0. now in fact left of space if you watch what i did i left a space um and i realized at the time i was doing it that's what i was doing which is why i decided to pick it up now so this says that 3 is bigger than 0 and three is less than zero you actually don't strictly speaking need parentheses around here because this is an atomic formula as we sometimes call it this is a basic building block out of which we're building more complex formulas this simply states a fact three greater than zero an atomic fact a single fact this states another atomic fact three lessons over so when you have basic facts about arithmetic or whatever they are they stand on their own the conjunctions and the kind of quantifiers are what combine these basic facts so you strictly speaking don't need to put parentheses around these um this is a case where i typically would just leave a bit of extra space in here to sort of make it clear that this is a unit and that's a unit on the other hand if you want to be safe and it's always wise to be safe if you're at all unsure you could put those things in okay and i did this here i went back and put them in just to make it clear okay um then um well some mathematicians would say that conjunction disjunction are more or less the same or conjunction disjunction implication we're getting down to a sort of a general grouping now where everything has roughly the same strength there are actually some arguments that say that conjunction should be tighter than disjunction but it's it's it's not particularly strong in any case i think in this this is a case where one should always use parentheses to disambiguate the point is that the conjunction applies to whatever's to the left of it and whatever's to the right of it and if you want it to apply to a whole bunch of things you put them inside parentheses likewise here you would have a whole bunch of things and the same is true for disjunction implications so regardless of whether you think that's stronger than those the issue should never really arise because you should always put things in parentheses to just say it's this guy or this guy and in here there could be a whole bunch of things and that whole bunch of things will be disjoined with this and likewise here if you have an implication of a conditional this whole thing would be the antecedent and this would be the consequent now in here there may be all sorts of conjunctions and disjunctions and stuff there may be quantifiers in here there may be quantifiers in here there could be all sorts of stuff in here negation signs inside this whole thing would imply that whole thing so whenever you're looking at i mean the basic thing with all of these is when you've got a quantifier or a negation symbol or a conjunction or a disjunction or an implication or equivalence actually i didn't talk about equivalence but equivalence is just a conjunction of two uh implications the biconditional so we could put that one in here as well strictly speaking if you did assume that that had more tight binding than this you could write something like this you could write a and b or c and d and if you put space in there that i would think is fairly uh fairly clear that it meant to be this or that so this is i would say at least the way i was brought up let's put it that way as a mathematician i was brought up to us to say that that actually is is okay and it's different it's not ambiguous but i would almost certainly now i think i've cured myself of that childhood sin i i would always put in parentheses and say it's a and b or c and d i mean you just have to be very careful about not making sure that things are nicely not ambiguous okay um so golden rule put parentheses in and then everything applies to whatever comes between the next parentheses it's kind of optional as to whether you have parentheses around the quantifiers but the parentheses that follow providing you always do it this shouldn't be any problem okay now let me illustrate those points with examples that are all related to that quiz at the end of lecture five okay so let's uh let's take these one at a time this was actually from that particular quiz i've got a fool and for all applies to everything that's adjacent to it now that parenthesis there teams up with that parenthesis there and i've actually written them not as parentheses but a square brackets to make it absolutely clear that is the thing that the fall applies to okay for all applies to all of this in particular what it's going to say is that pick any license and then it's going to say something about the licenses now the license l is going to appear on both sides of this conditional and it will be the same l because the l has been picked here and once you pick the l it'll apply to everything here so that l is bound by that quantifier okay so we'll say l is bound by the quantifier for all l okay quantifier binding okay so let's read it now in uh in english it says for any license l if there is a state in which l is valid if l is valid in some states at least one state then l is valid in every state okay for any license l if l is valid in some state the exists here simply applies to this the exist binds what's next to it so they exist simply binds this thing likewise the for all binds what's next to it which is this thing so what this says is for every license for any license or for every license held something happens what happens if that license is valid in some states then that license that same license is valid in every state so this is the one that actually says a license that's valid in one state is valid in every state which is true in the united states by the way um okay that's the first one let's look at the second one um what's the difference let's see what we've got um for all applies to something in the middle so for all applies to everything here because i've got the parentheses the only difference is that instead of having a conditional or an implication i've got a conjunction so let's see what that how we would read that um okay i mean the binding is the same the for all applies to everything here this exists applies to this thing this for all applies to this thing and the l is the same l here as here because once you've said the for all l within this expression here the l is determined by that the l is still bound so as as was the case there the l inside here is bound okay but what does it how would we read it we'd say for every license l there is a state in which the license is valid and the license is valid in all states so let me just write that down for any license l there is a state in which l is valid and l is valid in every state that doesn't mean the same as the previous one just think about this for any licence cell for any license there is a state in which that license is valid and early is in fact valid in every state but this is actually false this precludes the fact that there are invalid licenses for example if you go to california and you drive with too much alcohol in your bloodstream you will find yourself with an invalid license not every license is valid so this is actually a false statement the one above was true this is true in in the united states this is false this means something different okay in fact really uh it's the it's the first thing that was uh the problem for every license there is a state in which it's valid well that's simply not the case already the first conjunct makes it invalid didn't arise in the first one because in the first one the bar the part that said it's valid in a state was the antecedent if it's valid in a state then it's valid in all states so that said for every license if it's valid in a state this says for every license it is valid in a state well that's not the case not all licenses are valid okay um so there is a distinction between these two and in fact the distinction is a meaningful one in terms of validity of licenses and so forth um let's look at this one well here we don't have these brackets so let's see what's going on this means for all l for every license there is a state that the license is valid in that state well is that true is it true remember this is this is a unit the for all and the exist applies to whatever's next so there's a line here the four and the exist don't apply to that they apply to what's next and there was no bracket so it doesn't get included here so what this actually says is for every license there is a state in which that license is valid so what this really says is that all licenses valid somewhere well okay that's not true and it's the statement is if that's the case then for all s2 that would say l well this would say that l is valid in all states well now we've got all sorts of things gone wrong um as a conditional this guy um would look on the face of it as if it was going to be true because the antecedent is false it's not the case that all licenses are valid somewhere that can be invalid licenses so this is a false antecedent now you might be tempted to say since it's a false antecedent the conditional is true but not quite because this guy isn't even defined this is undefined this is meaningless what's l what is that l it's not governed by that quantifier this is just um an orphan it's just sitting there we don't know where it comes from we don't know what it means it's just a letter that has no internal meaning to this formula so it's not the case that this is a valid uh conditional it's actually undefined this is meaningless unless you know what l is if you know what that is you can assign meaning and once you know what l is then you know if l replied to my license if that l there was my license um then we would have a meaningful and in fact a true conditional but as it stands you've just got a completely an unbound variable so the l there is what we might sometimes call just an unbound or free variable unbound variable free variable okay and until you assign a value to it it doesn't have any real meaning and then uh let's finally look at this one for all l1 for this and what's the difference here somewhat similar to the one up here but not quite okay let's just read it so it says for every license and for all pairs of states the license is valid in one state and the license is valid in two states um well that really just means all licenses are valid in all states again this is not the case in the united states because you can have invalid licenses um so we've got something that's that's actually false and it's it's over i mean there's redundancy here because the second s adds nothing new it simply says for all licenses and for all states the license is valid in that state and it's valid in the other state so we could just scrap that and scrap that and we'd have the meaning without any of that stuff so there's nothing actually wrong with this it's just um i mean it's a false statement but it's uh it's got redundant clauses the second clause says adds nothing that the first one didn't already state okay finally let me just try to distinguish between four cases that uh beginners typically get found to be very confusing they're actually really very distinct um and if you find this confusion between these four that's a sure sign that you haven't yet mastered the the notations and the and what they mean okay let me just uh write down uh a transcription of what it means and then let's just ask ourselves uh exactly what that signifies so in english that would say for every x if p of x then q of x if okay for every x if p of x then q of x this is very common okay for every uh for every number for every real number if that number is non non-negative then it has a square root etcetera etcetera etcetera so um this occurs a lot in mathematics this kind of statement for every x if p of x then q of x okay very meaningful um and it's the same x here notice once you've got that for all the x here is the x here so whatever x providing the x satisfies p then it satisfies q so this establishes a relationship between p and q because if you've got an x that satisfies p then that x will definitely satisfy q so this is a very strong and very common uh statement to make this is also pretty common this says for every x p of x and q of x it says that every x satisfies p and q it's kind of strong um i mean it doesn't occur terribly frequently because that's really the same and i mean you could equally you could just as equally say for all x p of x and for all x q of x because you're basically saying everything satisfies p and q and that's equivalent to everything satisfies p and everything satisfies q um notice by the way that this is non-ambiguous because of the binding the fall binds what's next to it so the fall can only bind that therefore all bands what's next to it and so i don't need the parentheses here because in this case the four wall absolutely can't be confused with that so here's a case where you don't need extra parentheses i didn't even write the parentheses here you don't need them this is totally clear in this case okay and it's equivalent to that so you don't see this very often because it really is just saying everything satisfies p and everything satisfies q but it's okay if that arises don't worry about it it might in a context it might be sensible to write that down um but it doesn't have the same sort of logical force that this does this has real logical force it establishes that there's a relationship between p's and q's okay what about this one um this says there is an x for which p of x and q of x well this is again pretty common in mathematics this is quite a strong statement it says you can find a single x which satisfies p and satisfies q okay so you can find an x which has the property p and which has the property q so this is a strong statement that one was strong that one's strong um i mean this one's strong but it's only strong because each part is strong so the there's a there's almost a redundancy in the way it's written so this is maybe i would put strong in quotation marks to sort of say well yes it is strong but it's not strong because of the logical structure it's strong simply because it's making a statement about p and q both been satisfied by all x's okay what about this last guy um this is one that people often write down and this really means nothing in in any real sense it says what there is an x such that if p of x then q of x okay there is an x such that if p of x then q of x um it's pretty rare to have to ever need to say that actually this really doesn't arise particularly frequently uh that you would need to say something like this if you see yourself writing and exist with an implication um the chances are very high that you've sort of got confused um this this is it's really let me just put it says let me just say that this is weak okay it's not on the same strength as these guys because this says for every x if it satisfies p then it satisfies q that's making a strong statement for every x there's an implication this simply says there's one x for which there's an implication well in a sense the implication is almost vacuous then i mean one thing to say for example is that if you can find an x that does not satisfy p in other words you can find an x for which p of x is false if you can find an x anywhere for which p of x is false then you'll have a conditional that's necessarily true so this can be made true by finding an x that doesn't satisfy p okay that's all it would take to make this thing true so if you're trying to make a strong statement if you're trying to make an existence statement if you're using this to say there is an x with a certain property then you could make this guy true simply by finding an x does not satisfy p because if you can make that part false the conditional becomes true so that's one of the reasons really why this is weak um it's uh you know i'm sure there will be circumstances where this is uh this could have some significance but basically my message for you would be forget that one uh it's just uh if you see yourself fighting and exist with uh an implication after it the chances are very high that you've got confused um you know always be prepared to override what i say um you know all sorts of circumstances can arise but in general these guys are all quite significant um that's particularly significant that one's very significant this one is sort of less so because it really just reduces to the two separate things and this one is really pretty weak so exists combined with implications if you see that flag it and say do i really mean what i'm writing that okay well i hope that's helped clear up some of the basic issues about uh reading formulas but like many things at this stage uh really the only way to get rid of any confusions is to just do a whole bunch of examples for yourself okay good luck on sorting these things out as you work through the rest of the course okay problem set three so we've got a variable ranging over doubles tennis matches uh doubles by the way is when uh you have two people playing against two other people they're teams of two people playing against one team against the other uh t is another variable ranging over doubles tennis matches where rosario partners with antonio wx means it was arya and her partner whoever that partner is wins the doubles match x and we have to find the english sentences that mean the same as the symbolic formula exists at twt at least this kind of thing is always a stretch because we're taking everyday language which is ambiguous and vague and underspecified and can be sort of pummeled and stretched in different directions and we're going to capture it with a formula which is precise and well defined so when it says mean the same there's going to be some slack here and sometimes there's a little bit of arguments that one could have um i've picked an example where i think it's fairly clear at least from a mathematical perspective it's clear if you haven't got used to dealing with mathematical language you may have found this more difficult and this whole course is aimed at people who haven't got that familiarity so so i think you will find it difficult okay so let's just hit them one at a time rosario and antonio win every match where they are partners well that's actually every match where there are partners there's really uh a hidden for all t in this thing it's about a faulty um in fact basically what you're saying is in every match the way there are partners they win or she wins i mean one wins they both win they both win it's a bit of a partnership team okay whereas this is existing so it's not that one okay let's look at the next one rosario sometimes wins the match when she partners with antonio so there is a time there is a game when they they partner together when they win that's this one so that's correct now that really does capture it there is a time when they when they play together and as a partner there's a partnership and they win now you know you can looking some sources into to say that sometimes uh is is used exclusively to mean more than one i mean i don't think that's the case i mean you found people that say that um you know language is flexible in any case in mathematics we always interpret things like some uh sometimes as at least one you know that's the whole point about the way we set things up we we eliminate these ambiguities by being specific and we're specific to say that whenever you're asserting something exists sometimes some game or whatever you mean at least one okay in which case you've got existing existing quantifier that means there is at least one tennis game where they play together and they win okay so that one's okay um whenever rosario plays with antonio she wins the match well that's really um again that's for all t wt so it's not that one rosario and antonio win exactly one match where they are partners well that one didn't gonna work right because that says exactly one match there's no specification here of exactly one match if you wanted to do that there's a notation this notation exists a unique t such that wt you can say it and you can see it other ways i mean you can this is just an abbreviation uh for for an expression you know we've seen that in the problem since um actually one of the assignments so you can capture it but this doesn't capture it this just says there's at least one it doesn't say there's exactly one way they win okay rosario and antonio win at least one match when they are partners that's it that's another one that's fine at least one match when they are partners okay if rosario wins the match she must be partnering with antonio well um first of all there's a universal quantifier floating around here i think this is yeah because it's saying whenever she wins the match she must be partnering so there's a universal quantifier here but it's even worse because the universal quantifier is actually for all x for all matches okay for all doubles tennis matches where she wins something or whether so there's something else for us so there's actually here what's generally known as a scope problem in this statement the quantification is actually over something different it's over all possible double tennis matches not just the ones where she's partnering so not only does this not capture it there's actually another issue there's a scope issue involved okay because here the t ranges only over games where they play together in this case we're looking at games where rosario plays with whoever she's playing with okay so um there were a couple of things that prevent this one been there been the right answer okay well i'm you know as i said at the beginning from a mathematical perspective this is actually very clear it's definitely b and it's definitely e and the reason i can be so definitive about that is because i'm familiar with the way we've set up the meaning of this in mathematics to correspond to mean at least one and we interpret in mathematics we interpret anything that exerts an existence to mean exists one at least one and so things like sometimes some of these some of those they're all interpreted to mean at least one okay let's look at the next one well same setup as in question one the only difference is now we're talking about for all t w of t so let's say let's run through this one rosario and antonio win every match where they are partners so every match where they are partners that would be for all t because that's what t captures t is the doubles ten as much as where they partner and they win what is that so that one's okay what about this one i mean this has really got nothing to do with that has it this is rosario always has got nothing to do with winning it's just saying she always plays together that would essentially say that um did x and t are the same variable in fact i mean it's just saying that there's there's no distinction between x and t um so i think rather than cross say this is wrong i'll say this isn't even a candidate i mean it's got nothing to do with winning okay let's look at part c whenever rosario partners with antonio she wins the match whenever she partners that's for all tea she wins she wins there wins so it's all the same in doubles tennis so that's okay what about this one sometimes rosario wins the match oh no i mean first of all um it's it's not about t it's about x sometimes she wins with whoever she's playing with um and it's an existential one so it's essentially of the form exists x w of x that's really what it means sometimes rosario is in the winning team she wins well that's not that it's a different a quantifier and it's over x not t so um not that one i won't cross it out because at least it does talk about winning so it's a candidate but it's not the right candidate okay rosaria wins the match whenever she partners this is whenever she partners that's essentially for all t and whenever that happens she wins okay bingo that matches that that's correct and finally if rosario wins a match she must be partnering with antonio well um essentially you've got something like for all x here um etc so as before as in question one we've got a scope issue here um this is actually about all possible matches not just the ones where she's she's partnering and the conclusion is that she's partnering with antonia so so this doesn't i mean it does talk about winning but it's it's it doesn't come close because there's a scope problem the variables mean different things okay so it's not that one so in this case we've got a we've got c and we've got e um it's kind of unusual to have one of these multiple choice things where three of them are correct but there you go sometimes sometimes that happens okay let's move on to question three when question three if you look through these looking for something that seems to say there's no largest prime i think you quickly end up looking at uh this one d which says that for any number x there is a number y which is prime and bigger than x so that certainly says there's no largest prime now the question is do any of these say the same in a different way well let's just look at them in terms that says that don't exist in the x's and y's for which x is a prime y is not a prime and x is less than one we'll have plenty of pairs x and y that satisfy that so this is actually false so i mean we entered we weren't asked to say whether things are true or false but this is false and we do not we do actually know there is no largest prime that's euclid's theorem that the primes are infinite uh in in the list of primes goes on forever so it can't be this because this is actually false but but in any case it doesn't mean the same this just means something um just nonsensical um of course there exists pairs x and y with x sub prime and y not a prime and x less than y so to say it's not the case is it's clearly wrong what does this say um for every x there is a y such that x well first of all that would say for all x x is prime that would say every number is a prime so that's false as well so that can't possibly be that can't be it that can be it what does this one say four x and four y x's that says as well every number is a prime that's false wow why are we rattling through these these are these are just plain false um they're absurdly false if you like that was the one that was okay what does this say there is an x such that for all y what whoops that says all numbers are prime again forget that it simply says all y's are prime well they're not so that's also absurdly false scratch that one i mean i'm scratching them because not only don't they actually say that but they actually say something nonsensical what does this one say for all x there is a y exit whoa same thing again what this actually says is that everything is a prime which is false and it's absurdly false so it's not just that a b c e and f don't capture it they don't capture anything sensible whatsoever um this one on the other hand does say there's no largest pack okay let's move on to number four well i won't go through these in in detail as i did with the previous question because the same kind of considerations apply the only one that actually captures it is e let's just read what it says there is an x which satisfies phi and for all y's that satisfy phi x and y have to be equal in other words it's impossible to find a y that satisfies phi other than the x that you start with so there is one certainly the statement begins by saying there is something that satisfies phi and this part says it's the only thing that satisfies phi because if you look at all the possible y's that could satisfy phi the only one that does so is y equals x so this actually captures there is a unique x of 5x and these four um they're not only if you try to figure out what they say if they say anything vaguely sensible at all they turn out to be just nonsensical and they certainly don't capture that incidentally this symbol is a moderately common symbol in mathematics it's not something i made up for the exercise the exist with the exclamation point does actually mean there is a unique x you you'll find that quite a bit in mathematics often you need to be able to say there is a unique solution to something now this exercise tells you that you don't need to have a separate quantifier to mean there is a unique one because you can actually define it in terms of the standard existential quantifier and the universal quantifier so this is in fact just an abbreviation but it's a useful abbreviation and so you'll often see it okay well for question five uh let me observe that this symbol i mean you do see this symbol and sometimes in computer science um very rarely in mathematics uh except in a search in a situation like this where i'm using this to refer to some arbitrary but but unspecified binary operation so this isn't a particular operation i just mean there is some some operation which i'll call x or power y and we need to be able to express the fact that that's not commutative so this doesn't have a particular interpretation i'm just using it to mean any particular or any any unspecified binary operation and so what we need to do is ask ourselves which one of these means that it's not commutative well commutative let's write down what it means to be commutative commutative means for all x and for all y x our y equals y r o x so that's what being commutative means which one of these negates that well when you negate universal quantifiers you get existential quantifiers and things that are true become false or the equality becomes an inequality and so if you skip through these you find yup there it was c that says there is an x there is a y for which they are unequal there is an x there is a y for which they are unequal uh so that's certainly uh a negation do i do any of the other ones falling as as being a negation not really not even close because when you negate both the forwards become and exist they don't remain for all so it's not that one um they don't mean for walls and there's a fall there so for a variety of reasons uh none of these three qualify for that so there's no there's no possibility of having two possible expressions that's the only one okay okay question six uh evaluating this proof and this is a very typical of the kind of work you see from students who are beginning to look at proofs because what this person's done is they've identified the key idea this is absolutely the key mathematical idea behind this um the the the fact is that this um this is not prime um and in fact what i'll do let me just give the proof that the person should have done okay and then i'll then i'll discuss why why there's a problem with with writing this down okay so what the person should have done is something like the following you begin by saying the claim is logically equivalent to the following statement for any positive integer n n squared plus 4n plus 3 is not prime it's logically equivalent to that to say there doesn't exist an integer for which that's prime is logically equivalent to saying that for any positive integer it's not prime and the person then was going which should maybe prove this is this is true okay we prove this uh this statement so we're proving a logically equivalent statement okay so let n be a positive integer and i'm doing this one in gory detail because i'm trying to get maximum points for this one okay then by basic algebra uh n squared plus four n plus three equals n plus one n plus two n plus three but n plus one and n plus three positive integers each greater than one okay n plus one is at least two n plus three is at least four so these are positive integers greater than one so by definition n squared plus four in plus 3 is not prime because it's a product of two positive integers each greater than one okay so that's what a person should have done now let's go back to what was here this is the key algebraic uh heart of this thing already but it's not a proof and the reason students often do this kind of thing is they're used from high school they're used to the fact that algebra is all about algebraic manipulation and indeed it is um but we're talking about proofs here and a proof is much more than getting the the algebraic manipulation right uh if the algebraic mathematician's not right you don't have a valid proof but a proof is all about giving reasons and making a [Music] giving an explanation it's a story a story with a beginning a middle and an end uh you know most you know you know there's a there's a i mean there's a joke about a a woody allen in a woody allen movie where woody allen in the character woody allen character says he's been reading a book uh it's about it was war and peace and he summarizes it by saying it was about some russians well this is like saying it's about this um obviously war and peace is is just about some russians but there's much more than that and that's really what we're doing here we're looking for the the full story uh this one's actually i think going to be fairly difficult to grade um i'm going to put four for logical correctness here because logically this is the heart of it once you realize that that's the logical heart so that's that's fine okay um clarity i'm gonna have to give a zero i just there's just no clarity about this because there's no explanations nothing okay uh this is really nothing nothing valuable here um opening um there isn't an opening just jump straight in um let's see stating conclusion whether person did state the conclusion so i'm going to get them you have to give four marks the conclusion is stated um that's that is how you're supposed to end um okay as i did here that's not prime uh reasons there are no reasons given okay i mean that's that's just going to be a zero um and then let's see what have i got here um overall valuation zero i mean as a proof i can't really give anything for that okay well i don't know let me think um you know actually now i'm going to be a bit generous here i'm going to give two for that i think because this is this is key i mean this is i'm going to give some credit for this i mean i it is the key part of this um and it was the the setting that was wrong so i'm going to give two for that so that means i've got four eight i've got ten for that one um um yeah a little bit generous maybe as a proof um the person certainly has has the algebraic ability and seeing this is key i mean that really is uh yeah okay i was i was i think that's okay um i think that's actually a good mark to give for that one okay but these are these are not easy to do um you're making value judgments you're trying to sort of assess a whole bunch of different things um the way we structure this course is you'll be seeing a lot of exercises like this and the intention is that by the time we get to the end of the course you've got the general gist of of how to do this i mean all instructors develop their own methods and i'm just giving you mine as an example but with something that's sort of essentially qualitative as as grading proofs is and it's like grading essays you know people people end up with different you know there are stricter graders and less strict graders my goal is always when i'm grading this kind of a context is to is to is to look for reasons to give people marks because i want to give the credit for what they're doing but at the same time point out the things that still need to be done okay um alrighty well um that was the end of problem set three how did you get on with assignment five mastery of quantifiers is the one last ability you need to be able to cope with definitions of mathematical concepts and with mathematical reasoning i'm going to begin by looking at the way negation affects quantifiers but before i start let's see how you're progressing by way of a quiz okay let's see what we have is it the case that for every real number x x plus one is greater than or equal to x yes it is is it true there is a real number x so you said x squared plus 1 equals 3. well that would mean x squared equals 2 so we're really saying is it true that there exists a number x whose square is 2. well if we were talking about the rational numbers the answer would be no because the square root of 2 is irrational but for the real numbers it's true what about this one well for very large negative x x cubed is an extremely large negative number which will dominate all of these numbers so for large negative x this expression this cubic expression is negative so it's not the case that it's always positive so that one's false what about this one is it true that for all x there is a y x cubed plus y cubed equals zero that would mean y cubed is equal to minus x cubed the answer is yes given any x just let y be negative x when you cube a number you cube a positive number it's positive you cube a negative number it's negative so by taking y equals negative x in all cases you'll get zero so this one's true what about this one is there an x such that whatever y you add to it you get zero the answer is no if the quantifies were the other way around the answer would be yes for every y there is an x such that x plus y equals zero namely you let x be negative y but it's not the case there is one x that works for all y so that one's false what about this one well is it the case that for all x there is a y such that if x is non-negative then y squared equals x well let's look supposing we took an x if x was negative then game over because this conditional would be true it would have a false antecedent if that x was non-negative then there is a y whose square is x namely the square root of x so whatever x we take there's always going to be a why that satisfies this it's true you might have to think about this one a little bit because there are two cases involved given an x you can find a y that makes this thing true if the x you're given is negative any y will make that true because if the x is negative the antecedent is false but if the x you're given satisfies this condition if the x you're given is non-negative then the y you pick is the square root of x you see what's going on if you're given a negative x there is no y that satisfies y squared equals negative x not in the real numbers but that eliminates that case so this one's quite subtle i suggest you think about this one it's the kind of thing that you encounter quite a lot in mathematics okay well how did you do okay let's move on in mathematics and in everyday life you often find yourself having to negate a statement involving quantifiers yeah of course you can do it simply by putting a negation symbol in front but that's not enough at least often it's not enough you need to produce a positive assertion not a negative one the examples i'll give should make it clear what i mean by positive here but roughly speaking a positive statement is one that says what is rather than what is not in practice a positive statement is one that contains no negation symbol or else one in which any negation symbols are as far inside the statement as is possible without the resulting expression being on julie cumberson let's look at our first example of negating a quantified statement let a of x be some property of x for example x is a real root of the equation x squared plus two x plus one equals zero and i'll show that not for all x a of x is equivalent to the exists x such that not a of x for example it's not the case that all motorists run red lights is equivalent to there is a motorist who does not run red lights well in this case it's pretty obvious that these two are equivalent the proof that i'm going to give in the general case is actually the same reasoning you automatically do when you look at this specific case if it seems hard to follow the issue is purely one of the abstraction i'm proving an equivalence so i'm going to prove an implication from left to right and then from right to left so we begin with the left to right case i'm going to assume not for all x a of x well if it's not the case that for all x a of x then at least one x must fail to satisfy a of x so for at least one x not a of x is true in symbols there exists an x such that not a of x is true well that's the left to right implication i assumed not for all x a of x and i concluded there exists an x not a of x now we want to do the other implication so assume exists x not a of x well in this case there's an x for which a of x is false that's just expressing this in everyday english well if there's an x for which a of x is false then a of x cannot be true for all x in other words for all x a of x must be false expressing that in symbols it's not the case that for all x fx and now i've proved the second implication from exists x not a of x to not for all x a of x assume exist x not a of x conclude not for all x a of x if you found this reasoning hard to follow it's purely because of the abstraction the logic is exactly the same as in the example almost certainly you had no difficulty with this example and that's because you're familiar with this situation the human brain is very good at doing logical reasoning about familiar situations when we turn those situations into abstract versions the brain finds it difficult at least at first and in fact one of the things about becoming a mathematician is learning to take familiar everyday reasoning that we don't even think about would you do it automatically and reproduce it in an abstract situation and it's the abstraction that's causing you difficulty you can follow the logic but you find it difficult to follow it in an abstract situation and that's just because of the way the human being works okay why don't you try this one show that not there exists an x a of x is equivalent to for all x not a of x and before you start you might want to look at an everyday example this is coming up as an example in the assignment for this lecture but you might want to try it now while this example is fresh in your mind okay good luck on that one with that one under our belt now we can go back and look at that earlier example of the negation of all domestic cars are badly made let's see with the set of all cars d of x means that x is domestic and m of x means that x is badly made with this notation the sentence becomes for all x in c d of x implies m of x for all cars if the car is domestic then it's badly made well we've already seen what happens when you negate a universal quantifier the negation becomes there exists an x in c such that dx does not imply m of x and here i've actually abbreviated the following not the case dx implies m of x well it's not much of an abbreviation but instead of writing not dx implies m of x i've used this simpler notation dx does not imply m of x okay but in any case looking at our previous example the one that i illustrated with the motorist when you take for all x something and you negate it you get the exist sex with the negation of the thing inside just compare what's going on here with the previous example one comment uh why am i not saying existex not in c beginner students often do this kind of thing and the reason they do it is they are trying to go through a formula symbolically and negate things and that's not the way to go about these things you've got to think about what the symbols mean in this case it's talking about x's which are in c the only x's we're interested in are x's which are in c in other words the only objects we're interested in are cars but if the only objects we're interested in are cars the negation will only be talking about cars so the nc part simply tells us which kinds of objects we're dealing with and we don't negate that okay now let's look at this part we know that dx does not imply m of x is equivalent to d of x and not m of x in fact this was the key part of our reasoning to figure out the truth table for the conditional d of x does not imply m of x if d of x can hold and nevertheless m of x can fail so we looked at this when we did truth tables so the negation of the original sentence which is this is equivalent to there exists x in c this part is the same as this part d of x and not m of x well what does this mean in everyday language it means there is a car which is domestic and is not badly made and that's it so if you found it difficult to figure out the negation of this sentence when we first looked at this example now because we've got this notation and this little method for dealing with these uh negations you should be able to follow it through fairly straightforwardly moreover the previous time when we looked at this even if you got the answer you might not have been totally confident that you got the answer right this kind of reasoning makes it clear that this is the right answer and that's the whole point of introducing this formal notation and this logically precise reasoning we're no longer left not certain if we've got the exact best correct logically correct negation the mathematics leads us to the answer and that is the correct answer okay let's do another example let's look at the claim all prime numbers are odd now we all know this is false and we know it's false because we know that two is a prime number which is not odd but let me tease apart the logic that underlies that conclusion because in more complicated situations where we're not familiar with the with the domain or with the result all we'll have to use is the is the logic itself so i want to make heavy weather of this in order to understand the kind of reasoning that we'd use in situations where we don't know the answer okay so i'm going to let p of x mean x is prime and i'm going to let o of x mean that x is odd so the sentence can be written symbolically as for all x if x is a prime then x is odd when i negate this i get exists x not p x here's o of x or another way of writing it there is an x switch to p of x and not o of x the negation of not for all x p of x implies o of x is there exists an x shows the p of x and not o of x in words i e there is a prime that is not odd so in order to prove that this is false which is the same as proving that the negation is true the logic says find a prime that is not odd and we can 2 is a problem that's not odd so if i want to prove conclusively that this statement is false which is the same as proving that its negation is true what i need to do is demonstrate that there is a prime that's not odd and in this case that's immediate we just pull 2 out of the hat observe that 2 is prime and it's not odd therefore the negation is true the original statement is false well in that simple example that was making much ado about nothing but in more complicated situations going through the underlying logic is often the only way forward by now you should have noticed the pattern of behavior that's happening with these uh these symbolic negations when we negate a for all becomes exists and the for all sort of jumps inside and when it jumps inside we can manipulate it and get the result into a positive form where the negation is at the most innermost point you can actually turn these into formal symbol manipulation rules um and which means you can implement them on computer systems and there are computer systems that do this kind of reasoning for you i'm recommending very strongly that you don't do that because what we're trying to get at is mathematical reasoning and moving symbols around doing symbolic manipulation is not mathematical reasoning it's connected with mathematical reasoning it supports mathematical reasoning we can use it in mathematical reasoning but the real essence of what we're trying to do is to understand what the statements mean understand what the symbols represent and reason with the concepts it's reasoning with concepts that mathematical thinking is all about not symbolic manipulation okay we showed that this was false by finding a counter example in this case the counter example was 2. supposing i modify it as follows all prime numbers bigger than 2 are odd in symbols for all x bigger than 2 if x is a prime then x is odd okay now i'm going to give you a quiz what is the negation of this statement is it one there exists an x less than or equal to two such that p of x and not o of x or is it two there exists an x greater than two such the p of x and not of x okay which one of those two do you think it is well the correct answer is 2 it says there exists a number bigger than 2 which is prime it is not odd that's the negation of that statement why isn't it number one well because the original statement is about numbers bigger than two so the negation must be about numbers bigger than two we don't simply negate every symbol in sight and say well in terms of the ordering on the real line if we negate something being bigger than two then it means it's less than or equal to two that's a symbolic negation that's a sort of a detailed localized negation about the ordering on the real line but the sentence as a whole means something about all numbers bigger than two so it's negation means something about all numbers bigger than 2. okay let's move on let x denote a person let p of x mean x plays for sports team t and let h of x mean x is healthy and let me look at the statement there is an x such that p of x and not h of x what does that mean in everyday english there is an unhealthy player on team t now let me negate that i'm going to find the negation by symbolic manipulation now this remember is the method that i'm suggesting that we don't use i'm doing it that way in order to illustrate the mathematics and underlying logic if we negate this we know that when we negate and exists it turns into a for all and then the negation moves inside if you like and then when we move that negation inside we get not p of x or h of x so the negation moves inside we get p of x becomes not p of x when we negate a conjunction we get a disjunction and when we double negate something we get the original thing back remember this is not the way i'm recommending you do it i'm doing this as an illustration of the underlying logic this thing looks familiar remember when we looked at the conditional p conditional q means the same as not p or q they have the same truth table so this guy can be rewritten as p of x here's h of x in english all players on team t are healthy well is that the negation of that there is an unhealthy player on team t well if it's not the case that there's an unhealthy player on team t then all the players on team t must be healthy which is definitely the case that is definitely the negation of that and again let me stress this is not the way you should do this kind of thing i'm using this to illustrate how negation behaves with quantification next point suppose i write down a sentence like the following for all x if x is greater than zero then there is a y such that x y equals one is that true or is it false well there's no way of knowing it depends what the x denotes if x denotes a motorist or an automobile or a healthy team player it's completely nonsensical if x denotes a natural number it makes sense but it's false if x denotes a rational number it makes sense and it's true but the point i want to make is that a quantifier only tells you something if you know what the variable denotes associated with any quantifier we have what's called a domain of quantification which tells us what does the x denote if the domain of quantification is obvious if we've set it in advance or the context makes it clear what it is then we can use statements like that if there's any danger of misunderstanding if there's any potential ambiguity we can make the quantifier more explicit by stating the domain of quantification to make this more explicit i could write it as for all x in the set of rational numbers if x is positive then there is a y so it said x y equals one and now i've got a true that's unambiguous it tells us that for every positive rational well wait a minute wait a minute i haven't actually been explicit as to what the y is have i now arguably having made it clear that the x denotes irrational it's reasonable to assume that the y denotes are rational but if i want to i could be even more explicit and say for all x and q if x is greater than zero then there is a y in q such that x y equals one i think most mathematicians would agree that if you're explicit about the quantifier at the beginning of a statement then absentee indication to the contrary the other quantifiers denote the same thing but if there's any possibility of misunderstanding or if there's any potential ambiguity it's better to be explicit everywhere as to what the quantifiers denote something else you need to be aware of with regards to quantification is that mathematicians sometimes omit the quantifier we write things like if x is greater than or equal to zero then square root of x is greater than or equal to zero if you see an expression like this what that means is the following for all x perhaps in r or perhaps in something else for all x in our x greater than or equal to zero and plus square root of x greater than or equal to zero or something like that depending on what the x denotes i've assumed from the context and this is a reasonable assumption i've assumed that the the variable is meant to range over the real numbers even though there's no explicit quantifier here this would be read as meaning something like that this is what's known as implicit quantification the mathematician is simply leaving out of the formal expression any explicit mention of the quantifier it's left implicit in the way this is written now this is a fairly advanced point i'm certainly not recommending you do this but the professionals do this all the time so you need to be aware of it in case you come across it in something that you read but please avoid doing this i realize by saying this that i'm saying do as i say not as i do because like all professional mathematicians i do write things this way all the time but while you're still learning about these ideas it's best to avoid doing it because if you don't use this in the right kind of context all kind of difficulties can arise and having given you one caution let me give you another one this is about combining quantifiers with conjunction and disjunction let me just do this by way of an example let n the set of natural numbers be the domain of quantification let e of x mean x is even and let o of x mean x is odd look at the formulas for all x e of x or o of x and for all x e of x or for all x of x i guess i don't need that last bracket right okay for all x e of x or o of x and for all x e of x or for all x of x notice that the first one is true it says for every natural number it's either even or it's odd what does the second one say it says every natural number is even or every natural number is odd well that's false well the point i'm trying to make here is you can't just take a for all and take it inside a bracketed expression if you do that you like it to end up turning a true statement into a false statement or vice versa similarly if i take there exists x e of x and o of x we do it with an n this time and i'll take exists and x e of x and exists in x of x again i've put that extra bracket in there can't stop myself doing that what have i got this says there is an x which is both even and odd which is false what does this one say there is an x that's even and there is an x that's odd that's true so the same thing happens with exists and conjunction say you can take an existence statements like this and in this case it's false but if you pair them up in this way it's true in other words be very careful when you're reasoning with quantifiers conjunctions and disjunctions you can't simply take things inside the way you sometimes do with arithmetic if you think about what these things mean you're not likely to run into that difficulty but if you start treating these as symbolic expressions to manipulate then things could go badly wrong time for a quiz okay well we've just seen that this expression something like 4x ax or bx is not equivalent to 4x or 4xbx the example we looked at was whether numbers are even or odd and it was the case of all natural numbers the number is either even or odd but it's not the same as saying that all natural numbers are even all natural numbers are odd in the case of even a mod for natural numbers this is true but both of these disjuncts are false or the disjunction is false okay what about this instead of having disjunction we've got conjunction are these equivalent or not what do you think well the answer is yes they are equivalent now you could argue this in in an abstract form but let's just look at an example that's pretty illustrative um let's take something like all athletes are big and strong okay so x denotes athletes a means big b means strong so that says all athletes are big and strong this would say all athletes are big and then we'd have all athletes are strong it's pretty clear that in the case of the example that is equivalent to the conjunction of those two if all athletes are both big and strong then in particular they're all big and they're all strong and conversely if they're all big and they're all strong then they're all big and strong in other words when you've got universal to quantification if it's combined with a conjunction then you get the equivalence of these two forms but if universal quantification is combined with a disjunction they're not necessarily equivalent i mean this if this isn't a proof this is just an example but if you take that example you should be able to come up with a simple little logical argument showing that that implies that and conversely that implies that so the answer is yes in this case let's look at another variant well again in this case when we had even and odd numbers that showed that these are not necessarily equivalent but what about this where we have an existential quantifier combined not with a conjunction but with a disjunction do we have equivalents in this case what do you think again the answer is yes these are in fact equivalent what would be a good example oh let's take something like um there is a player let's let's keep within with athletic examples they're a player who um or let's say who i know who is a good attacker or a good defender okay so there was a player who was a good attacker or a good defender is that the same as saying let's just simplify it a little bit there's a good attacker or there is a good defender okay and we're going to destroy those so if there was a player who's a good attacker or a good defender one or the other then whichever one it is will be here if the player who is referred to here is going to either be a good attacker or a good defender so one of those two is going to be true which means that the disjunction is true so if this is true then that's true and conversely if there's a player who's a good attacker then that player is a good attacker or a good defender in fact a good attacker or if it's this one then that's true so we have an implication that way with a disjunction and we have an implication that way with the distinction the equivalent again this isn't a proof this is just an example to to illustrate the fact but you shouldn't have any difficulty taking that example and turning that into a simple little argument at least i hope you won't have any difficulty doing that turning that into a simple argument to show that that implies that and that implies that so in the case of existential quantification if we have it with a conjunction then we don't get equivalence but if we have it with a disjunction then we do have equivalence and if you think about what's going on here the fact is that for all he's like and let me put quotes on here because this is a sort of specialized use for all he's like conjunction and exists is like disjunction because for all says it's something's true for all and conjunction means that all of the conjuncts have to be true this simply says at least one and this says it's at least one so that's all about all things and that's about at least one so we get good behavior of what whether you call this good or not you get this nice behavior if we have exists with disjunction because exists is essentially a disjunction sort of thing and for all he's done that was on the previous slide for all is really a conjunction sort of thing and it's when they're mixed up when you get the disjunctive thing here and the conjunctive thing here that things fall apart okay well there you go how did you get on with those two parts of the quiz and this is the last point i want to make about quantifiers suppose we're having a discussion about oh let's say the real numbers so if we have variables x y and z they're assumed to denote real numbers the domain of quantification will be the set of real numbers and in the course of the discussion about the real numbers we want to talk about rational numbers or particular rational numbers you know we might want to mention that there's a rational number x well there's one way we can do that we can talk about the rational number in this context we can say there is an x in q or we can say for all y and q or if i want then to talk about a natural number i can say for all n in the set of natural numbers so by introducing explicit domains of quantification in this way in the course of discussing real numbers i can restrict attention to particular rationals or natural numbers or whatever and so in the course of an argument i can actually use multiple domains of quantification well in examples like this where we're talking about different sets of numbers this is fine but in other situations it really doesn't make sense to introduce multiple domains of quantification for example suppose i'm going to be talking about oh let's say animals so suppose the domain of quantification is the set of animals and then i might want to say something like every leopard has spots well how would i do that well i suppose i could say for all x in the set of leopards x has spots but in the next sentence i might want to talk about tigers and then giraffes and then who knows what and very soon i could have a whole range of different domains of quantification floating around and that's not very clean it's not very nice and it's certainly not very sensible because the discussion isn't really about leopards or tigers or giraffes or whatever it's about animals and if the discussions about animals then the set of animals should be the domain of quantification so instead of writing something like that what i should write is for all animals if that animal is a leopard then it has spots that allows me to say something like there is an animal which is a horse and has spots there's a spotted horse or i could say for all x if x is a tiger then x doesn't have spots the point is if i'm talking about animals then when i refer to specific animals i'm not changing the domain of quantification i'm still talking about animals i'm just describing properties of particular kinds of animals the discussion is still about animals so the domain of quantification should be animals it's different here because the domain of quantification arguably is the set of real numbers or the set of all numbers and then because we have these standard subsets it's okay to introduce things like this you don't have to you could actually use something analogous to this but in this kind of situation yes if you're talking just about real numbers then you really don't want to be introducing particular subdomains you should use the analog of these kinds of expressions and talk about if it's a rational and this if it's a natural number than this but if this is really a discussion about all numbers not just the real numbers then these are just particular categories of real numbers and natural numbers and so forth and you can have sub domains you can have different domains of quantification this isn't an issue of right or wrong it's an issue of whether it's sensible and helpful to work in a certain way and using separate sub-domains of quantification is fine if there are natural domains and it makes sense to talk about it that way but the idea of a domain of quantification is that that tells us what it is we're talking about if we're talking about animals then don't introduce different domains of quantification that are not animals even if they're subdomains it gets complicated okay that's enough about quantifiers for now i suggest you go away and see how you get on with assignment number six well for assignment six i'm just going to do the last three questions number seven eight and nine uh i think you should be able to do the first six uh on your own and check it on your own or with other students so in number seven we have to negate statements and put them in positive form part a was for all x and n there was a y in n x plus y equals one and the negation for that is uh and i think you've got the same answer as i do if you get it right there is an x in n such for all y in n x plus y is not equal to 1. you you could have written that last part as not x plus y equals 1 if you like and that's that's equally correct i'll just use this this way this way of writing it okay and in this one for all that for x greater than zero there is a y less than zero x plus y equals zero the negation of that in positive form is there is an x greater than zero and it's greater than zero okay for all y less than zero these don't change around they stay the same way because these simply tell us what x's and y's we're looking at and uh we get x plus y not equal to zero here okay for part c there is an x uh greatest there is an x so it's a four epsilon greater than zero negative epsilon less than x less than epsilon and when you negate that you get for all x there is an epsilon greater than zero and in this case um you're going to have to sort of split this into two with one of two if x is not between negative epsilon and plus epsilon then either x is less than or equal to negative epsilon or x is greater than or equal to epsilon i didn't put parentheses around either of these i mean i could have done if i wanted to be also clear but this takes precedence over the logical operations arithmetic expressions or inequalities and so on take precedence over over logical connectives because these are connectives they connect statements about mathematics and this is a statement about mathematics and that's a statement of what i mathematics but if you wanted you could have put parentheses around that and you could have put parentheses around that that would have been there that would have been another way of doing it moving on to the last part for x and then for y and n there's a z and n x plus y equals z squared we're not talking about whether these are true or whatever we're just writing them down and negating them so you could exist an x in n exists a y and n for all z and n and i've just written this again x plus y not equal to z squared you could simply put a negation sign in front of that expression okay well that's that's number seven it's fairly straightforward uh let's move on to number eight question eight is our old friend uh abraham lincoln and the uh this uh famous statement of his or allegedly famous statement of his and uh we need to negate that famous sentence so let's let fxt mean you can fool person p at time t and then his statement is this one um you can fool all the people some of the time there are sometimes when you can fool everybody you can fool some of the people all of the time so there are some people that can be fooled for all at all times but you cannot fool all of the people all of the time okay let's just mechanically go through and negate that um looking at the formalism and then we'll try to interpret the answer and express the answer in terms of english okay so negate exists it becomes for all when it for all it becomes exists fpt becomes not fpt okay uh conjunction becomes disjunction again i'm not putting these in parentheses and i'm not going to do it here because this is a this is a is a hole the quantifier is binding they're very tight binding and then we have disjunction uh conjunction here and then disjunctions here they're they're less tightly binding so looking at this one they exist becomes for all four becomes exist we get negation conjunction becomes disjunction and the negation here just disappears so i've got a positive statement okay so in terms of the formalism this is fairly routine um in fact this is such a routine thing that this is essentially algorithmic i just went through and applied the the patterns the rules that i've observed happening the interesting part of this i think and you weren't asked to do it but let's do this let's see how we can express this um in english language and um we're not going to get something quite as as nice as this i don't think because we need to try and be to avoid ambiguity as much as possible so the first part would be the best i can think of at the moment is let's see at any time there is someone you can't fool and whoops or because we've negated or okay let's have a look at this one um i can't really think of anything better than to say the following for every person you can't always fool them um well you might be able to say well i think if you try and swap it round you run into sort of an american melanoma type problems so to try and avoid that i think i would i would want to write it this way and then the final clause um but that's easy of course let's just say you can fool all the people all the time okay i'm i'm moderately happy with that you might have different uh opinions on on the best way to write them um but number two is a bit ugly but i i'm not sure we can make a better job of that one okay um in any case we've we've negated the thing and and this was nice and clean okay now now let's look at number nine well number nine involves one of the most uh important and most famous uh some might say infamous formulas of advanced mathematics of university level mathematics uh it's this definition when students entering mathematics university to study mathematics for a math major it's this definition that that usually causes them the most problems and during their first year um in fact i think it's this definition that probably is responsible for more math majors giving up uh mathematics in their first year at university than anything else it's a really tricky thing to understand as as many of you know this and i've seen film discussions on the on the forum um figuring this out is really hard this was uh hundreds of years of effort uh starting with the invention of the calculus by newton and libras in the in the 17th century it took a long time uh you know several hundred years before mathematicians were able to figure out the notion of continuity and come up with this definition this was late 19th century that this was done and it was uh it was a tricky thing negating it is relatively straightforward actually because we've we've got rules for doing them and so when you negate it um what you get is that the for all becomes exists and they still are greater than we're still talking about positive numbers epsilon the exist becomes a for all the for all becomes and exists and then there's a bunch of stuff that's in a bracket here this is a conditional an implication and so when you negate it you get the antecedent conjoined with the negation of the consequence so here's the antecedent again i haven't put parentheses around that because this is a mathematical statement and that that takes precedence over the conjunction of disjunction these are just connectives they're operators that bring things together so that's a piece on its own that guy conjoined with the negation of this guy okay and the negation of it being less than epsilon is it greater than or equal to epsilon okay so so that's the negation um and that's uh relatively mechanical to do that as long as you sort of pay attention to preserving things that need to be preserved changing for walls to exist and negating a conditional the interesting question uh is what on earth does this thing mean you know you can read it through as i just did for all epsilon greater than zero there's a delta greater than zero such that four x yada yada what is it what does it mean well this is capturing in in in a symbolic language in an algebraic form formalism it's capturing something geometric so let's see what it's capturing let's look at the original definition of continuity it's about functions so let's look at a function this way i'll draw the rail line vertically this is a real line and then i'm going to draw the real line vertically here so this is the real lines instead of writing horizontally as we usually do i'm going to write it horizontally and the function f is going to take numbers here to numbers here okay so somewhere here we've got a f applies to that and it gives me f of a okay now we're trying to capture the notion of continuity at a that means when we go slightly to the left or right of a and left or right is up and down the way i've represented it then the numbers don't sort of have a discontinuity and the way to capture that it turns out and chances are very very high that you're not going to follow this the first time it's going to take you weeks if you need to to to master this but but here's what this formula says it says the epsilons are going to work on here it says let's take an epsilon here and let's look at f of a plus epsilon and f of a minus epsilon so i'm going to take an epsilon interval around f of a and what this definition says is that given an epsilon and one of these intervals i can find a delta so here's a plus delta and a minus delta so starting with an epsilon which gives me an interval here for any one of those i can find a delta which gives me an interval here such that now let's look at this any x in this region because this says that x is within delta of a so any x in this interval gets sent to an image f of x in here so it's saying in order to make sure that all the values of the function are in this interval i can find an interval around a such that they're all sent into here so if i want to hit the target imagine this as hitting the target like like throwing darts at a dartboard if i want to hit the target within a specified accuracy of a of f of a i can always do it by starting out within an interval of a so to get within a given interval around f of a i can always find an interval around a that does it so everything from here gets sent into here and if you think about it long enough you realize that what that means is that the function is continuous at here there's no no no jumps and to try and understand that let's look at what this guy means in terms of a diagram i'll do the same thing again i'll draw the real line and i'll draw the real line and here's a and here's f of a now the negation says there is some epsilon in in the previous case this was happening for all epsilons you could find a delta in this case there's a fixed epsilon that we can find and we look at the interval around there f of a minus epsilon okay what it says is that there is an epsilon such that no matter what you take here no matter which one you take here here we found one of these guys we said take any interval here we can find an interval here here we're saying there is an interval here for some epsilon such that no matter what delta we take here no matter how small you make this delta no matter how close you hear was a point that gets sent to that you can always find a point in here that gets sent outside of there maybe sent that way it may be sent that way so the sum epsilon here that no matter what you do here no matter how close you are to a something gets sent out here in other words there are points really really really close to a that gets sent outside this region so there's no way that you can you can get all points in here a gets sent to that but arbitrarily close to where there are points that get sent away so there's a discontinuity because only a gets sent close to a when you get well some of the points may be but no matter how close you get to a you'll find points of descent further away so that's a discontinuity things jump there's a sort of jump from there to things here or here now i you know the chances are high that this if this is the first time you've seen this that explanation uh is going to be hard to follow this is really difficult to understand this is extremely extremely hard and the goal of this course actually isn't for you to understand this the goal of this course is to give you the machinery so that you're now able to take a whole semester course on real analysis during which you should be able to understand that so i wouldn't worry too much if you can't understand this the point is if you could understand the first part and understand how to deal with the formalism you now have the machinery you need to understand this this very deep um conceptual definition from within mathematics okay and to have reached that stage in four weeks i think this is pretty pretty remarkable um the hard part is to come and for those of you that want to go on and study more mathematics sooner or later and hopefully sooner you're going to need to master this definition of what it means okie dokie well as far as we're concerned we're done with this one well for question one on problem set four we have to choose which one of the following is equivalent to uh to this expression okay we're negating a universally quantified statement in which there's a conditional and we're told that there's only one of these uh one of these answers correct the correct answer let me just jump to the correct answer and then see what's wrong with some of the others is is part c the f not for all becomes and exists and the negation goes into this part when you negate a conditional you end up with the antecedent together with the negation of the consequence and when you negate a disjunction the negations filter in and the disjunction becomes a conjunction so if you follow the rules about well they're not rules they're i mean they're sort of our rules but we you know i i recommend you not to think of them as in terms of rules because that's not really getting at what this course is about um but there are certainly patterns of activity that you can get to uh get to recognize then what happens is universals become exists you have the truth of the antecedent in place of an implication or conditional you have a conjunction and then when you negate these guys and the negation applies to each one and that becomes a conjunction okay but as i just indicated i just referred to really what i want you to do is concentrate always on why it is that you get that behavior why does this give you px and not that why does negating this give you a conjunction there so it's all about understanding um if you if you simply learn to apply the rules you really don't have a useful skill okay computers are good at applying rules that's what they do that's all they can do people uh can go much beyond that okay um what's wrong with some of the others well the first one is just hopeless i mean there's just nothing remotely like that if you've got that as your answer um then either you are having a temporary aberration or you really really really haven't got the the issue of this uh the others um you know there were reasons why people could go wrong and the others were put in um because of the mistakes that people frequently make okay in the case of this one um the negation is in the wrong place um when you negate a conditional the negation doesn't come together with this one the negation should come in front of here and in front of there so it's just um mixing up where the negation comes in okay but otherwise you apply so so it's like applying you're basically applying the right sort of let me call it rule um you're doing the right thing but you you've got one step out it's like getting a negative sign a minus sign in the wrong place in an equation um looking at this one um well everything went fine except you forgot to change disjunction to conjunction but everything else was fine and in the case of this one you forgot to that that should be conjunction and everything else was fine okay so in cases b d and e there was just one thing wrong and so it's even possible you just did that by a slip i mean heaven only knows you've seen me make slips often enough in the uh in the lectures in the tutorials where i write the wrong thing down or whatever it's uh you know we have to keep a lot in our minds when we're doing these things and frequently what our hand does isn't what we're thinking it's doing and sometimes what we say isn't what we think we're doing mathematics is like that when you're really focusing on the mathematical concepts and the heart of it you can mix slips with the writing and with the words you use i do it all the time and uh we all do that's that's just part of thinking mathematics it takes a lot of concentration to focus on the mathematics and the the everyday things like writing and using words tend to miss out on that because our mind is focused on the content okay let's go on and look at number two now so for number two let me just begin by reading this formula out in everyday english or at least stilted everyday english it says for every person p there's a person q and a game t such that p beats q in game t so more colloquially everybody beats somebody in some game everybody beats somebody in some game everybody beats somebody in some game but which ones say the same as that that one certainly does that means the same as everybody beats somebody in some game um it doesn't mean the same as that so it's not that one that's not a possibility and for every player there's another player they beat all the time that's not quite the same if there was a fall there for all t then whenever they play p would be that person that person q so that's that's not quite you would need a different quantifier that's just some game if there was a fall there you would have it in all the games there is a player who loses every game no no that's about losing that's a player well actually while we're on this one what would this look like if i wanted to put that in a formula i would start with the following i would say there is a q such that for all p and for all t w p q t well in a minute i'll say why i should start because i was looking ahead to what would be wrong with this but let's just read it so it says there is a player such that for all players and for all times p beats q in that game so it sort of says the player loses every game the problem is when you take for all p here one of the p's that are going to crop up is cue him or herself okay and a player can't beat themselves so this doesn't quite work because the p could include the the queue itself in fact that one of the p's will be the q okay and it's also well it depends so this is up there's gonna yeah i think that would be okay now you've got to be a little bit careful actually it depends how you interpret the thing um for all t t just goes over games of tennis whereas we're just talking about games of tennis in which people uh people win so you you'd have to sort of think about it and decide whether you need to put clauses in here to make it clear that you're only looking at t's where they play together there'll be similar issue here again you could start with if you wanted to do this one you would start with saying there is a p let's see for all q for all t w p q t the only difference is here we had q p and here we've got p q but you'd have the same issue among the queues is the possible p and you'd have to be careful to take account of the fact that the t itself ranges over all possible games and you only want to be making a statement about the games which they play together um and you could have a madness you could put clauses and there were various ways out of that um if i thought this was a big deal i'd have thought about it before i started to record this little piece but i think we've we've solved the problem in any case that one came up the first time that was the one that uh that has that meaning none of the others do so um put those as no's and we've solved that one okay let's go on to number three now normally when i'm going through examples in in classes i don't make deliberate mistakes i don't need to because i make plenty of real mistakes when i'm going through mathematics in any case but this time i'm going to make a deliberate mistake um and i'm going to do it to emphasize the issue that's going to come up here okay so what i want to do is go through these three and give you answers to all of them but one of them's actually going to be a wrong answer and then we'll correct it so this one actually if you remember the previous question this is the one that we saw that means everyone wins a game okay for all players p there's a player q and a time t so it's a p beats q at that time what does this say everyone beats well it's almost the same right except instead of playing every player beats every beats one player you say everyone beats everyone else at some time in some game okay because it's all it's all pairs of players p and q and this one it's essentially um well actually what's the difference that was for all p there exists a q this is for all q to exist so this isn't doesn't say everyone wins a game this one says everyone loses a game okay seem plausible well one of these is actually uh is actually not possibly true let's just see could that be true we're talking about whoops there's a typo there that should be possibly okay let's correct the typo um i've got it right up there okay let's see which one cannot possibly be true okay that's what the issue is here well that could possibly be true okay that's okay that could be true everyone beats everyone else all the time at some time well everyone loses a game well that's certainly possible that's okay come into this one can this i mean is it is it possible for this to be true it's certainly possible for that to be true it's certainly possible for this to be true we're talking about the real world here when people are playing tennis is this one the the one that's that can't possibly be no there doesn't have to be one you could have said that none of these are the case in fact if you got this far you probably did say they're all okay but actually this one's not okay because when you say for all p and for all q that includes the case of q and p being equal that would mean everybody has to beat themselves at some time amongst everything else so a player cannot beat every other player because the problem is no player can beat herself or himself whichever you want okay no player can beat herself so that can't possibly be true so this is the guy that can't possibly be true and it can't possibly be true because the q and a p have to be equal what you would have to say if you wanted to make that something that could be true is you'd have to say for all p for all q um if p and q are different then there is a t such a w p q t then you'd be okay because you'd say for all pair pairs p and q providing their difference then at some time at some game p does beat q in that game so that would be a way of making it possible but as it stands that cannot possibly be true because players cannot beat themselves okay so the issue was whether you can t the formula doesn't correctly tie them into what they say about the real world it wasn't mathematics it was deciding these it was the real world and in the real world players cannot beat themselves well actually in a figurative way players beat themselves all the time but in the sense we're talking about here that's not the case okay now let's move on to question four okay now this expression is is a colloquial expression and so when i set this question up i i realized that uh um for non-native speaking english students this would be uh this would be somewhat challenging so what i did was i only gave you options that you should be able to dis to distinguish between by by knowing the logical structure the reason i like to give these kind of examples is because they capture an awful lot of social and cultural knowledge and there's an interesting challenge in capturing that kind of thing uh in in formalisms but to help you along i i gave you three options that you should be able to sort out just on the basis of logic if you know what um what being a lover means in this case so being a lover means you're in a mutual relationship which means you've got um you love someone and that person loves you but if you look at this one this says um for this person x so these are all about a person x they all talk about some person x this doesn't say that person's a lover this this this part says that person is in a love in a relationship with everybody else so this part because there's a universal quantifier that says l love x love z and z loves x so this person x uh is in a loving relationship with everybody well that's nonsensical so we can forget that one so it comes down to these two because in each of these it says the person is in a loving relationship x is in a relationship with some z and it's mutual x love z and z loves x the same clause here so the choice is between a and c well let's look at what a what a says that says for all x and uh for all y if the x is in a loving relationship then y loves x now the y doesn't come in here so that the for y is to do with this part so it says take any person x if that person is is a lover then every person loves them if they know them so that actually is the correct one okay all people love a lover of these three that's that's the one you know you could argue about whether that's the the absolute best interpretation but out of these three um it's certainly a correct one and so it's it's the one here uh let's look at this one this one uh is uh it's a little bit different because it's got full y in here so it says for all x if x is a lover then but it doesn't say then it says and for all y l y x so the for all actually applies to this part as well so what this really what follows from this is that for all x and for all y l y x in other words everybody loves everybody well that's not the case um because this isn't conditional on being a lover this just really says that's the case and that's the case so part of this is saying that for all x and for all y l y x well that's not the case i mean it's not the case that everybody loves everybody the world would be a nice place i guess if that was true but it's not true so it can't be that one so we can we've eliminated this one um because it's uh it doesn't capture it doesn't use the fact that uh being a lover and we've eliminated this one because it basically just boils down to saying uh everybody loves everybody and that leaves this one and and this is definitely one one good interpretation of everybody loves a lover okay so we were able to reason that one out and i would hope that even without the detailed understanding of what this means in english um there's only one of these that that will stand up to our to analysis and after all the idea is to sort of look at how the formalisms capture relationships from the real world well for question five we have to find which statements are false okay so let's just hear what they say um for all x four y four z if x is less than or equal to y and y is less than or equal to z then x is less than or equal to z that's true it's actually known as the transitivity of the of the order relationship okay if x is to the left of y and y is to the left of z then x is to the left of z okay so that one's true what does this one say for all x for all y if x is less than or equal to y and y is less than or equal to x and x is equal to oh that's true okay um the only way you can have x less than or equal to y and y less than or equal to x is if they're actually equal what about part c for all x there is a y well that's certainly the case because given an x you can take y equals x and then you'll have x less than or equal to y and y so that one's true let's begin as look as though they're all true let's look at the last part there is an x such that for all y y is less or x or x is less than y well you might be tempted to say yep i'm given any x it's the case that every other y is either less than x or bigger than x but wait a minute among the y's governed by a universal quantifier is the x you start with so if there was an x with that property how could this happen because when you look at all the y's among those y's would be x itself so you would have x less than x which is impossible so that one's false and it's false because the universal quantifier includes the x itself given an x any x that you find um when you say when you universally quantify otherwise you include that x and then that fails and that fails so there's the one that's false there is a false one and it fails because universal quantifiers go over everything and that means the the the y willing among the wires that you're looking at is the exit you start with okay well for the last question on problem set four which is a question six we have to look at this uh this piece of reasoning that a student gave this was actually when i first gave this course in the fall of 2012 and a student was trying to understand the euclid's proof that the prime numbers are infinite in number and there was a key step where you form that product and you add one and the student wanted to verify that in fact m plus one was not divisible by p and came up with this argument so i thought this would be a good one to look at it raises a number of interesting issues so um first of all let's just see what a student does suppose n is divisible by p well arguably the student you could say the student doesn't begin by saying um what n is and doesn't say what p is but in this context there's no need to it it's already been stated that n is an integer it doesn't matter whether it's positive or negative by the way um and it's already said that p is a prime so this in this context i don't think there's any need to demand that the student repeats it here what's key is that the student is is writing down the assumption on which the argument is going to be based okay so i'm going to say that the oh i'll come to the various other issues in a minute the opening i think is fine okay um that's good um what does the student do then um then there's an integer q she said a equals p q yes i said that's the definition of divisibility um i'm not going to demand that the student spells that out because it's fairly clear what's going on so m plus one is people so that's okay then dividing through by um by q um and by p you've got numbers one over p now the students now have written a lowercase q there and that's obviously meant to be enough because q i'm not going to worry about that that's just a typo in writing it ditto here um so i'm not going to deduct anything for that those are just typos i mean just using the wrong lowercase and uppercase confusion so n plus 1 is not divisible by p okay logical correctness well in a sense this is okay right i mean everything steps fine everything's true so i'm gonna give three why aren't i giving four i'll i'll come to that in a minute what about clarity um i've got to get four this is absolutely clear what's going on um stay at the conclusion yep conclusion stated that's four for that one um a reasons given absolutely reasons are given um yeah i mean as i observed at the time the student sort of doesn't say um this is by definition of divisibility and so forth but you know there's a limit to how much you can write down uh within the context of this class and the intended audience which is the students in this class then this is fine and you know it's fine to put more details in the reasons if you if you want to wear on the side of caution but given the fact that this is absolutely clear i think there's enough reasons given okay and uh that brings me to the overall evaluation and this is where i'm gonna put a zero down and this is also why i didn't give a full four for this because this is logically correct in terms of doing mathematics but this is an argument about integers and the integers is a number system in which you can add subtract and multiply what you can't do is divide division is not an operation on the introduced it's an operation on the natural on the rational numbers and on the real numbers but it's not an operation on the integers so in in in going into division this argument goes beyond the realms that we're permitted to use so i'm giving i'm being generous here actually and saying i'll give you three because it's this is mathematically correct in terms of the rational numbers um i mean i could have been tougher and i could have reduced it to two i could have even put a zero down but as i've said before i'm looking for reasons to give people marks not to take them away we're trying to make people better thinkers not to make them feel bad about themselves okay so i'm going to give credit where it's due but strictly speaking it's not right here's what the student should have done okay so we've got the first part we've written n equals p q okay that was okay okay now what i'm going to do is i'm going to argue by contradiction i'm going to say suppose n plus 1 where divisible by p then there is an integer call it r so it said n plus one equals pr then we have n plus one minus n equals pr minus pq which is p into r minus q but n plus 1 minus n is 1. so what i've shown is that p times r minus q equals one that means that p that means that one is divisible by p but that's a contradiction p is a prime number so it's at least equal to two so it can't divide into one one isn't divisible by any by two or three or anything so there's a contradiction hence the original assumption here was false hence m plus one is not divisible like p notice that this is almost the same in in a sense this is this is equivalent to this now it's not the same as this but it's equivalent because the uh the r here is n plus 1 over p so this is the r so i've said i mean i'm sort of doing the same thing except here i'm not using division i'm doing everything in terms of divisibility which means i'm doing it in terms of addition subtraction and multiplication all i'm using here is addition subtraction and multiplication i'm not using division i'm getting around it by introducing this r if you like and that's a significant difference because we simply don't have division as an operation in the integers we have divisibility which is a property that may or may not hold between two two integers but we can't divide one integer by another because division is not an operation in the integers okay um so i give a reasonable amount of credit for this um i think i was generous with this one because i tend to be generous and that means the total grade is 19. okay well that was the end of problem set for how did you get on with assignment six if you've followed everything so far and managed to do fairly well on all six assignments you should have a good idea of the kind of linguistic precision required in mathematics now we can start to put that precision to use improving mathematical statements in the natural sciences truth is established by empirical means involving observation measurements and the gold standard experiment in mathematics truth is determined by constructing a proof a logically sound argument that establishes the truth of the statement the use of the word argument here is of course not the more common everyday use to mean a disagreement between two people but there is a connection in that a good proof will preemptively counter explicitly or implicitly all the objections the counter arguments that a reader might put forward when professional mathematicians read a proof they generally do so in a manner reminiscent of a lawyer cross-examining a witness constantly probing and looking for flaws learning how to prove things forms a major part of college mathematics it's not something that can be mastered in a few weeks it takes years what can be achieved in a short period and what i'm going to try to help you do here is gain some understanding of what it means to prove a mathematical statement and why mathematicians make such a big deal about proofs first what is a proof and why do we use them i'll answer the second question first since their purpose dictates what they are proofs are constructed for two main purposes to establish truth and to communicate to others constructing or reading a proof is how we convince ourselves that some statement is true i might have an intuition that some mathematical statement is true but until i've proved it or read a proof that convinces me i can't be sure but i may also have to convince someone else and that's the second purpose of a proof for both purposes a proof of a statement must explain why that statement is true in the first case convincing myself it's generally enough that my argument is logically sound and i can follow it later in the second case where i have to convince someone else more is acquired the proof must also provide that explanation in a manner the recipient can understand proofs written to convince others have to succeed communicatively as well as biologically sound there's actually not as much of a distinction here as my words might imply for complicated proofs the requirement that a mathematician can follow his or her own proof a few days weeks months or even years later can also be significant so even proofs written purely for personal use need to succeed communicatively the requirement that proofs must communicate explanations to intended readers can set a high bar some proofs are so deep and complex that only a few experts in the field can understand them for example for many centuries most mathematicians believed or at least held a strong suspicion that for exponents n greater than or equal to 3 the equation x to the n plus y to the n equals z to the n has no positive whole number solutions for x y and z that was conjectured by the french mathematician pierre de ferma in the 17th century but it wasn't finally proved until 1994 when the british mathematician andrew wiles constructed a long and extremely deep proof over the centuries it became popularly known as fermat's last theorem since it was the last of several mathematical statements fairmat announced that remain to be proved most mathematicians myself included lacked the detailed domain knowledge to follow wildsprout ourselves but it did convince the experts in the field the field by the way is analytic number theory and as a result fermat's ancient conjecture is now regarded as a theorem fermat's last theorem is an unusual example however most proofs in mathematics can be read and understood by all professional mathematicians though it can take days weeks or even months to understand some proofs sufficiently to be convinced by them i've chosen the examples in this course to be understood by a typical student in a few minutes or possibly an hour or so examples given to college mathematics majors can usually be understood with at most a few hours effort proving a mathematical statement is much more than gathering evidence in its favor to give one famous example in the mid-18th century the great swiss mathematician leonard euler said he believed that every number beyond two can be expressed as a sum of two primes this property of even numbers had been suggested to him by christian goldbach and became known as the goldbach conjecture it's possible to run computer programs to check the statement for many specific even numbers and to date 2012 it's been verified for all numbers up to and beyond quintillion most mathematicians believe it to be true but it's not yet been proved all it would take to disprove the conjecture would be to find a single even number n for which it could be shown that no two primes sum to n incidentally mathematicians don't regard the goldbach conjecture as important it has no known applications or even any significant consequences within mathematics it's become famous solely because it's easy to understand was endorsed by euler and has resisted all attempts at solution for over 250 years whatever you may have been told at school there's no particular format that an argument has to have in order to count as a proof the one absolute requirement is that it is a logically sound piece of reasoning that establishes the truth of some statement an important secondary requirement is that it's expressed sufficiently well that an intended reader can perhaps with some effort follow the reasoning in the case of professional mathematicians the intended reader is usually another professional with expertise in the same area of mathematics proofs written for students or laypersons generally have to supply more explanations this means that in order to construct a proof you have to be able to determine what constitutes a logically sound argument that convinces not just yourself but also an intended reader doing that's not something you can reduce to a list of rules constructing mathematical proofs is one of the most creative acts of the human mind and relatively few are capable of true original proofs but with some effort any reasonably intelligent person can master the basics and that's my goal here euclid's proof that there are infinitely many primes which i gave in the first lecture is a good example of a proof that requires an unusual insight let's look at it again here's what we did the idea is to show that if we list the primes in increasing order then the list can be continued forever so we imagine we've listed the primes p1 is 2 p2 is 3 p3 is 5 etc all the way to some stage pn and we're sure that we can always add another prime to the list and to do that we look at this number n which we obtained by multiplying together all the problems in the list so far and then adding one now this number n consists of the product of all those numbers p one two p n plus one so it's certainly bigger than all of those numbers so it's bigger so n is bigger than all the problems in the list well if n is prime then we know that there's a prime bigger than p n namely n in which case we can continue the list probably not by adding n itself n because it's the product of all these plus one is going to be a lot bigger than p n so n is almost certainly not the next prime but that doesn't matter if n is prime it shows there is a prime bigger than the one at the end of the list and that means we can continue the list the alternative is that n is not prime in which case there's a prime q less than n that divides it but none of the primes in the list can divide n since if you divide n by any of those primes you're left with that remainder 1. if you try to divide p1 or p2 or any of these primes into this number n it gets swallowed up by this part and then there's a remainder of one okay so q has to be bigger than p n those are the first n primes so if q is not equal to one of those it must be later on in the list in which case we've shown again that there's a prime bigger than p n and the list can be continued again this particular queue that divides n is not necessarily the next prime but as before that doesn't matter showing that there is another prime is all you need to do because then you can take the next prime whatever it is and add it to the list either way either if n is prime or if n is not prime either way there's another prime to add to the list it follows that there are infinitely many primes and the theorem's proved there are two creative ideas in this proof the first one is here to show that if we list the primes in increasing orders p1 p2 etc then the list can be continued forever so the first creative idea is to think about listing the primes and showing that the list can always be continued the second creative idea was this one defining this number n in such a way that it guarantees that we can always find another prime i would say that this idea is one that most mathematicians would come up with sooner or later it's a fairly obvious one this one is genius okay this is true genius let me give you another example and this time i'm going to prove that result i promised earlier that the square root of 2 is irrational and i'm going to write it the way mathematicians typically do when they write up results for publication in in professional journals or in books namely we we call it out by calling it a theorem so in mathematics a result that's sufficiently significant or important that it's worth mentioning as such is called a theorem in this context let me mention there's another word we often use called lemma and a lemma is a result which is worth calling out for some reason but doesn't quite merit the the status of uh of being called a theorem it's actually if you like a little theorem okay the next thing a mathematician typically does is indicate that we're going to going to begin the proof okay so this is just part of the way mathematicians lay things out we we specify the theorem and then we say we're going to give the proof you don't have to do it this way it's just a convention the essence of being a proof is what comes next proofs are about their logical structure not the way we write them down i'm going to begin by assuming on the contrary that the square root of two were rational now if you've never seen the proof that the square root of two is irrational before this first step is going to seem pretty mysterious why do i begin by assuming the opposite of what i'm trying to prove well by the time i get down here you'll see why the reason this is a great example is in about six or seven lines i can make it clear why i'm doing something right in the in the first step okay well in that case if square root of two were rational then there are natural numbers p q with no common factors such that the square root of two is p over q remember a rational number is one that can be expressed as the quotient of two integers in the case of a positive number it would be two natural numbers and we can always pick those natural numbers or those integers to have no common factors in other words when we write a rational number as a quotient we can always cancel out any common factors and express it as a quotient where the two numbers themselves have no common factors again it might seem a little bit mysterious why i'm being particular about counting out common factors but as uh as with the the first step by the time i get down here it'll be clear why i'm doing this well squaring that equation gives me 2 equals p squared over q squared rearranging i get 2q squared equals p squared i multiply both sides by q squared that gives me a 2q squared on the left and then when i multiply the right by q squared it cancels that q squared and i'm left with a p squared so p squared is even it's equal to twice something hence p is in why because the square of an even number is an even number the square of an odd number is an odd number so the only way i could get the square of a number p to be even is if the number p itself is even even squared is even odd squared is odd so p is 2r for some r i'm now going to take this equation p equals 2r and use it to substitute back in this equation so i take this equation i've got 2 q squared equals p squared and p equals 2r so i've got 2q squared equals 2r all squared which is 4r squared we'll forget the middle term now i've got 2q squared equals 4 r squared i can cancel the 2. well if q squared is 2 r squared then q squared is even but exactly has happened before with p squared if q squared is even then q is even aha see what's happened i've deduced here that p is even i've deduced here that q is even so p and q are both even but they can't be because we assumed p and q had no common factors if they're both even then they have two as a common factor but this is impossible since p and q have no common factors well the logical reasoning here the algebra the arithmetic is all sound that's absolutely everything's perfectly sound how can we have arrived at an impossible conclusion by a piece of sound reasoning well the only thing that can possibly have gone wrong is we began by making a false assumption remember we began with an assumption the only way we could have reached a false conclusion by a valid argument and this is valid if you don't believe me go and check it at least for yourself there's not many steps see if there's anything wrong with in any of these steps there isn't if we reach a false conclusion by a logical argument then we must have started with a false assumption hence the original assumption that the square root of two were rational must be false hence square root of two must be irrational and when i was at school uh teachers used to insist that we write qed at the end of a proof quarter demonstrandum which is latin it is actually not a bad idea to indicate when a proof ends and mathematicians have different uh different ways of doing it sometimes they write a little box at the end how you express it i mean how you lay it out on the page is not that critical um i mean the idea is to lay it out in a way that's uh that can be followed what makes a proof of proof isn't the fact that you call it a proof it isn't the fact that you end in a qed it's the logical flow of the steps it has to be logically precise and you have to be able to follow it well there's the proof the reason this is a good example to give is it's short it's concise and all of the steps are simple arithmetical steps it's very easy to follow every step and yet when you follow this small number of simple steps you've proved a significant result in fact this result when it was first discovered by one of the pythagoreans in ancient greece was dramatic it changed the course of greek mathematics because until then they felt that quotients of integers were sufficient to measure any length but square root of two is the length of the diagonal of a right angle triangle whose sides measure one and when this result was proved by one of the pythagorean mathematicians it showed that quotients of integers were not sufficient to measure all lengths in in geometric figures and that changed the course of greek mathematics and and subsequently the rest of mathematics it was extremely dramatic incidentally there's a story you'll read about in books and on websites that say that the this was discovered by a young mathematician and the the greek mathematicians the pythagoreans were so annoyed and so scared that this would uh kill their career and their profession that they they threw him overboard there's absolutely no evidence whatsoever that that was the case it's a great story but like many stories it's probably not true okay but in any case we've now shown that the square root of two is irrational and there we go i mean this is really a remarkable result very short very elegant incidentally when mathematicians talk about aesthetics when they talk about an elegant proof or a beautiful proof this is the kind of thing they have in mind not that it looks beautiful the way it's written out in fact i've just worked through it and it doesn't look particularly beautiful to look at but the logical structure is beautiful every step counts the result is established and every step can be understood fairly straightforwardly the complexity doesn't come because there's a deep results involved deep deep facts deep concepts the proof works because of the structure the logical structure okay so now you know why root two is irrational let me say a little more about that proof it's an example of what mathematicians call proof by contradiction and proof by contradiction is a general method that works as follows you want to prove some statement phi you start by assuming not phi you reason until you reach a conclusion that's false often by deducing both psi and not psi for some statement psi for example in the proof that the square root of two was irrational we proved that p and q have no common factors and yet we knew that p and q were both even so we have a statement and it's negation or not strictly speaking it's negation but but close enough uh for these purposes this is not an exact rendering of psi and not psi but the point is we've deduced two contradictory things it cannot be the case that they have no common factors and that they're both even but a true assumption cannot lead to a false conclusion hence the assumption not phi must be false remember we began by assuming not phi we reach a false conclusion a contradiction a true assumption can't lead to a false conclusion so the assumption must be false in other words phi must be true if not phi is false then phi is true well we begin by wanting to prove some statement phi and we're going to end up with phi and in the middle we reason by assuming the contrary so here's the contradiction proof going on in here there's phi there's phi in here we're reasoning with not phi we start by assuming not phi and we reach a conclusion a false conclusion and conclude that that's false let me say a little bit more about this we can look at proof by contradiction in terms of truth tables what can we conclude from a proof of theta yields psi where psi is false in the case of the proof of root two being irrational the theta was the statement square root of two is rational and the psi was the false statement that we never actually wrote out in full that p and q are both even and have no common factors which is false because there's no such that you can't there are no pairs of integers which are both even and have no common factors so in terms of the of proofs by contradiction the theta is the assumption you make at the start of the proof and the psi is the false conclusion you reach at the end so i'm going to use truth tables to understand how it is that starting with an assumption and reaching a false conclusion leads to concluding that the the assumption was false and since the assumption was counter or contrary to the thing we're trying to prove that would amount to proving the theorem okay so i'm going to write down the truth table for theta yield psi or theta conditional psi right the truth values down t t t f f t f f and we know what the table looks like here it's t f t t well if we've carried out a proof of this that means this thing is true so we can forget that line we're only interested in what happens when this thing is true so we've carried out a proof that this implication this conditional is true so it's one of these three in this case however the psi is false where's the sign false was here psi is true here's psi is true here size false so this is the only line in the truth table which fits this we have this being true which means it's one of these three and we have psi being false that's the only possibility in other words theta is false so when in a proof by contradiction you make an assumption a counter assumption an assumption counter to what you're trying to prove and you carry up some reasoning to deduce a false conclusion a contradictory conclusion or a contradiction so you've established in your argument that this thing is true this says that the argument is valid then the conclusion from having reached a false conclusion the the the big conclusion the global conclusion is that your original assumption is false okay it's a little difficult to explain this in words because we have to keep talking about truth and falsity and assumptions and conclusions and the assumptions and sub-assumptions and so forth but if you go through the original proof that square root of two is irrational once more and then think about it in terms of what i just said a moment ago about proof by contradiction in general and then think about it in terms of this truth table analysis uh and hopefully you should be able to understand why proofs by contradiction work uh they've it's a very clever idea and uh they're used a lot in mathematics and i've just noticed i've misspelled truth table so let me put the l in there okay there we go that was a fairly lengthy discussion of such a short argument but i know from many years of experience the beginners find the root 2 proof hard to really understand you may think you understand it but do you really let's see if you can produce a similar one try to prove that the square root of 3 is irrational you should definitely try to do this exercise but be prepared to spend some time at it go on give it a try i'll say it again this course is about the process of thinking not about getting results you can use the thinking abilities you develop in a course like this to get results in other courses and in other situations in your life it doesn't matter if you don't get the root 3 proof you'll have benefited from trying proofs by contradiction which we use in the root 2 theorem are a common approach because they have a clear starting point to obtain a direct proof of some statement phi you have to generate an argument that culminates in phi but where do you start the only way to proceed is to try to argue successively backwards to see what churn of steps ends with phi there are many possible starting points but just one goal and you have to end up at that goal that can be difficult but with proof by contradiction there is a clear starting point and the proof is complete once you have deduced a contradiction any contradiction with such a wide target area that's often a much easier task the proof by contradiction approach is particularly suited to establishing that a certain object doesn't exist for example that a particular kind of equation does not have a solution you begin by assuming that such an object does exist and then you use that assumed object to deduce a false consequence of a pair of contradictory statements the irrationality of the square root of two is a good example since that states the non-existence of two numbers p and q whose ratio is equal to root two even though there is no cookie cutter template approach to constructing proofs there are some guidelines we just met too proof by contradiction is often a good approach when there's no obvious place to start proof by contradiction is a useful method to prove non-existent statements of course you still have to construct a proof you've simply replaced a narrow goal post with an unclear starting point by a much wider one with a known starting point but like robert frost's fork in the trail that choice can make all the difference there are a number of other guidelines i'll tell you some but do bear in mind that these are not templates as long as you continue to look for templates to construct proofs you're going to encounter significant difficulties you have to start each new problem by analyzing the statement that you want to prove what exactly does it say what kind of argument might establish that claim let's look at another guideline how might you go about proving a conditional statement one of the form a implies b we want to prove a conditional phi yield psi well we know this is true if phi is false so we can assume phi is true why do we know it's true if phi is false well that was part of the definition of the conditional that we developed using truth tables so to prove it we assume phi and deduce psi this of course confirms the point i made earlier that despite its strange definition it's counterintuitive definition perhaps the conditional really does capture genuine implication because in all actual practical examples when you try to establish a conditional what you do is you assume phi and you deduce psi and this is genuine implication this says that psi is following from phi for example let x and y be variables for real numbers and prove the following if x and y are rational then x plus y is rational okay this is not a surprising result this is nothing deep i'm focusing not on the results but on the method i use to prove it and so i've deliberately chosen an example that's extremely simple so that we can look at the process of reasoning that's involved improving the conditional so step one assume x and y are rational in that case there are integers p q n and m such that x is p divided by m and y is q divided by n where in that case x plus y is p divided by m plus q divided by n which equals p n plus q m divided by m n hence x plus y is rational okay as i mentioned a moment ago there's nothing surprising here it's almost not a proof at all it's just really adding two things together but it actually is a proof because it has the right structure of a proof here's what we did we began by assuming x and y irrational we concluded that the sum is rational and in the middle there was an argument to demonstrate that fact the argument was actually fairly typical in what we first did was take the assumption and then unpack the assumption in terms of some useful information and once we'd done that we reasoned with that information to get a conclusion that in fact was the thing we were aiming for so we write down the assumption we carry out some reasoning and we reach the conclusion all three steps are important declare the assumption carry out the reasoning in a clear understandable fashion and then state the conclusion when you've reached it remember that proofs have two purposes one is to convince yourself and two is to convince other people and you may know what you're doing if you don't mention what your assumption is or what your conclusion is although i'll guarantee from experience that a week from now you'll forget exactly what you were doing so it really is good even for your own purposes to write down what your assumptions are but certainly from a communicative angle it's important to begin by stating the assumption then to lay out the reasoning in a simple understandable fashion and then to state the conclusion that you've reached okay well that's that's the most basic method of proving a conditional let me give you a quiz let r and s be irrational numbers see which of the following are necessarily irrational and let me stress that word necessarily number one r plus three number two five times r number three r plus s number four r times s and number five square root of r now this is a quiz format well i'm just asking you to select the ones that you think are necessarily irrational but unlike most of the quizzes where i expect you to be able to answer very quickly i'd like you to think a little bit before you answer each of these because the focus isn't really on getting the answer right i mean obviously i want you to get the answer right i'm sure you do too but that's not the focus the focus is on the reasoning that you need to carry out in order to get those answers in each case you're probably going to have to carry out one or two lines of simple reasoning in order to answer the question that's the focus of this particular quiz in fact in the assignment that's coming up in assignment 7 i'm going to ask you to write out proofs of each of the five answers so let me stress that the focus is on the logical reasoning okay see how you do well the ones that are necessarily irrational are one two and five in each case you use the fact that a rational number is one that can be expressed as a quotient of two integers and an irrational number is one that cannot be so expressed and then you carry out a couple of lines of reasoning to show that r plus 3 has to be irrational if r plus 3 was rational then r would be rational likewise you would carry out a couple of lines of reasoning to show that if 5r was rational and could be expressed as a quotient of two integers then so could have and similarly you carry out a symbol of argument for the square root of r the two that are not necessarily irrational three and four in order to show that r plus s is not necessarily irrational what you would need to do to find examples of irrationals r and s for which the sum is rational well how about taking r is the square root of two and s equal to uh let's say 10 minus the square root of 2 r plus s is equal to 10 which of course is rational it's an integer we know that the square root of 2 is irrational and a very simple argument shows that 10 minus the square root of 2 must be irrational in fact the argument you use to show that this number is irrational is a combination of the two arguments you used in parts one and two okay so r and s are irrational but their sum is rational in the case of this guy we could take r equals square root of 2 and s equals square root of 2. r and s are both irrational and yet r times s is 2 which is rational okay how did you do remember then in assignment seven i'm going to ask you to give proofs of each of the five answers well conditionals involving quantifiers are sometimes best handled by proving the contrapositive what's the contrapositive well we met that in assignment four so to prove a conditional phi yield psi what you can do is prove not psi yields not phi that's the contrapositive of phi yields psi you reverse the phi and the psi and you put negations in front of them and in assignment 4 if you look back we've proved that those two are equivalent we prove that using truth tables so let's look at an example of how this might work and the example i'll take is this one i want to prove that if the sine of an angle theta is not equal to zero then for all n in the natural numbers theta is not equal to n pi okay there's a conditional and i'm in the middle of some argument we'll imagine and i want to prove that if the sine of heat is not zero then theta is not a whole number multiple of pi well the statement's equivalent to not the case that for all n and n theta dot equal to n pi implies not that sine theta is not zero well there's lots of knots in here so let's clear them all out and put this into a positive form so in positive form this is not for all n and n we're going to get there exists n in n and the knot is going to move inside and it's going to negate this knot it's going to wipe that one out and then i'm going to have theta equals n pi yields not that sine theta is not zero means that sine theta is zero so this is the contrapositive of this and that's the thing we're trying to prove well we know this we know that whenever you've got a whole number multiple of pi its sine is zero so this proves the desired result obviously this is a highly contrived manufactured example again the focus is not what i'm actually proving it's the method i'm proving i'm picking simple examples so that we don't have to worry about the mathematical content we can just look at the logical reasoning and in this case we start out we want to prove something we replace it by the contrapositive and then we prove the contrapositive in this case the proof of the contrapositive we just pull on well-known knowledge about the sine function that the sine crosses the x-axis whenever you were to hold multiple of pi okay let me tell you about one more thing involving proofs of conditionals to prove a biconditional phi equivalent to psi we generally construct two proofs one of phi yields psi the other of psi yields phi and since the biconditional is just a conjunction of the two conditionals that clearly amounts to a proof of the bad conditional occasionally it's easier to prove the two conditionals far yields psi and not far yields not psi and i'm going to leave you to find out why this is enough why does this work if you look back at the assignments you should find a clue as to why it's enough to prove these two in order to prove the biconditional okay now i'd like it to complete assignment seven as usual completing an assignment means at least you should attempt all of the questions you may not be able to get them all out at least not at first but remember the the benefit from working on the assignments in this course is actually trying them whether you get them out or not is is is not that important remember my favorite example of learning to ride a bike the fact that you don't succeed in riding a bike for many days or weeks when you're learning doesn't mean you're not making progress the same with learning to swim you actually get the benefits during the process when you're trying and failing and then suddenly you find you can either ride the bike or you can swim and it's the same with this kind of material in this course okay good luck on assignment seven well i won't do all of the examples in uh in assignment seven um but uh let's do a few of them uh starting with number one uh prove or disprove the statements all bears can fly well we'll we'll find a counter example i mean that's false and to show it's false what i need to do is to find a counter example and an obvious one is the ostrich that's a bird that can't fly so over the counter example okay well now let's look at number two prove or disprove the claim for all x in our vertex and y in our x minus y squared is greater than zero uh that's also false and again to prove to something that's a universal quantified just like the one above it's a universal quantifier in this case two universal quantifiers and to prove that a universally quantified statement is false what you need to do is find a counter example well that's one way of doing it at least and the most obvious one here well anything that takes x and y equal will will do it so let's give a specific counter example let's take x equals y equals one and in that case x minus y squared equals zero and zero was not strictly greater than zero okay um came close right um if you if we'd excluded x and y being zero um then we then would have then we'd have a positive result but uh it says it's true for all x y and r and a single counter example is all it takes and in this case any pair of equal numbers gives us the gives us a counter example okay number three um prove that between any two in equal rationals there's a third rational so let's let um x and y be rationals x less than y okay then because they're rationals x is p over q we can write y is r over s where p q r and s are integers and we have to show there's a number between what the obvious thing is to take the the mean let's just take x plus y over 2 and if that's rational then that would have proved the result well here's the proof that it's rational x plus y over 2 is equal to p over q plus r over s over two okay which is p s [Music] plus q over q s all over two which is p s plus q r over 2 where was it qs can't see underneath my hand here which is rational because it's a quotient of two integers but of course x is less than x plus y over two is less than y and we're done okay that's one two and three knocked off very very quickly let's go on to do under number seven the one about square root of three well just as we did with the square root of two we're going to prove it by uh contradiction so i'm going to assume square root of 3 were rational okay then i could write 3 root 3 is p over q where p and q natural numbers and i can always assume that they have no common factors okay because if you pick a pair of integers a pair of natural numbers that do have a common factor you can cancel it out so you can always express a a rational number in that form then if we square that i get 3 is p squared over q squared so i can multiply across by the q squared and get 3q squared equals p squared so 3 divides p squared but three is prime and if a prime divides p squared that means uh three divides p if the prime divides a pair of numbers multiplied together it divides one of the numbers so 3 divides p so that means p is of the form 3 let's call it 3r okay so that means p squared equals 9 r squared so i can take p squared and substituting it back in here to get 3q squared equals p squared equals 9r squared so if i'll forget the middle term now 3q squared equals 9r squared let me factor the 3 out i've got q squared equals 3 r squared so that means 3 divides q squared but if 3 divides q squared then just as before with p that means that 3 divides q and now we've got a contradiction since p and q have no common factors and yet we've just shown that 3 is a common factor so there's a contradiction almost exactly the same as the one for square root of 2 so if we compare the 2 cf the proof for square root of 2 in the case of the square root of 2 we talked about numbers being even but if we talk about something like p being even that's just another way of saying 2 divides p so really all i've done here is i've taken the previous proof the one for root two and instead of talking about it in terms of even an odd i could recast it in terms of whether it's divisible by two or not and then the facts the fact about two that we use there we said if 2 divides p squared then it divides p it used the fact that 2 was prime so i might just as well use 3. so it's exactly the same proof as before except instead of talking about divisible by 2 i'm talking about been divisible by 3. i didn't express it that way before i talked in terms of even or not but even just means you it's a multiple of two okay well we've uh we've moved on um let me decide which one to do next i think i'll do number eight um that shouldn't take too long so we'll do number eight next okay so the converse of us of an of an implication of a conditional is uh the implication in the in the opposite direction where you swap around the antecedent and the consequence but everything remains more otherwise unchanged so the the converse of this is if the yarn rises the dollar falls let me just shut them around same here if negative y less than negative x then x is less than y and in the case of c if two triangles have the same area then they are congruent uh now in terms of what we're doing this is this is totally trivial this is just swapping things around without doing anything um what i'm really trying to get at with this uh this exercise is is in part to sort of contrast it with the with the contrapositive and also to observe that truth and falsity can change in this case we start out with something that's true this is a truth to implication that's a true implication too so sometimes the converse of a true implication is a true implication let's look at this one if two triangles are congruent they have the same area that's true but this one if two triangles have the same a they're congruent that one's false so sometimes the converse of a true statement is true and sometimes the converse of a true statement is false now that's a different situation from from from what we find with with the contrapositive but when you swap down the order uh you can sometimes get truth going to false and you can sometimes preserve truth okay well that's really all there is to that it was just to sort of give and give an opportunity to to reflect on these things and and to recognize that truth and falsity plays its own game when you're dealing with converses okay let's go and do number 11. okay that was that one about the the rational the irrational numbers but we've already looked at some of these in the lecture where we've we've observed that some of them are rational let me see we observe that this one can be rational we observe that this one can be rational and um there's an issue about this one being rational okay these were easy we did those in the lecture uh that was that was a later issue um that leaves one two and five uh they're the ones where uh uh where the irrational and so we have to prove them this time okay and so starting with number one say um what we're trying to show is that it is irrational so yes it's irrational okay and uh let's see suppose um uh uh suppose that uh r plus three we're rational okay that's a plus sign there then r plus 3 would be of the form p over q where p and q are integers where then r would equal p over q minus 3 which is p minus 3q over q which is rational and that's a contradiction because r is assumed to be irrational and we've shown that if our plus three were rational then i would have to be rational and the others are similar you simply assume the contrary you express it in terms of p over q and then you do a tiny little amount of manipulation and you end up showing that the uh the number r is rational here and the square root of r is uh that r is rational in the square root example okay very straightforward um nothing really much to be said about this one okay the only one left now to do on uh on assignment 7 is question 12. so let me just quickly go through question 12. well the key facts that you use in in dealing with all of these are that if n is even then an actuation if and only if then n is 2k for some k some integer k okay and that n is odd if and only if n equals two k plus one for some k okay the even numbers are the ones that are multiples of two and the odd ones are the ones that are one more than a multiple of two okay they're in between the multiples of 2. so for example if we do part a if m and n are even then we would have m equals 2k for example we'd have n equals 2l so m plus n would equal two k plus two l which is two into k plus l which means it's even it's two times something else um do i need to do one more well let me just do one more anyway just just uh because i've i mean i'm on a roll now let me do a number d part d if one of them is even it's 2k and the other one's odd it's 2l plus 1 then m plus n equals 2k plus 2l plus 1 which equals 2 into k plus l plus 1. okay so it's 2 it's twice something plus 1. etc for all the others okay nothing terribly deep about uh about these things but in terms of providing other proof it's basically a case of just coverage for hours and evenings and odds in this way and doing a tiny amount of algebra okay that was a very easy assignment i think well it was meant to be an easy assignment it's easy for me to say that right i've been doing this for years um if you haven't met this kind of thing before then it probably isn't an easy assignment um i fell into the trap of mathematicians there of of saying something's easy in the same way we talk about trivial that's a sort of a term of art that we use in the game so uh you know everything's easy when you know how to do it right okay but i was looking ahead to assignment eight which was decidedly difficult uh for me and for you guys okay assignment coming next is difficult well question one in problem set five asks us to judge whether this is a valid proof or not well certainly the algebra looks correct right if m and n are odd then by definition um an odd number is simply one that's not even and an even number is one that's a multiple of two so an odd number has to be of the form two p plus one so that's okay for some p and for some q so if the audience just says that in which case multiplying them together you get that the object was correct so m n is odd everything so far until this part because it doesn't complete the proof what we've shown is if we've shown that mn is odd if m and denote in other words we've proved the implication that way okay if m and n are odd then m n is odd so in this statement we've proved a right left implication and we need to show the others part we need to prove the only if part in other words the implication from left to right it's if and only if so we've proved the if part from right to left we need to prove the only part from left to right ie we need to show let's write it out in full if it's not the case there's m n o odd then mn is on he's not odd yeah he's not odd okay there are various ways you can talk about this okay um i a well if it's not the case the term ender odd that should have been a comma there okay if it's not the case that m and den are odd then m times n is not odd okay i got the tens correct the the plurality that was i how that is okay i if at least one of m and n is even then m times n is even well if m equals 2k it doesn't matter which one we assume is going to be even because multiplication is going to be commutative if m equals 2k then m n equals 2k times n which is even so this was hard this is how they're going to set the world on fire right i mean there's nothing terribly deep about this it's extremely elementary and it's not that you don't know this and understand it the question is can you write it down in a proof and the point here is that we have to start with the definitions the definitions are n is even if and only if there's an integer k such that n equals 2k and n is odd if it is not even so starting from these two definitions this is the proof we give the first part that was actually given to us that was fine from those definitions we started with the definitions going the other direction we took the definition so the question is simply important though all this really involves is reducing this to the definition and then the argument itself is very simple that was a slightly more complicated bit of algebra but not much more so it's not about the complexity of the algebraic manipulations you've been able to do that for years i know that the question is can you reduce uh the statement to something coming from the the definitions um because in many cases you simply would this would be so simple compared with most proofs in mathematics you wouldn't go into these details but the focus here isn't on the at this stage it isn't on the complexity of the argument the focus is on the structure of the argument are you able to get the logical structure right and experience tells me that it takes most of us and it certainly took me a long time to get the sense of what do i have to prove in order to prove something that's not something that's not cookie cutter you've got to build up experience in what constitutes a proof and being able to judge whether proof is correct or not okay let's go ahead and look at number two now let me jump first of all straight to the uh to the correct answer and the correct answer is a um this says there are people and there are times at which you can fool those people so you can fool some of the people some of the time remember existence quantifiers are what we use in mathematics to capture the word sum or at least one a little bit atypical in terms of what the word sum means in in everyday language okay and here the way i've expressed it is there's a person and there is a time such that you cannot fool that person at that time okay there's a person and a time such that you cannot fool that person at that time which is actually equivalent to saying you can't rule all the people all the time formally you could take that part and you could rewrite it as it's not the case that for all x and for all the x t it's not the case that you can fool all the people all of the time okay but i better write it this way because we're only asking for the equivalence i'm not saying which is the closest way of capturing it i'm just saying which is one's equivalent okay um let's look at part b what does this um what is the first part we're the same in all of these three so the distinction is in the second part so let's see what this one says can we express that in english um well this is it's tricky to express in english because of these american melanoma foundation type issues i mean you could say something like oh let's say you can't fool everyone let's try something like this at some time or other well for everyone it's the case that you can't fool them at some time or other um i mean really what it's saying is it's not the case but for every person uh there's a time when you can fool them okay so um you can't fool everyone at some time over there um yeah i'm that's sort of i'm i can't think of a way of saying this that that really makes that that reads well in english and yet which really captures this um but hopefully it's clear what the thing means you know you it's not the case that for every person there's a time when you can fool them um okay um to me that says the same but you know you may interpret that one different because this is one of these things like american melanoma foundation that it's ambiguous as its natural language so often is okay um let's see again in the case of this what does this one say it's not the case that there is an x and there is a t so you shouldn't this one i think is easy to say in english okay it basically says you can never fool anyone okay it's not you cannot find a single person in a single time so you can fool that person at that time you can never fool anyone so this one i'm very happy with okay that one really captures it um and the smiley doesn't indicate that i think that's true it indicates that i think that's a really clean interpretation of what that one means when then number four doesn't arise because we've already found something correct well questions three four and five were just truth table arguments that was essentially a revision material so i won't go through those those here um let's just take a look at question six which when you first meet it looks as though it might be terribly deep especially since we've we've seen this now after we've looked at things like square root of root 2 and square root of 3. but actually this one when you step back and think about what it says turns out to be a very simple we simply observe that whenever n is a perfect square in other words it's the square of some integer any integer i know infinitely many integers so there are infinitely many perfect squares then of course the square root of n is just k which is rational and that's all there is to that one um i mean not only is it rational it's a whole number the point is there are infinitely many numbers n for which the square root of n is a whole number and hence rational namely all the perfect squares okay um that's always to that you don't really need to say any more that's it that shows that it's true and it tells you why it's true and that's the proof because you only asked really to sort of say whether it was true or false but it was this that i was interested in can you construct a proof of its negation of vitoria's negation that was really what it's about it's all about proving things here okay well finally question seven is one of these uh fallacious proofs there's a whole range of these things uh quite amusing um usually they depend upon uh dividing by zero in some disguised form this one's different okay so let's follow it through because normally the point is just to find the mistake we're clear that there's going to be a mistake because one doesn't equal two uh and and indeed one of the things i asked you to do was to find a mistake but the focus here isn't so much on filing the mistake it's on how do you grade this as a proof and remember when we're grading proofs uh logical correctness is certainly part of it so there's going to be some losses of grid here but also it's about communication okay so let's go through this one and uh just see how we would grade the thing okay well for example is it clear it's absolutely clear so we're gonna have to give four marks for clarity this is very clear there's a good strong opening we start with the identity one minus three equals four minus six okay both sides are equal to two so uh full match we're giving the opening um the conclusion is stated so it's formax for stating the conclusion and um reasons given yes um absolutely adding nine plus four nine over four to both sides is completing the square to give you a perfect squares then we're seeing it factors which it does that's why we added down from 4 to both sides i'm taking square roots of both sides absolutely reasons reasons reasons so in terms of the structure this is wonderful now overall i'm going to have to give it zero right i mean that's clear because the thing is playing false um the question is you know what do i give for logical correctness well first of all we need to identify the ident where the error is occurring right and the error is here it's in this line because when you take square roots then of course there are positive negative square roots that we had and in fact the correct solution is to say the the minus left hand side the negative root on the left-hand side minus one minus three over two and it's the positive root you take on the right hand side when you take square roots so that's what you should have in other words three over two minus one equals two minus three over two in other words a half equals a half okay so no big deal there i mean that's uh that's nice the world still exists and the half does equal a half even though one doesn't equal two so the issue here was when you're taking square roots um there are two two possible signs and then the one that makes it valid is uh is the negative on the left and the positive on the right okay so this is where the mistake is the only question is um do we give partial credit for the fact that there's logical correctness everywhere else this is a judgment call i would say that these steps are so simple it's just a very elementary algebra arithmetic that um at this level i'm not going to give particular credit for getting these bits right and i'm going to say this this is a big mistake this is huge okay when you take square roots you have to take a plus or minus that's a mathematical mistake realizing that not recognizing that you need to worry about the signs when you take square roots so that's a big error from a mathematical point of view so this grading that i've given it which means i'm actually giving 16 means i'm giving no points at all on the mathematical correctness i'm giving lots of credit on everything else now if this was a mathematics course i wouldn't be using this rubric this way i would take account of these but they would have less weight i mean these have all got the same weight maximum of four and that's because the focus of this course is on on proofs and reasoning and and communication and these are important parts of that you know the assumption in this course is that you already can do some mathematics and so i'm not really grading you on that i'm grading you on mathematical thinking and mathematical communication so there's zero marks here on the mathematics but there's formats and everything else and this is the reason why i did this one this is where everything is well laid out but the thing is playing wrong so it distinguishes between mathematical correctness and uh the mathematical thinking and the communication and then writing a proof out correctly having said that in this case it's blatantly obvious that something's gone wrong but very often in mathematics mathematicians professional mathematicians make mistakes buried in proofs and because they're professionals they can usually write things correctly we know how we learn as part of becoming mathematicians we learn how to write things out we give reasons so in fact this although it looks absurd here you often find that because mathematicians know how to lay things out well results get published even though they're absolutely wrong and whenever it gets published basically what that means is that the referee of the journal where it's published has graded it and said this is this is correct this is absolutely correct so you know this is not an unrealistic situation in principle when false results get proved and when when false results get published that is because they never really proved the false results get published when what's going on then is that the the referee has gone through it and think everything and thought everything's okay okay here it's dramatic because the answer is absurd it's patently false but this is actually not an unrealistic scenario and and so giving a high grade um isn't isn't you know it's not a bad thing i mean it is i'm doing this to emphasize the distinction between the various things we're looking at here and this one i think makes it clear that there are other issues involved other than other than logical correctness but as i say if this was a regular mathematics course i wouldn't be giving 16 out of uh out of 24. in fact i i'm not even sure i'd give any marks for this if a student came up for this because it really um this is really a big mistake okay but within this context within the context of this course um this is the kind of thing we're looking for in looking at mathematical thinking and mathematical proofs and communication okay there we go well that's the end of the problem set questions but let me leave you with one more tantalizing little puzzle we're probably familiar with the story of archimedes and lived in greece about 250 bce or their boats who was asked by the king to determine whether a crown had been given uh was actually made of pure gold or not and that involved calculating the density and uh he knew how to calculate to find out the the mass of the the weight of the crown but the question was how do you calculate its volume no archimedes knew lots of mathematics for calculating volumes indeed he'd invented a lot of that mathematics he was able to calculate areas of circles and volumes of spheres and various other shapes like boxes and rectangles and pyramids and so forth he knew all of that stuff so he had a lot of mathematics at his disposal that he could have applied but it didn't seem to work for something irregular like a crown or at least not easily but then one day when he's taking a bath this is the sort of story goes when he's taking a bath he has this this amazing insight he says to himself if i immerse the crown in water it will displace some water in fact the amount of water it will displace is exactly equal to the volume of the crown so if i collect the water when it spills out of the bath but when i put a crown inside it's into the into the water then i'll be able to just measure the volume of the water in a standard way and i'll know the volume of the crown and as that story goes he was so impressed and tickled by his solution that he jumped out of the bath and ran stark naked through the streets crying out eureka eureka which is greek for i found it i found it um now i've known lots of mathematicians i certainly haven't known archimedes but i doubt if even a mathematician deep in the throes of solving a problem would run naked through the streets however i can imagine him being extremely pleased with himself and having a great adrenaline rush when he had that insight because that's a great example of thinking outside the box he knew lots of techniques for calculating volumes he invented many of them but on this occasion he thought outside the box and found a really elegant solution that was different and the puzzle i'm going to give you is very much along those lines and it actually is about taking a bath and here it is if it takes half an hour for the cold water for set to fill your bathtub and an hour for the hot water force it to fill it how long will it take to fill the tub if you run both photos together now this looks like one of those those frustrating little word problems you get in in in high school okay where you you end up you sort of say that the the rate of flow of the cold water be f um and then the t and you you write down some equations and you you you figure something out okay um and i'm sure you know how to do that you can apply a standard technique for doing this kind of thing involving rates of change and you get the answer but you don't need to do any of that you don't need to do any calculations at all in order to solve this one if you think outside the box you can answer it without doing any of those calculations based on the rates of flow or anything like that you simply have to think of the problem a different way if it helps if you're inspired by the archimedes story you might want to run about and get in the bath and see if you can come up with the idea then i i must admit from personal experience that a lot of my best ideas in in solving mathematical problems and improving mathematical theorems throughout my career have actually been obtained when i've been sitting in a bath or shortly after taking a bath there's something so relaxing about getting in a bath that it frees the mind to come up with these out of the box uh solutions so this isn't this isn't really about do you know the standard methods you probably do this is about can you think about this in a different way that allows you to solve it without really doing any of his music whatsoever okay well with that hint i'll leave you to it enjoy how did you get on with assignment seven in this lecture i'll say something about how to prove statements that involve quantifiers which is true for almost all mathematical theorems universally quantified statements of the form for all natural numbers n some property a n holds where the quantification is over all natural numbers are often proved by a method known as induction i'll explain the method of induction with some examples from number theory number theory is one of the most important branches of mathematics studies the properties of the natural numbers 1 2 3 etc we look at some elementary parts of the subject in later lectures but for now it provides good examples of induction proofs but first let's look at how we might prove an existence statement we want to prove there exist texts such as a x the obvious way is to find an object a for which a of a for example to show that there's an irrational number just prove that square root of two is irrational which we did unfortunately this doesn't always work sometimes we use indirect proofs for example i worked through that problem about the cubic equation where we showed that there was a point on the x-axis where the curve crossed it and hence that the cubic had had a solution we never found the solution we just showed that one existed let me give you a really fascinating example of an indirect proof i'm going to prove that there are irrationals r and s such that r to the power s is rational that actually came up in assignment seven if you remember and i promised then that i would look at this later well now is that later the proof involves considering two cases case one if the square root of 2 raised to the power square root of 2 happens to be rational we can take r and s both equal to square root of 2. then by the assumption that square root of 2 to the square root of 2 is rational we've shown that there are numbers r and s which are irrational with r to the power s rational well if case 1 isn't the case then we must be in case 2 which is where root 2 to the root 2 is irrational in which case take r equals square root of 2 to the square root of 2 and s equals square root of 2. and if we do that then r to the power s is root 2 to the root 2 all to the root 2 which by the way that exponents work is root 2 to the power of root 2 times root 2 which is root 2 squared which is 2 and 2 is rational so again we've proved that there are irrationals r and s such that r to the power s is rational taken together case one and case two prove the theorem i really like this example uh it's not just clever but it's cute and it's cute because of the way it works we really don't know whether this number root 2 to the root 2 is rational or irrational it's probably irrational actually but the point is we don't need to answer that question we just look at the two alternatives if it does happen to be rational seems weird but if it does happen to be rational then we can find irrationals and s with out of the s rational this way on the other hand if root 2 to the root 2 is irrational which probably is the case but again we don't need to know that then we take r to be this number has to be that number and then r to the s is two which is rational so we don't know officially which one of these is the case um and more to the point we don't need to prove which one of these is the case we simply look at the two alternatives and depending on which alternative we're in we take different rationals and we take different irrationals hour and s without of the s being rational so the choices of our and s are different in the two cases but because these two cases exhaust all possibilities this number is after all either rational or irrational then this proves a result by the way this is known as the method of proof by cases and it's a method that's used a lot in in advanced mathematics pretty neat huh well now let's take a look at how we might prove a statement involving a universal quantifier suppose we wanted to prove a statement for all x a of x one way is to take an arbitrary x and show that it satisfies a of x for example to prove that for all n there is an m m bigger than n squared where the variables m and n range over the natural numbers well this is a pretty trivial example but i want to focus on the method that you would use to handle the two quantifiers in a statement like this where in particular the first quantifier is a universal quantifier let n be an arbitrary natural number set m equal to n squared plus one then m is bigger than n squared and this proves the statement there is an m m bigger than n squared and it follows that the statement for all n there is an m m bigger than n squared is true okay well as i said this is pretty trivial uh let's just see what's going on in terms of the logical reasoning you want to prove a statement involving two quantifiers of the form for all n there is an m and bigger than n squared what we're going to do is eliminate one quantifier by replacing it by an arbitrary natural number now by arbitrary i mean we make no assumptions at all other than the fact that it is a natural number the argument would have to hold for any natural number whatsoever because we're trying to show that this is true for all n so we pick an arbitrary one then we carry out some reasoning to verify the rest of the formula okay when we eliminate that for all n we're trying to prove that there is an m bigger than n squared which is this statement now in this case it's a trivial argument we just square in add one and let them be that number so we explicitly find an m that satisfies this this property in practice uh you're probably gonna have to do a lot of work here with a more significant example there could be several lines or even pages of arguments involved at this stage but the point is you eliminate the first quantifier by picking an arbitrary end then you carry out some reasoning to prove the rest of the formula which is this thing here and when you've done that because the argument works for an arbitrary natural number it follows that you've proved it in this form the for all then has been handled by with an arbitrary natural number so we've taken away one of the quantifiers carried out some reasoning and proved the result but let me stress that this works because the n that we pick is arbitrary well another approach would be to use method of contradiction to prove for all x a of x assume not for all x a of x this is equivalent to existex not a of x well let c be an object such that not afc notice that this isn't an arbitrary object it's a specific object admittedly an object whose identity we may not know that is guaranteed to exist by this the point is to to prove this universal statement we began by assuming its negation its negation gives us an existential statement and that tells us that there will be some object a specific object for which not afc now we've got a starting point to carry out a proof to derive a contradiction now reason with c together with the fact that not afc to derive a contradiction now the exact reasoning you use will depend upon what the statement a says but the general pattern is this you want to proof for all x a of x and you're going to use contradiction so you assume not for all x a of x that's equivalent to exist x not a of x and that tells you you can find an object or at least you know that an object exists for which not afc and then if you can you reason with that object using the fact that not afc did you have a contradiction okay again uh we'll see examples of this later on i'm just giving you the general uh overview of the way these arguments work so as in the previous case uh if this looks a little bit mysterious uh it'll become clearer i hope when we look at some specific examples later in fact at this stage you might want to look back at assignment 7 and see what was going on there in the light of of what i've just said here and before i let you go let me give you a little quiz well the correct answer is no this is not a valid proof the point is picking a positive rational p arbitrarily is not the same as letting p be arbitrary the choice of p may indeed be arbitrary but once you've made it you've got a specific p and we've even said what it is that's actually different from saying let p be arbitrary an arbitrary choice of p leads to a specific p which you may have chosen arbitrarily but you might have been unfortunate in your choice you might have chosen a p that just happens to make something happen um without there been any specific reason it's a subtle point i know but you you you it's a it's an important point so you really need to to sort this one out letting p be arbitrary is not the same as picking in an arbitrary fashion a specific p the point is that p is specific its choice may have been arbitrary but once you've made it it's specific in contrast when you argue with an arbitrary p you don't know anything about the p you certainly wouldn't know that it was equal to 0.001 or whatever you may need to think about that one a little bit longer or talk it over with some of your colleagues now let's look at proof by induction this is the special case where we're proving the universal quantified statements where the quantifier ranges over the natural numbers for example suppose i want to prove that 1 plus two plus etc etc etc plus n equals a half n n plus one so the sum of the first natural numbers is a half times n times n plus one well if you're meeting this problem for the first time a smart thing to do is to begin by checking the first few cases just to see first of all to see if it's true for the first few cases and to try and get a sense of of what's going on so if n equals one what do we get well then we've just got one here and putting n equals one into the formula we have a half times one times one plus one which is a half times one times two which is one and indeed one equals one so it's true in the first case in the second case n equals two what do we have we have one plus two is the left-hand side equals put n equals two here we get a half times two times two plus one which is a half times two times three i have to so we've got three and one plus two does equal three so the formula is correct for n equals two um we might want to try n equals three what do we get then we get one plus two plus three equals a half three three plus one which is a half three four which is six which is correct so the formula checks out for the first three cases and you know we might want to do a couple more cases but at least that gives us some sense of what's going on we we by doing this we understand what the formula we understand what the equation's saying now so far this isn't a proof this is just checking the first three cases in fact you've got to be very careful about jumping to conclusions based on the first three or four cases so let me just stress that this is not a proof okay let me stress it again beware of jumping to conclusions and to drive this point home let me give you an example where if you jump to conclusions you'll come to the wrong conclusion consider the formula p n equals n squared plus n plus 41. if you start working out values for this formula you found something rather interesting you find that all values of p n for n equals 1 2 3 etc are prime numbers well not quite that's true until you reach n equals 41. and when you get to 41 you found that p of 41 works out at 1681 which is 41 squared so the first series of values for n equals 1 2 3 et cetera all the way up to 41 which is which is well beyond the number of cases most people would try out by hand you find you've got a prime number but if that leads you to believe that this formula generates primes for every n then you're going to be wrong because when you get to n equals 41 it's no longer prime this example by the way is due to the famous swiss mathematician leonard euler in 1772 and it's a cautionary tale uh telling us not to jump to conclusions based on on numerical evidence nevertheless it's always a good idea it's often a good idea to begin looking at a problem by working out a few cases not only do you get an understanding of what the formula is about but you can sometimes when you work out a few cases you can find a pattern in this case there actually isn't much of a pattern to find you work out the values and you just get get numbers and it turns out that the first 40 of them are primes but in this case if you look at what's going on after even if you if you did maybe a couple more n equals four n equals five you'd begin to see a pattern of what's going on how the two parts of the formula are combining what's happening to the half the half is dividing into the second one here the first one here the second one there and after you've done a few cases you can often find a way of of of capitalizing on the pattern that's there in order to produce the proof that i'll show in a minute so induction proofs can be can often be derived by looking at the first few cases seeing what the pattern is when you start plugging the numbers in and then actually building a correct rigorous proof okay so the next step now is to to begin with this numerical evidence and our understanding of how the formula works in order to come up with a rigorous proof of the fact that this formula holds for every n the method of induction depends on a principle known as the principle of mathematical induction okay and here's what it says suppose you want to prove a statement for all n aven establish the following two statements one a of one that's known as the initial case or the initial step two prove the following statement for all n for all natural numbers n if a of n is true then a of n plus 1 is true that's known as the induction step now intuitively this gives the statement for all n a of n as follows by the initial step step one a of one if we now apply the induction step in the special case where n equals one we have a of one by the initial step and the that yields if n plus one which is a of two so by two or by step two a of two a one yields a of 2 so from f1 we can conclude okay we begin by proving the initial step so if we've proved a of one then we know that one's true we can now apply the induction step in the special case where n equals 1 and having proved this if we have a 1 we can conclude af2 then we can do the same thing again once we've got a of 2 we can put the 2 in here and conclude af3 so by af2 and the induction step we can conclude af3 etc all the way through the natural numbers so it's a little bit like taking a row of dominoes and you knock the first one down and providing the dominoes are close enough so that when one gets knocked down the next one gets knocked down then once you start to topple the row of dominoes the first one not goes down then the first one knocks down the second one the second one knocks down the third one the third one knocks down the fourth one the fourth one knocks down the fifth one and so on and so on and so on all through the natural numbers so you get a dominoes falling down so this basically says knock the first domino down and put the dominoes close enough together so that when one falls the next one falls okay that intuitive explanation is is fine as far as it goes um but that's certainly not a vigorous demonstration of this method the uh the fact that proving these two together does yield that universal distance that university quantified statement that's actually quite a quite a deep result of mathematics okay and and let me just stress that you need an axiom oh a principle to make this work called the principle of mathematical induction the principle of mathematical induction is what tells you that one and two above steps one and two above yield for all n a of n okay now we have the the method of induction uh laid out before us i'll apply it in order to prove that identity that we looked at a moment ago about the sum of the first natural numbers so i'm going to prove a theorem now that for any natural number n one plus two plus three plus yadda yadda yadda plus n is a half n n plus one i'm calling it a theorem because i'm going to give a vigorous proof i'm going to use a method of mathematical induction and since one of the purposes of a proof is to explain why something is true it's usually good form to indicate to the reader the method you're going to use it orientates the reader to what's going to come next and i'm using a standard method here namely mathematical induction so it's good form to indicate that that's the method i'm going to use well i have to do two things i have to show that this is true when n equals one and then i have to show that if it's true at n it follows at n plus one so the first step is to say that for n equals one the identity reduces to well n equals one on the left-hand side i've just got one put n equals one on the right hand side i've got a half times one times one plus one which is a half times one times two and indeed the left-hand side does equal the right-hand side so that's true since both sides equal one so we've proved the first step incidentally you may have noticed that i referred to this not as an equation but as an identity and the reason is this an equation is something that you solve to find the value of a variable either x in the case of a real number equation or n in the case of a natural number equation but this isn't an equation that you solve for n this is an identity it's something it tells you that the two sides are always equal and we use the word identity to refer to an expression like this that claims or that states that two things are equal for all for all values of n so these two sides are identically equal so the correct terms are an identity for something like this an equation for something that you solve having said that like many mathematicians um in practice i'm sometimes not particular in the words i use if you caught me talking to my colleagues you might find me using the word equation where i mean identity possibly even using the word identity well i mean equation although that one's less likely but strictly speaking an equation is something you solve an identity is something that states that two things are always equal okay let's move ahead we've proved that the result is true at n equals one the next step is the induction step so i'm going to assume the identity holds for n i e i'm going to assume 1 plus 2 plus n equals a half n n plus one and let me know now and i'll put this in brackets i want to deduce one plus two plus plus n plus one equals a half n plus one n plus one plus one okay that's just a note for myself to indicate the target i'm aiming for and the challenge now is to start with what i know and deduce where i want to get well how can i get this from this well one way would be to add n plus one to both sides of that identity which means that i'll get the left-hand side here and then hope that when i work out the right-hand side having added n plus 1 i end up with this expression okay so let's add n plus 1 to both sides of let me call it star so star is the identity i'm assuming let me add n plus one to both sides of that that gives me one plus two plus plus n plus n plus one is a half n n plus one plus n plus one so i've now got the left hand side of what i want to prove i need to show that the right hand side is the right hand side of the thing i want to prove so i need to show that that is the same as that well let's just see what happens i can take the half out as a common factor so i've got a half into n n plus one plus two into n plus one and that equals a half n squared plus n plus two n plus two which is a half n squared plus three n plus two and this factors this is a half n plus one n plus two but that's the same as that i can write it explicitly if you like is a half n plus one n plus one plus one which is the identity with n plus one in place of n hence by the principle of mathematical induction the identity holds for all n and we're done we've proved the initial step a simple matter of observation and we've proved the induction step by a fairly simple algebraic argument and that's it that's mathematical induction in practice where this screen came up about a quarter of an hour into the lecture did you spot anything wrong with it at the time can you see anything wrong with it now well the problems in here this isn't the polynomial i meant to write down what i meant to write down here was the polynomial p of n equals n squared minus n plus 41. i've got the sign and for this it is the case that for all values of the polynomial for n equals 1 all the way up through n equals 40 it's prime okay all the way up and then p 41 if you put that in it's 41 squared minus 41 plus 41 which is 41 squared which is this number here 1681. let me call that piece of minus to indicate that there's a minus sign here because the polynomial i wrote down i'll call p plus of n was this one plus n plus 41. that one is actually prime for n equals 1 through 39 and if i put 40 in there let's put the minus in there to distinguish it if i put 40 in there i get what i get 40 n squared plus n is n times n plus 1. so it's forty times forty plus one plus forty one which is forty times forty one plus forty one so i've got forty times forty one and then another 41 that means i've got 41 lots of 41 which again is 41 squared equals 1681. these are sort of the same polynomial when you look at it in terms of the behavior here um if you just see what's going on they were both actually due to well actually i'm not sure historically whether i would have brought them both down but clearly once this had been discovered you're just manipulating them and what i did when i when i was giving the lecture this is spontaneous by the way almost all of these lectures are spontaneous they're not meant to be polished lectures i mean i'm familiar with the material so it sometimes looks as i'm just sort of reading it from a book but um if i am reading it from a book it's a book in my head if i'm having taught it for many years basically what i'm recording is spontaneous and this just i just pulled this off from my memory and i confused two things i knew that it didn't really matter what the sign was but it does make a difference as to where it stops being prime if you pick the negative sign it's prime all the way up to up to 40 if you pick the positive sign it's prime all the way up through through to 39. so it was a technical error but in principle it was correct and it certainly whichever one you pick gives a wonderful example of the fact that just because you have a repeating pattern it doesn't mean to say something's true for all n and remember that i i introduced this as an illustration of caution when i was talking about induction proofs okay well um what happened was the first time i gave this course um many students found the mistake and um i thought it was good to leave it in the second time i gave the course so that students would have a further chance of of finding the mistake themselves it's always fun to find mistakes this time i decided to leave it in but put in a quiz at the end just to prompt anyone that hadn't noticed a mistake that there was actually a mistake okay so now you know what's going on here and and you've seen me make a wonderful mistake in in public which is something that all good mathematicians should do okay well that brings us to the end of uh eight point a lecture eight point eight let's move on to lecture eight point b now where the correct answers are this one this one and this one if you check back the description i gave of the method of induction you shouldn't have any difficulty realizing why this is the correct answer why this is the correct answer and this is the correct answer so if you've got any of these wrong you definitely need to go back and look at the lecture again and make sure you really understand induction it's a very important technique in mathematics well let me give you another example of an induction proof now and this time i'll lay out the proof in a way that makes explicit the connection between the argument i give and the description of induction i gave a moment ago and the example i want to give is the following theorem well calling it a theorem is actually a bit of a stretch but here's what i'm going to show that if x is a positive real number then for any natural number n one plus x to the n plus one is bigger than one plus n plus one x okay this result is not going to set the mathematical world on fire but it is a good example to illustrate the way mathematical induction works and in particular to connect the arguments that i'm going to use to the early description i gave for mathematical induction so here's the proof well i'm going to begin by stating the method i'm going to use it's mathematical induction since one of the purposes of proofs is to explain why something is true it's always a good idea to begin by stating the method you use that orientates a reader to the argument that's about to come and to connect it to the description i gave of induction before i'm going to let n let a of n be the statement one plus x to the n plus one bigger than 1 plus n plus 1 x so by induction i'm going to prove 4n here n well the first step is to prove a of 1 and a of 1 is the statement one plus x squared is bigger than one plus two of x plus two x we put n equals one here we'll get one plus x squared greater than one plus two x is this true well the answer is yes it's true by the binomial theorem which tells us that 1 plus x squared is equal to 1 plus 2x plus x squared which is bigger than 1 plus 2x because x squared is strictly positive notice that x is positive so in particular x is non-zero hence x squared is strictly positive hence we have a strictly positive term here hence that expression is strictly bigger than one plus two x so a of one is true the induction step involves proving for all n a n implies a n plus one or how do we typically prove a statement like this we pick an arbitrary n and prove a n implies a n plus one and how do we prove an implication like this we assume the antecedent and deduce the consequence well a n is the statement one plus x to the n plus one greater than one plus n plus one x and a n plus one is the statement one plus x to the n plus two is greater than 1 plus n plus 2x so the induction step boils down to assuming this and deducing that format so i'm going to focus on this now and i'm going to begin by looking at the first expression 1 plus x to the n plus 2. in order to use the induction hypothesis i'm going to pull out one of these terms and write that as 1 plus x times 1 plus x to the n plus 1. and by the induction hypothesis one plus x to the n plus one is bigger than one plus n plus one x and that equals one plus n plus one x plus x plus n plus one x squared which equals one plus i've got m plus one x and another x so all together i've got n plus two x plus n plus one x squared this is positive so that expression is strictly bigger than one plus n plus two x and that part of it in the part here connected with this is this in other words this this proves a n plus one and the theorem follows by induction okay i've laid this out in gory detail here but my purpose was to connect the kind of argument that you give when you do an induction proof with the description of induction that i gave a moment ago in terms of predicates a n okay so this is so what i've been doing here is really emphasizing how my earlier description of induction connects to an actual proof if you take out all of the pedagogic stuff that i've put in here you've really got a two or three line argument i mean it really boils down to that step and then this step here those are the two key steps of the of the induction proof okay well there we are let me summarize how induction works you want to prove that some statement in is valid for all natural numbers n you first prove a1 that's usually a matter of simple observation the next step is give an algebraic argument to establish the conditional a of n yields a of n plus one usually you assume a of n and then you deduce a of n plus one from it and typically you do that by looking at the form of a of n plus one manipulating it into such a form that you can apply a of n in order to carry out an argument in other words reduce a of n plus 1 to a form where you can use a of n in conclusion by the principle of mathematical induction this proves for all n a of n occasionally you might have to work pretty hard just to establish the first case generally this step is easy and this one is where you have to put in the effort but neither of these two steps on their own constitutes an induction proof it's the whole package that constitutes induction my illustration of induction as as knocking over a row of dominoes while it's intuitively helpful obscures the fact that this principle of mathematical induction is actually a pretty deep principle of mathematics the reason it's deep by the way is because this is talking about an infinite collection it's talking about all of the natural numbers no infinitely many of them and the moment you start talking about infinity in mathematics there are subtleties and complexities lurking on the wings incidentally that's why we usually complete a proof by writing a sentence something like by the principle of mathematical induction the result follows because this is heavyweight and it's usually good practice and a shrewd move to mention a heavyweight if there's a heavyweight in the room okay well that's induction i'd better tell you about a fairly common variant of induction we sometimes need to prove a statement of the form for all n greater than or equal to n zero here then where n zero is some fixed number not five or ten or twenty or a million or something in this case the first step is to verify a of n zero a of one may not be true in fact it's usually the case that when we are trying to prove something of this form the reason we have an n zero as the starting point is precisely because the result doesn't hold at the beginning we have to go out somewhere through the natural numbers until the result starts to hold general induction starts at one this variant starts at some point beyond one but other than that the argument's essentially the same in particular the induction step is to prove for all n greater than or equal to n zero a of n implies a of n plus one in fact the example of induction i want to give you next uses this variance of induction and that example is part of a famous result of mathematics called the fundamental theorem of arithmetic before i do let me give you a quiz is one a prime number okay i know this isn't the focus of the course but there's a lot of confusion about this and if we don't clear it out of the way now it's going to keep haunting us uh as we go through the material that's coming up now you know i i i could of course simply give you the answer um and in previous times when i've given this kind of course both in a mooc and elsewhere i've given students the answer but it's one of these things that the people just don't sort of really pay attention to and then it comes up again when when there are homework exercises and discussions so what i'm going to do is give you this as a quiz so i want you to answer whether this is a prime number or not whether one is a prime number or not okay well the answer is one is not a prime number and the reason is the definition here's the official definition of a prime number it's a positive integer n greater than one whose only exact divisors are one and then that's all there is to it it's a definition so from now on please be very careful that you uh you really know the definition of a prime number one is not a prime number incidentally whilst we're on the topic of getting definitions cleared out let me make another remark zero is not a natural number again this is by definition zero is an integer but it's not a natural number historically the natural numbers are the counting numbers and people historically started counting at one two three etc uh zero was very much at johnny come lately okay so these are these these come around all the time uh they cause confusion it's it's unnecessary confusion so i thought i would get it out of the way now and i would draw your attention to this definition by putting it in as a quiz okay now let's go on to uh to what i was trying to do here's the part of the fundamental theorem of arithmetic that i want to prove and it's the following theorem every natural number greater than one is either prime or a product of primes now we have to use that variant of induction for this one because it's not true at n equals one the theorem only holds for n equal to two or more in fact one is the only number for which it doesn't hold so we're only just in the variant we've just we just missed one point but we have to start at n equals two in other words the the first step of the induction is going to involve looking at n equals 2 not n equals 1. but with that one change it's going to be an induction proof so proof will be by induction well the induction statement here n is well you might think it's going to be n is either prime or a product of primes okay based on the examples we've seen so far you might imagine that a of n the induction statement is that n is either prime or a product of primes well that turns out not to be the case that just doesn't work we have to pick something else and this is what we pick for all m if two is less than or equal to m is less than or equal to n then m is either a prime or a product of primes so the induction statement a of n talks about all the numbers between two and n being prime or products of primes so it's a sort of cumulative statement if you like as we go through the natural numbers we iterate on the cumulative fact of things being primes or product of primes and the induction proof establishes that a n is true for all n bigger than one so we have to start with the first step namely n equals 2 a of 2 says what well if n equals 2 there only is one m between two and then so the universal quantifier here uh is is essentially vacuous i mean it's just a special case it's a degenerate case there is just n equals two so this statement when n equals 2 simply says 2 is either prime or product of primes which is true two is a prime now we're going to assume f n and deduce a of n plus one well a of n plus one is going to talk about all numbers m between two and n plus one so let's look at one of those m's let me be a natural number two less than or equal to m less than or equal to n plus one well if m is less than or equal to n plus 1 m is either less than or equal to n or m is equal to n plus 1. if m is less than or equal to n then by a of n m is either a prime or a product of primes the only other possibility is that m equals n plus one well if n plus one is prime their name is prime so what happens if m equals n plus 1 and n plus 1 is not prime then the natural numbers p q such that one less than p q less than n plus one and n plus one equals p q if n plus one is not prime then it can be obtained as a product of two smaller numbers not being prime means you can find two smaller numbers whose product equals the number two smaller numbers other than one that is so this is just a definition of not being prime well since 2 less than or equal to p and q less than or equal to n by a n p and q are either primes or products of primes these two numbers p and q are less than n plus one so they're less than or equal to n that means these numbers are m's that fit into the induction hypothesis they're less than or equal to n so by a of n each of these is either a prime or a product of primes but n plus one is the product of them so n plus 1 is a product of two numbers each of which is either a prime or a product of primes hence n plus one is a product of primes the theorem follows by induction if you look back through this proof now you'll see why we took our induction statement to be this somewhat more complicated looking statement we wanted the income statement to talk about all the numbers up to and including n being either prime or products of primes so that when we went through the proof and we came down to a pair of numbers like this we could immediately apply the induction hypothesis a of n in order to conclude that each of those two were either prime or products of primes and therefore that the endless one was either a prime or a product of primes incidentally formulating induction statements that are cumulative in this fashion is quite common it's often the case that in induction proofs you end up dropping down to some number or some pair or group of numbers smaller than the one you've got and needing to apply the induction hypothesis to those smaller numbers as we did here with p and q okay how did you like that one okay so now you know about induction and with induction we're starting to get into some serious university level mathematics though the general idea is simple enough if you can always go one more step then you will eventually reach every natural number in practice it can require a lot of effort and ingenuity to construct an induction proof see how you get on with assignment eight well assignment 8 was really quite difficult i think on the other hand looking at the discussions on the forums i think many people found it difficult because they hadn't yet made this transition to doing mathematical thinking they were trying to to do something that wasn't really required this was very definitely the case with this one uh remember the the things i keep repeating are the one of the essences of mathematical thinking is that you uh you you first of all you ask yourself what is this asking me what is what do i know what does it tells me what kind of objects is it talking about and what do i need to do in in in terms of what's my target so you have to stop you pause you reflect you think about it what you don't do at least at the beginning is say does this remind me of a problem i've solved before that i can just instantly apply that previous technique uh because that can lead you in in in in the completely the wrong direction for example there are various words here you know in high school a good strategy was look for key words and and and try to map them into techniques but it really isn't a good strategy in terms of advanced mathematics okay there are many questions many problems we have about perfect squares that are about the natural numbers okay so what this isn't about the natural numbers this is about the integers completely different set of numbers so we're not a completely different set it's a large set of numbers so this is about the integers and the fact that some of the words are often used in discussions about the natural numbers well so be it i mean words get used for all sorts of contexts okay so this is a question about the integers and if it's about the integers then why don't we just take m equals n equals zero then m squared plus mn plus n squared equals 0 which is 0 squared and we are done folks this is not a trick question this is a question that gets at the heart of mathematical thinking and the heart of mathematical thinking isn't knowing a ton of techniques from mathematics the heart of mathematical thinking is what do i know and what do i need to do with what i know and what am i talking about stop slow down reflect think mathematical thinking at this level is not a sprint all of the techniques you taught in highs were taught in high school were very good for succeeding in high school and they certainly leave you with a lot of valuable skills but we've been there done that you know if we're beyond the high school if we've mastered that then we're trying to do something else high school sort of teaches you how to lay bricks what we're doing now is saying how can i take all of those bricks which are valuable things to have and use them to build a house we're now going from being bricklayers to architects uh we certainly need all of those bricks you know i am definitely not knocking high school right we use the stuff all the time but that was providing us with the basic tools now we're learning how to make use of those tools okay that's what mathematical thinking is about and sometimes we need a lot of tools and sometimes we can get by with something very simple because we just ask ourselves what am i trying to do now and we're not getting seduced into thinking that this is just another variant of something we've met before think of every problem as a new problem then you'll find that things get much more doable because you'll be focusing on thinking not applying techniques okay that's the end of my sermon well it's not really the end of my seven because that's what the whole course is about but let me just move on now to question two and in this case um you probably haven't seen anything quite like this before um so you have to start by asking yourself how could it happen that the answer be was a perfect square i mean just how could that come about what kind of thing must happen well um let's just write it this way how can we have m n plus one equal to p squared for some p well at this point one of those bricks that i was taught in high school becomes really useful because if m n plus one equals p squared that means m n equals p squared minus one and one of the bricks i had drilled into me in high school was that p squared minus one is p minus one p plus one so what we're about to do is take something that was drilled into it in high school and make clever use of it okay because if mn is p minus one p plus one then we could have m equals p minus one and n equals p plus one in that case p would be m plus one remember m is the number we're given we're trying to find an n um and i'm just using this to say what could that n be what must that end look like so given an m we'd be able to take p equals m plus 1 and then from this one n equals p plus two and now we found the n that is m plus two whoops m plus two okay p plus one p is n plus one and n is p plus one so n is m plus one so given the m we've already found the n so given n let's just summarize it now given m check n equals m plus two then m n plus one equals m m plus two plus one which is m squared plus two m plus one which is m plus one squared so we've found without we've answered the question we've said that given an m take n equals m plus 2 then m n plus 1 is m plus 1 squared which is a perfect square now if i hadn't gone through this it would have appeared that this was a rabbit out of a hat trick and unfortunately it's a consequence of the way that mathematicians often write their papers that they don't include all of the reasoning they just give you the conclusions um this would be an easy a simple embedded simple example but if i'd simply said given m while it would take n equals m plus two you would have said how on earth did he come up with that what made him think of that you know and and i look as i've got some kind of magical ability no i don't i just went in and said well how could i possibly get to that answer and then it was just something i learned in high school okay so you you know one of the techniques is just say how could i possibly get the answer that i'm asked to find now i'm couching this in terms of of classroom questions but the same the same issue arises when you're dealing with mathematical problems in the real world look at the problem what does it tell you what do you have to do how can i get the answer that i'm that i'm going for um so this might be a simple classroomy type example but it has many of the elements of good mathematical thinking and in this case it was how can that answer possibly arise and as soon as i sort of start to look at that it just drops out okay recognizing that was the entire key to the thing once you've done that straight forward okay let's move on and look at the next one question three well question two is another one of these things that when you first meet it you think wow this is asking me for something really deep uh how could i possibly come up with a with a quadratic all of whose values are composite how could i guarantee that all of the values are composite uh you know sounds like it's going to be really deep but now if you if you sort of just take a breath sit back and say you know how could it happen that numbers are composite well if they're composite they have to be the product of two or two other numbers okay wait a minute why don't we just take if that's going to always be the product of two whole numbers let's make it the product of two whole numbers and lo and behold that is indeed of the form n squared plus b n plus c where b and c are positive that's all there is to it we just wrote one down instantly we didn't have to prove that one exists which i mean we didn't prove it we did it by just writing one down and all we had to say to ourselves is no this sounds complicated but all it means is that the values of products and quadratics are products of things so we just write it we explicitly make it a product of two things the value will always be the products of two numbers n plus one and n plus two and it's for all positive integers so we'll have maybe two and three or whatever so these are always greater than one so this thing is composite n squared plus c plus two is always composite that's it didn't involve any advanced machinery just thinking about what the problem asks us to do very similar with this one you know it sounds deep or it's about the goldbach conjecture a problem that's been around for hundreds of years hasn't been solved um how on earth can we do anything with the goldbach conjecture well the answer is just look at what it tells us and ask us what we can do with it and first of all we observe that this is talking about numbers n bigger than 5 that are odd well if n is an odd number bigger than five then n is of the form two k plus three where k is bigger than one i mean normally we think of odd numbers as being of the form uh two k plus one but because i'm i'm looking for prime numbers here uh i'm gonna write it as two k plus three um and i can do that because n is bigger than five so for numbers bigger than five any odd number bigger than five is of the form 2k plus 3 where k is bigger than 1. well in that case since 2k is bigger than 2 because k is bigger than 1 2k is an even number bigger than two so by goldbach's conjecture 2k equals p plus q where p and q are primes in which case n which is two k plus three is p plus q plus three the sum of three primes we're done actually wasn't difficult at all once we sort of thought about what it says all of the complexity is in this unsolved problem goldbach's conjecture if we assume goldbach's conjecture which we're allowed to because it says if that's true then every even number bigger than five is the sum of three primes one of which in the case of our proof is three that was it okay actually it wasn't difficult at all it just looked as it might be when we first met it so another lesson we can learn is don't get put off because something looks complicated until you think about it you don't really know whether it's complicated or not okay a couple more on this uh on this assignment sheet and then we're done with it well questions five and question six the last two questions they're both induction proofs so what i'm going to do is i'm going to do uh example five and then i'll leave you uh to do number six if you've already tried it and failed and you came here looking for an answer you're not going to find it explicitly but hopefully by by watching me go through another example namely number five you'll be able to go back and do number six um because these things are all very similar at least the ones i'm giving you all very similar not all induction proofs are similar but these uh the ones i'm giving you from from number theory are all very similar okay well the first thing we have to do is express this is for some kind of an equation um you know with a formula in it because uh otherwise we're going to deal with this expression the sum of the first and odd numbers so we're going to write down a formula for the sum of the uh the first n odd numbers so what it asks us to do is show that one plus three plus five plus and then the nth one is two n minus 1 and we have to show that equals n squared okay arguably the hardest part of this whole thing is figuring out what the last term is okay so um that's what we've got to prove and we're going to do it by induction and we're told to use by induction so um we have to begin by looking at the first case for n equals one this becomes just one equals one squared which is true so it's true for n equals 1. okay so now let's assume the result let me call that star so let me now assume star and now i need to prove the same the corresponding result with an n plus 1 in place of n so why don't i just add the next term to both sides the next term will be two n plus one so both sides of star in which case i get one plus three plus five plus plus two n minus one plus two n plus one this is the first n plus one okay that's the first n plus one odd numbers this is the next case in the induction and now when i add 2 n plus 1 to the right hand side it becomes that wait a minute i can now use another of those bricks that i learned in high school thank you to my high school math teachers because now i can just get this one straight out this was drilled into me boy it's good to have these things at my fingertips that's n plus one squared we're done it's the same formula with n plus one in place of n wasn't difficult at all in fact all the machinery i needed to solve this one i learned in high school the one thing i didn't learn in high school not at least not very well was how to strategize about a novel problem um and that's what this course is about this is how do you take all of that great stuff you learned in high school and strategize and use it and get new results and think about new problems but before you do that you have to ask yourself what is the problem i'm given what do i know about it and what am i having to prove and if you misread a sentence you're going to end up doing the wrong problem you know there were a lot of a lot of people writing on the forums where what they were really saying was you know i demonstrated all my high school skills and great virtuosity and solved a problem that i wasn't asked to solve um well you know in this course there's a there's very little penalty for doing that except you feel bad um you know if this wasn't a course if you were working for a large corporation you might be out of a job or demoted or moved to a less interesting job so this is a a low penalty experience for how to think mathematically and which is why these moocs are great okay well as i said i'm going to leave you to do question six um and having seen another example uh i hope you're not put off by the formulas and the complexity of question six it's really just logical thinking okay mathematical thinking okay so much for assignments eight um seemed hard at the time actually was hard but hopefully it doesn't seem quite so hard now it's not hard if you become a good mathematical thinker and that's achieving that is what the whole thing's about okay how did you get on with assignment eight though the focus of this course is a particular kind of thinking rather than any specific mathematics the integers and the real numbers provide convenient mathematical domains to illustrate mathematical proofs those domains are number theory and elementary real analysis the principal advantage is example domains is that everyone has some familiarity with both systems yet very likely you won't have been exposed to their mathematical theories i'll take the integers first most people's experience with the whole numbers is by way of elementary arithmetic yet the mathematical study of the integers looking beyond mere calculation to the abstract properties those numbers exhibit goes back to the very beginnings of recognizable modern mathematics around 700 bce that study has gone into one of the most important branches of pure mathematics number theory most college mathematics majors find that number theory is one of the most fascinating courses they take not only is the subject full of tantalising problems that are easy to state but require great ingenuity to solve if indeed they have been solved and many haven't but some of the results turn out to have applications crucial to modern life internet security being arguably one of the most important unfortunately we'll barely scratch the surface of number theory in this course but if anything you see in this lecture arouses your curiosity i would recommend that you look further you're unlikely to be disappointed the mathematical interest in the integers lies not in their use in counting but in their arithmetical system given any two integers you can add them subtract one from the other or multiplying them together and the result will always be another integer division is not so straightforward and that's where things get particularly interesting for some pairs of integers say 5 and 15 division is possible 15 divides by 5 to give the integer result 3. for other pairs say 7 and 15 division is not possible unless you're prepared to allow fractional results which takes you outside the integers if you restrict arithmetic to the integers division actually leads to two numbers a quotient and a remainder for example if you divide nine by four you get a quotient of two and a remainder of one nine equals four times two plus one this is a special case of our first formal theorem concerning integers the division theorem the division theorem says the following let a b be integers with b greater than zero then there are unique integers q and r such that a equals q b plus r and zero less than or equal to r less than b well there are two parts to this theorem there's an existence path there are integers with this property and there's a uniqueness part those integers are unique with this property in proving the theorem we're going to take those one at a time i'll first prove existence and then i'll prove uniqueness so proof as always it's a good idea to begin a proof by stating the method you're going to adopt in this case what i'll say is i'm going to prove existence first then uniqueness i'm not using any standard method so i'm not able to state that i'm using something like induction or whatever but it will orientate the reader to say that the first part of the proof will deal with existence and the second part will deal with uniqueness okay so we're going to handle existence well look at all non-negative integers of the form a minus kb where k is an integer ensure that one of them is less than b what the theorem says is that among the integers a minus q b namely the integers r there is one with r between zero and b so i'm going to look at all possible candidates for a minus kb or a minus qb if you like and i'm going to show that among those candidates one of them satisfies that condition and the k for which it satisfies that condition will be the q that i'll take with the theorem well first of all i need to show that there are such integers for example take k equals negative absolute value of a then since b is greater than or equal to one a minus kb equals a plus absolute value of a times b which is greater than or equal to a plus absolute value of a which in turn is greater than or equal to zero well having shown that there are such integers by producing a k that gives me such an integer i can choose the smallest one let r be the smallest such integer and let q with the value of k for which it occurs i e will have r equals a minus q b so with r and q defined in this way i will indeed have a equals qb plus r if i can show that r lies between zero and b in this fashion then i'll have proved the existence so to complete the proof we're sure that r is less than b well suppose on the contrary that r is greater than or equal to b in other words i'm now going to use proof by contradiction to show that r is less than b well if r is greater than or equal to b then a minus q plus one b equals a minus q b minus b which equals r minus b which is greater than or equal to zero by virtue of this assumption thus a minus q plus one b is a non-negative integer of the form a minus kb but r is the smallest such and yet a minus q plus 1 b is less than a minus q b which is r and that's a contradiction if r is the smallest such we can't find a smaller one there's a contradiction hence r is less than b because we obtain that contradiction on the assumption that r is greater than or equal to b and that proves existence if you haven't seen a proof like this before it almost certainly looks very complicated it's actually not complicated or deep it's just intricate in its structure the idea is to look at all of the possible candidates to make this true well first of all we have to show there is a possible candidate and we actually do that by exhibiting one and then to get a candidate which satisfies this in this additional requirement we pick the one where the r is is the smallest is that it's the minimum value of r that gives us one of these things and then we use the fact that it's minimal in order to show that it satisfied that's to satisfy this requirement each individual step is just elementary algebra it's just somewhat intricate to push the argument through there's a lot of arguments in mathematics like this uh there's no deep mathematics involved there's nothing more than elementary arithmetic and a little bit of algebra but there's a bit of intricate structure to make the proof work so if you haven't seen this kind of argument before it'll probably take you several reads through uh you need to look at it several times before it begins to make sense okay well that proves existence then we need to turn to uniqueness to prove uniqueness we show that if there were two representations of a a equals q b plus r and a equals q prime b plus r prime with zero less than or equal to r and r prime less than b then r equals r prime and q equals q prime so the first part of the division theorem the existence part that we've already proved shows that there are representations of this form what we're now doing is showing that if there are two such representations then in fact they're identical the the r is the same and the q is the same incidentally you probably realize by now that the the letter q is going to denote quotient and letter r is going to denote remainder um we haven't officially defined quotient and remainder yet so i haven't written anything like that down but i'm using the the familiar notation because what we're doing is we're giving the mathematical underpinnings of the familiar knowledge that you probably have about about arithmetic dividing one integer by another okay so we have an equation here qb plus r equals q prime b plus r prime let's rearrange it rearranging i get r prime minus r equals b into q minus q prime and i've labeled that equation 1 for later reference now i'm going to take absolute values in this equation and when i do that i get absolute value r prime minus r equals b times absolute value of q minus q prime which i've labeled equation two b is positive remember so when you take absolute values b remains the same but negative b is less than negative r which is less than or equal to zero and zero is less than or equal to r prime which is less than b so negative b is less than i prime minus r is less than b well if r prime minus r is between negative b and plus b that means that the absolute value of r prime minus r is less than b so by two b times absolute value of q minus q prime is less than b well i can divide by b now to give me absolute value of q minus q prime is less than one but everything here is an integer and the only way you can have a pair of integers the absolute value of the difference between them being less than one is that that absolute value is zero which means q has to equal q prime and then by equation one now has to equal our prime and that proves uniqueness and with it we've proved the division there well again if you haven't seen arguments like this before it's going to seem pretty daunting but when you go through it step by step this is really just very basic arithmetic arguments with inequalities take it step by step and you should be able to follow the arguments through right to the end okay good luck with that if this is the first full-blown rigorous proof of a theorem like this that you've encountered you'll probably need to spend some time going over it the result itself isn't deep it's something we're all familiar with our focus here is on the method we use to prove conclusively that the division property is true for all pairs of integers time spent now making sure you understand how this proof works why every step was critical will pay dividends later on when you encounter more difficult proofs by gaining experience with mathematical proofs of simple results like this one which is obvious mathematicians become confident in the method of proof and can accept results that are not at all obvious for an example of a result that's not obvious in the late 19th century the famous german mathematician david hilbert described a hypothetical hotel that has a strange property hilbert's hotel as it's become known is the ultimate hotel and it has infinitely many rooms as in most hotels the rooms are numbered using the natural numbers one two three etc one night all rooms are occupied when an additional guest turns up i'm sorry says the desk clerk all our rooms are occupied you'll have to go somewhere else the guest a mathematician thinks for a while before saying there is a way you can give me a room without having to eject any of your existing guests before i proceed with this story you might like to stop the video for a moment and see if you can see the solution the mathematician guest has seen the clerk is skeptical but he asks the mathematician to explain how he can free up a room without ejecting anybody already in the hotel where it's simple the mathematician begins you move everybody into the next room so the occupant of room one moves into room two the occupant of room two moves into room three and so on throughout the hotel in general the occupant of room n moves into room n plus one when you have done that room one is empty you put me in that room the clerk thinks about it for a moment and then has to agree that the method will work it is indeed possible to accommodate an additional guest in a completely full hotel without having to eject anyone the mathematician's reasoning is totally sound and so the mathematician gets a room for the night the key to the hilbert hotel argument is that the hotel has infinitely many rooms indeed hilbert formulated the story to illustrate one of the several surprising properties of infinity you should think about the above argument for a while you won't learn anything new about real world hotels but you will come to understand infinity a bit better and the significance of understanding infinity is that it's the key to calculus the bedrock of modern science and engineering and when you're satisfied you understand hilbert's solution try the following variants first variant the hilbert hotel scenario is as before but this time two guests arrive at the already full hotel how can they be accommodated in separate rooms i should add without anyone having to be ejected second variant this time the desk clerk faces an even worse headache the hotel is full but an infinite tour group arrives each group member wearing a badge that says hello i'm n for each of the natural numbers n equals 1 2 3 etc can the clerk find a way to give all the new guests a room to themselves without having to eject any of the existing guests and if so how examples like the hilbert hotel demonstrate the importance of rigorous proofs in mathematics when used to verify obvious results like the division theorem they may seem frivolous but when the same method is applied to issues we are not familiar with such as questions that involve infinity rigorous proofs are the only thing we can rely on now back to the division theorem well the way i've stated the division theorem it only applies to division by a positive integer b there's a more general version that i'll give you now so theorem general division theorem let a b be integers with b non-zero then there are unique integers q and r such that a equals q times b plus r and zero less than or equal to r less than absolute value of b so the general division theorem is almost exactly the same as the previous division theorem we proved except that instead of demanding that b is strictly positive we're saying that b is non-zero and then here we have the absolute value of b and the proof follows fairly straightforwardly from the previous result then since the absolute value of b is greater than zero the previous theorem tells us there are unique integers q prime and r prime such that a equals q prime times the absolute value of b plus r prime and zero less than or equal to r prime less than absolute value of b well now we simply let q equal negative q prime and r equal r prime then since absolute value of b in this case with b negative is negative b we get a is qb plus m 0 less than or equal to r less than absolute value for b and that's this theorem so we simply threw it back to the previous result and with the general division theorem now established we can formally give names to the the two numbers q and r okay and officially we say the number q is called the quotient of a by b and r is called the remainder and so we've now cycled right back to something that you learned in elementary school about dividing numbers and that there were sometimes remainders there were quotients and remainders okay only now we've done it with some sophistication and we've shown that everything is well defined how about that well now that we have the division theorem available we can look at the important mathematical property of divisibility if division of a by b produces a remainder r equal to zero we say a is divisible by b hence a is divisible by b if and only if there's an integer q such that a equals b q for example 45 is divisible by 9 but 44 is not divisible by 9. the notation that we use for divisibility is this b vertical line a denotes a is divisible by b and let me give you a warning at this point b divides into a or a is divisible by b is not the same as b divided by a with a slanted line this guy is a relationship between a and b it's true or false this guy denotes a rational number the result of dividing b by a in the rational numbers so we have a property a relationship which is either true or false and we have a notation for a rational number now these are quite distinct concepts this is something to do with two integers tells you whether two integers are in a certain relationship this is a notation for a specific number the reason i'm issuing this as a warning is that beginners in particular often confuse these two things they both after all do have to do with division but in this case it's a property has nothing to do with uh with fractions with rational numbers this it's a specific number it's a specific rational number that we're referring to now there wouldn't be as much of a problem if the notation wasn't very similar and it gets worse when people with sloppy handwriting and you've probably noticed that i'm one of them often don't distinguish clearly between a vertical line and the slanted line in fact this one is already almost vertical uh maybe it would have been better if i'd written it like that okay that's better but because the notation is similar it's often very easy to get confused between these two things the context does disambiguate if you understand what's been discussed you shouldn't have any problems but bear in mind especially when you're meeting this material for the first time that there is an important distinction here and in fact i've got a quiz coming up in a moment which i hope is going to help cement your understanding of this notion so that you'll be able to distinguish it from this one okay and once we've got divisibility lined up we can define the all important notion of a prime number a prime number is an integer p greater than 1 that is divisible only by 1 and p notice that we explicitly exclude 1 from the prime numbers so the first prime is two the next one is three then five then seven and eleven and so forth now for that quiz about divisibility okay how did you do well remember the condition we have to check is this one b divides a if and only if there is a q such that a equals b q well let's see what we have the correct ones a b d f g and h why don't i have a well remember the whole definition of divisibility was under the assumption that b is non-zero we can't consider divisibility by a zero number so that one doesn't work and likewise that one doesn't work and of course the reason this one doesn't work is because 7 simply doesn't divide into 44. okay do be careful about this though division bar zero is problematic under any circumstances and so we we rule it out and in this case it was ruled out in the very statement of the divisibility theorem okay how about this one well again the condition we have to check is that b divides a if and only if there is a q such that a equals bq and again this is under the assumption b non-zero everything is a is an integer here but the correct ones and that that and that so why the others false well this one is false because that simply doesn't divide that however if you sort of got out a calculator and actually tried to divide it then you were missing the obvious point this is an even number and that's an odd number and we cannot have an even number dividing an odd number so if you had to do any arithmetic with this uh then you were making the mistake that i've been warning you against of acting as if you were in high school this is a course about mathematical thinking and what i hoped you would have done is to think about this mathematically and realize that you can't have an even number divided into an odd number so this is a cautionary reminder that this is not about jumping into calculations and applying procedures this is all about thinking about a problem okay so that's false but i would hope that you would recognize it's false without doing any work whatsoever than a little bit of thinking okay i occasionally like to slip things like that into into quizzes just to keep people on their toes um well this one is false of course because that's simply not the case that to 2n divide n squared the reason uh the others are false well this is false for various reasons i mean it's the same i mean 2n simply doesn't divide n squared for various reasons the reason this is false is because if n can vary over z then that will include n equals zero and so we can't have this because we can get a zero coming in that includes n equals zero likewise for this one okay same thing so this one fails because zero was included if i excluded zero it would be fine the only difference between these two is that this is the positive integers one two and three and so on and so forth this one because it's all of the integers includes zero okay well let's move on now i want to prove a theorem now that gives the basic properties of divisibility so theorem let a b c and d be integers with a non-zero then the following all hold one here divide zero and a divides a two a divides one if and only if a equals plus or minus one three if a divides b and c divides d then ac divides b d this is under the assumption c is non-zero because we can't have divisibility by zero the division theorem itself excludes the case of division by zero four if a divides b and b divides c then a divides c and this is under the assumption b non-zero again we have to exclude the possibility of division by zero five a divides b and b divide j those two hold if and only if a equals plus and minus b six if a divides b and if b is non-zero then absolute value of a less than or equal to absolute value of b you can only divide smaller numbers into bigger numbers or at least numbers that are no bigger than you can't divide a number b by something that's bigger than it in absolute value one more seven if a divides b and a divide c then a divides b x plus c y for any integers x y okay you prove all of these by going back to the definition of divisibility that what i'm going to do is just i'll give you i'll prove two of them uh as examples um let me just uh pick number four let's prove four that's this one here all the proofs are essentially the same idea if a divides b and b divides c then that means there are integers d and e such that b equals d a and c equals e b that's the definition of divisibility in which case c equals d times e times which again by the definition of divisibility means a divides c let me do one more let me do six okay so a divides b so that means since a divides b it means there is a d such that b equals d a in which case taking absolute values absolute value b equals absolute value d times absolute value a and since b is non-zero we know that the absolute value of d will have to be greater than or equal to one so the absolute value of a is less than or equal to the absolute value of b okay see b is non-zero so d can't be zero so its absolute value is greater than or equal to one and if the absolute value of d is greater than or equal to one then a has to be less than or equal to b the other statements in the theorem are proved similarly you just take it back to the definition of divisibility and uh when you you know that's uh remember the definition is b divides a if and only if there is a q such that a equals b q and in all cases you go back to the definition and then the thing drops out in a couple of lines okay well that's the that lists all of the basic properties of divisibility okay it's time to prove the fundamental theorem of arithmetic theorem every natural number greater than 1 is either prime or can be expressed as a product of primes in a way that's unique except for the order in which they are written for example four is two times two or two squared six is two times three eight is two cubed nine is three squared 10 is 2 times 5 12 is 2 squared times 3 and so on and so on 3 366 equals 2 times 3 squared times 11 times 17 and so forth i'll work that one out in advance by the way okay so 2 itself is prime 3 is prime 4 is a product of primes 5 is prime 6 is a product of primes 7 is prime 8 is a product of primes 9 is a product of primes 10 is a product of primes 11 is a prime 12 is a product of primes and so on and so on and so forth the expression of a number as a product of primes is called its prime decomposition and when you know the prime decomposition of a number you know an awful lot of information about that number and how it behaves with with relation to other numbers but we proved part of this a little while ago prove the existence part though i'm going to give you another proof of existence in a moment but the new part is to prove uniqueness the uniqueness proof will require euclid's lemma that says if a prime p divides a product a b then p divides at least one of a b the proof of view consumer is not particularly difficult but it would take me outside the scope of this course remember the the focus of this course isn't to isn't to teach you number theory and number theoretic techniques is to develop mathematical thinking and it would be too much of a detour from that goal in order to prove euclid's lemma but if you just google euclid's lemma you should be able to find some references that would lead you to a proof let me give you the existence proof more precisely let me give you a new proof of existence remember this is the statement of the theorem any natural number greater than one is either prime or can be expressed as a product of primes in a way that's unique except for their order and we're going to prove existence the previous proof of existence that i gave you use a method of mathematical induction and i gave it as an illustration of the method of induction but i'll give you a different proof now i'll prove it by contradiction suppose they were a composite number that's a non-prime that could not be written as a product of primes then there must be a smallest such number call it n since n is not prime there are numbers a b strictly between one and then such that n equals a b if a and b are primes then n equals a b is a prime decomposition of n and we have a contradiction because n was chosen so as not to have the prime decomposition it was actually the smallest number that didn't have a prime decomposition well if it's not the case at and be a primes then at least one of them must be composite but if either of a b is composite then because it's less than n it must be a product of primes so by replacing one or both of a b by its prime decomposition in n equals a b we get a prime decomposition of n and again we have a contradiction that proves existence that's the first part of the proof of the fundamental theorem of arithmetic now let me prove uniqueness so what i have to prove is that the prime decomposition of any natural number n greater than one is unique up to the ordering of the primes and i'll prove it by contradiction so assume there is a number n greater than one that has two or more in fact different prime decompositions let n be the smallest such number let n equal p1 times p2 times da da da times pr equals q1 times q2 times da da qs be two different prime decompositions of n so we have n as a product of r primes and we also have n as a product of s primes some of the primes being different well p1 is a prime factor of n so p1 divides n so p1 divides this product since p1 divides q1 times q2 through qs and let me stop at this moment and see what i've done i've said p1 divides n so p1 divides this and i'll split that into two that way and notice that p1 divides q1 times that and i've done that in order to apply eukary's lemma so now let me continue that sentence by euclid's lemma either p1 divides q1 or p1 divides the product q2 through qs remember euclid's lemma says that if i have a prime number p and if p divides a product a times b then p divides at least one of a b so euclid's dilemma was that if p divides a product a b it divides at least one of those two numbers so from here p1 divides this product i've now written that as a product of two numbers and i can now apply euclid's lemma and say that if p1 divides that times that number it must divide at least one of them either p one divides q one or it divides the other part or maybe both of them okay hence either p1 equals q1 or else p1 equals qi for some i between two and s okay if p1 divides q1 then they're both prime so the only possibility is that p1 equals q1 or else p1 divides this product of primes which means that p1 actually is one of those primes so first case i know that p1 equals q1 second case another p1 equals qi for one of the i's between two and s but then we can delete p1 and qi from the two decompositions in star which gives us a number smaller than n this is two different prime decompositions contrary to the choice of n is the smallest such okay having established that p1 is equal to one of these queues we don't know which one but it's going to be one of them i can delete the p1 from here i can delete the appropriate queue from there and when i delete p1 from there and some q from there i still have equality i still have a prime decomposition of a number but having deleted that common prime the number that i get is smaller than n so i get a number smaller than n which has two different prime decompositions the number smaller than n will be p2 times delta times pr though so that will be the number smaller than n and i've deleted a number from here and what's left is the prime decomposition here so the proof involves simply identifying the fact that we're recognizing the fact and the shooter's euclid's lemma that you're going to have the same prime on both sides you can delete it and get two different decompositions of a smaller number and that proves uniqueness okay well now we've proved the fundamental theorem of arithmetic cool huh so did that make sense chances are you're going to have to spend some time trying to come to grips with this though i know you're familiar with arithmetic especially whole number of arithmetic this is probably the first time you've tried to really analyze numbers see how you get on with assignment nine well as i stressed in the lecture you've got to be careful in distinguishing between these two notions you have a notation for a relationship something that's true or false a is divisible by b or b divides a and you've got another notation for a number the result of dividing a by b this makes sense when you're talking about the integers this actually is not defined for the integers i mean for some integers you get an answer but this is this i mean division actual division as an operation is an operation not on the integers it's an operation on the rational numbers or the real numbers um in in the integers all you have is addition subtraction and multiplication you don't have division what you can do is say whether you have divisibility and divisibility is divided defined in terms of multiplication okay um so it's all about the distinction between the two you've got a property something that's true or false and you've got a notation for a number this is an actual number okay so um the actual answer i'm going to end up with is the thing here at the bottom to the question as concisely and accurately as you can the relationship this is a relationship b divides into a or a is divisible by b if and only if a divided by b happens to be an integer and sort of summarizing what i just said above there that's a notation that denotes a rational number rational number that's a different set of numbers this denotes a relation b divides a i e there is an integer q such as a equals q b and whenever you're dealing with divisibility um with this notion and you're working with the integers you have to reduce that abbreviation to this there is an introduction to that that's what that means okay that means that that no thing means that okay and then to get down to here in the case where you do have divisibility then of course the queue that's here is the quotient i mean we use q to denote the word to stand for the word quotient anyway but notice that this says nothing about division division doesn't arise here it's all about the result of multiplying two numbers so this makes perfect sense when you're talking about the integers okay so we're not doing sort of arithmetic in the sense of calculating here you know obviously everything that's involved here about dividing one whole number by another you know you learned it in elementary school it is just division of whole numbers the focus here however is on what you're doing within certain systems of numbers we have two systems of numbers here we have the integers and we've got the rationals they're just two separate systems of numbers in the case of the integers you can add subtract and multiply in the case of the rationals you can add subtract multiply and divide but they're different systems of numbers and so the focus here is on what you can do in the integers that's what number theory is about and then in the very last lecture uh lecture 10 we'll be actually looking at the rationals and the reals but that's a different system of numbers you can do different things with it so we're taking a more sophisticated look at elementary arithmetic but it still is after all elementary arithmetic okay let's look at numbers two and three where the issue with all problems like this is you have to express the divisibility property in terms of multiplication remember divisibility is a is a property of pairs of integers you can't divide integers all you can do with integers is add subtract or multiply them and divisibility arises when you take this definition a divides b if and only if there is a q an integer such that b equals q times a so the general strategy for dealing with the divisibility actually is really the only strategy is you replace issues like this you replace a statement like that with a statement about multiplication because the point is that there is no operation here there's nothing that that's not an operation to deal with it's not an arithmetic operation on the integers remember this is all about the integers so you have to express it in the in the language of the integers and the language of the integers allows you to talk about addition subtraction and multiplication but not division okay so how do you assure that uh i mean how do you answer this one what's the proof well this one is is sort of immediate because the the very definition of divisibility explicitly excludes uh a not been in in zero a not not equal to zero it excludes air being zero okay um so it's false and that's the reason nine divides zero well that's definitely true and to show that it's true you simply express [Music] the definition express it in terms of the definition so you would have to show that there is a queue look at the definition you have to show there is a q so it's at zero equals q times nine well of course there is zero itself is one of those things okay so it's false it's true this one's false for the same reason a was false you're not allowed to to have a equal to zero in in the notion that includes that requirement this one is definitely true um and the proof is just write it in terms of this this is basically what you end up having to write in each case if you look at these that's what i'm going to end up writing i'm going to end up reformulating it in terms of the definition that's really all it involves just reformulate the statement in terms of the definition sure that the definition is true okay well in this case it's true because q equals one makes it true um in this case um we know that there's no such q i mean i mean you could argue it just by since uh any possible q would have to be less than less than seven say um you could actually explicitly if you wanted to prove that even more detail you would just let q be all of the possibilities that have a chance of being that q equals one two three four five six seven and seven sevens is already 49 so you you're out of it so you actually only need to go to six of course so you could explicitly list all of the possible queues if you wanted to but that would be so trivial i think you could just leave it like that you know at this level you know if this was if we were talking to kids in the elementary school we would ask them to probably list all of the possible cues and make sure that none of them give you the answer 44 but at this level you you you can just take that has been obvious um this one's certainly true you exhibit the q namely q equals and minus six q equals negative six uh ditto here you exhibit the queue and again it's uh it's negative seven um here you exhibit the q and the q is eight here you need to show that for all n one divides n well that's certainly true and the reason is that for any n in z n equals n times one right attribute right one divides everything four and in z fall n and n n divide zero that's true because again for any n and z zero equals zero times n and this one oh this this is what we've got to be careful with because if we're quantifying over all of the integers that includes zero itself and you're not allowed to have zero dividing anything okay that's excluded from the definition so this is the one you have to be careful with because it includes zero it's the only case where it goes wrong but you only need one counter example to make a uniform to make a universally quantified statement false and that one contour example is all it takes to get rid of that one okay so that one's false okay and we've done them all notice it was just the same pattern express divisibility in terms of the definition of divisibility and then in each case it just drops right out because this is after all just elementary whole number arithmetic you know it's not that there's anything deep going on here it's just that we're looking at it in a somewhat more sophisticated fashion than you did when you were in the elementary school everything you need to know to solve this you learned in elementary school it's just that we've now got a little bit more of a sophisticated gloss on it okay well as with the previous example what you have to do is reduce each of these to the definition of divisibility remember divisibility is a property of pairs of integers this isn't a division it's obviously related to division but you don't have division in the integers what you can do with the integers is you can add them multiply them and uh and subtract them i mean subtraction just being you know the inverse of addition um but you can't divide them okay but you kind of have a property of divisibility but to discuss divisibility within the integers you have to reduce it to to a discussion of essentially multiplication okay so how would you show that a divides zero that that you've got divisibility of zero by a well you observe that actually because of the properties of zero zero is equal to zero times a so in particular zero satisfies the requirement for divisibility okay there is a q in z such as zero equals q a namely q equals zero so by definition a divides zero okay similarly in the case of a dividing a and because of the properties of one a equals one times a so again the definition of divisibility is satisfied and it's satisfied in this case by letting q be equal to one so by definition a demands a okay so that was that one and the rest are essentially the same idea um a divides one if and only if a equals plus or minus one okay well we've got two implications to prove first of all let's assume that a equals plus or minus one then again all you have to do is show that there is something such that one equals q times a well if a plus or minus one that certainly is right conversely um ah that should be if little typo there conversely if a divides one then for some q one equals qa by definition of divisibility but if one equals q times a then the absolute value of one is the absolute value of qa which is the absolute value of q times the absolute value of a and if one equals that then the only possibility because these are positive integers now is that the absolute value of q is absolute value of a is one that's the only way you can get one and so if the uh if the absolute value of a equals one then a has to be plus or minus one uh let me just do one more and then let you to do all of the rest if a divides b and c divides d and a c demands b d okay well we know that there were q and r so it said b is qa definition of divisibility d equals rc definition of divisibility hence multiplying the two together you've got bds qa times rc which is when you rearrange them qr times ac so by definition ac divides into bd and the others are essentially the same idea in each case you just reduce it to the question of uh of multiplication through the definition of divisibility so you never actually do any dividing you express division in terms of multiplication and you can do that because division um is the inverse of multiplication okay so it the whole thing is going to work out so these proofs are always typically just one or two lines they really are just a matter of translating what it is you're having to prove into divisibility so the first line of any of these arguments really is just it's a matter of re-expressing what it is you're having to prove in terms of divisible in terms of multiplication via the definition of divisibility okay well that's it how did you get on with assignment nine there was a lot there so you might not have been able to finish it yet don't worry you can finish that later you won't need any of it for this final lecture where we're going to apply our mathematical thinking to studying the real numbers incidentally if you're not familiar with elementary set theory you should look at the special course reader on the subject before you proceed with this lecture numbers arose from the formalization of two different human cognitive conceptions counting and measurement based on fossil records anthropologists believe that both concepts existed and were used many thousands of years before numbers were introduced as early as 35 000 years ago humans put notches into bones and probably wooden sticks as well but none of those have survived or at least haven't survived and been found to record things possibly the cycles of the moon or the seasons and it seems probable they use sticks or lengths of vine to measure lengths numbers themselves however abstractions that stand for the number of notches on a bone or the length of a measuring device appear to have first appeared much later around 10 000 years ago in the case of counting collections these activities resulted in two different kinds of number the discrete counting numbers used for counting and the continuous real numbers used for measurements the connection between these two kinds of numbers was not finally put onto a firm footing until the 19th century with the construction of the modern real number system the reason it took so long is that the issues that had to be overcome were pretty subtle though the construction of the real numbers is beyond the scope of this course i can explain what some of the problems were the connection between the two conceptions of numbers was made by showing how starting with the integers it's possible to define first the rationals and then use the rationals to define the real numbers starting with the integers it's fairly straightforward to define the rational numbers a rational number is after all simply a ratio of two integers well it's actually not entirely trivial to construct the rational numbers from the integers you want to define a larger system the rationals that extends the integers by having a quotient a over b for every pair a b of integers where b is non-zero but how will you go about defining such a system in particular how would you respond to the question what is the quotient a over b you can't answer in terms of actual quotients since until the rationals have been defined you don't have quotients if you're interested you can find an account of the construction of the rationals from the integers in many books and on the internet but i'll mention once again that you should be cautious about mathematics found on the internet the point is constructing the rationals from the integers while it has some subtleties is fairly straightforward constructing the rails from the rationals however is a lot more difficult in in this final session i'll look at some of the issues involved in constructing and using the real numbers we'll start by looking at some properties of the rationals with the rationals you've got a system of numbers that's adequate for all real world measurements this is captured by the following property of rational numbers and i'll give it as a theorem if r and s are rationals with r less than s then there's a rational t such that r is less than t is less than s okay before i prove that let me give you a little note this property is called density the rational line is dense so if we have the rational line what density says is that whenever you have a rational number r and another rational number s we can find a rational number t in between r and s okay let's prove that well why don't we take t to be the mean of our and s well clearly r is less than t is less than s and the only question is is t irrational number well it obviously is but let's prove it let r be equal to m over n s equal to p over q where m n p and q are integers then t is equal to a half m over n plus p over q which equals mq plus np over 2 and q of course mq plus np and 2nq are integers so that proves that t is in q which was pretty obvious anyway but uh since we are focusing on how we prove things in mathematics at this stage i thought it was a good idea to actually go through a formal proof that one could give and density means that the rationals are good for doing measurements because what density tells us is that we can get rational numbers as close as we wish to any particular length you know if this is slightly too small if this is slightly smaller than a particular length and if this is slightly larger then we can always find another rational that's even closer on on the other hand density does not mean that there are no holes in the rational line let me just emphasize that the proof finished here so that was the end of the proof and with this remark we're on the edge of an abyss of understanding at least it's an abyss if you haven't seen this before this is uh there's some interesting stuff coming up uh it's stuff that wasn't fully worked out until the late 19th and early 20th centuries there's an example of a hole in the rational line there's root two and let me just make this a little bit more precise suppose i define a to be the set of all those rationals x which are either negative or non-positive or x squared is less than two and let me define b to be the set of all those rationals such that x is positive and x squared is greater than or equal to 2. let me draw a little diagram okay we'll have 0 here and a is a set that's going to go it's going to go out to the to infinity at the left and it's going to go to some point here that's going to be a let's delineate things here so a is going to be a set here b is going to be a set also it's going to go up to infinity and we're going to have every element of a to the left of every element of b and here is where root two would be if root two existed in the rational line this is the rational line q okay so we've got the rational numbers and i've split the rational numbers into two sets a which is everything negative and anything whose square is uh is strictly less than two and b is everything which is positive and whose square is greater than or equal to two and this splits the rational line into two pieces okay so clearly a union b is the rational line but and here's the kicker a has no greatest member and b has no smallest member now it was by considering situations like this that mathematicians in the late 19th and early 20th century were finally able to figure out a rigorous construction of the real number system a theory of real numbers but it took a couple of thousand years to get to that stage from the time when the ancient greeks discovered that the square root of two was irrational what this tells us is that the rationals are inadequate can't even spell inadequate now inadequate do mathematics why do i say that because in q we cannot solve the equation x squared minus two equals zero so the rationals are fine for measuring things that means they're fine for doing carpentry and and for doing various kinds of geometrical things and for building things and for tracking the stars and doing astronomy and so forth that's fine so long as we can get by with say the ten decimal places of accuracy or whatever we need but if we want an actual solution of equations then the rationals aren't good enough because even a simple a simple quadratic like x squared minus two equals zero cannot be solved the way out of this problem and this was the work that was done in the late 19th early 20th centuries was to say well we've got several different systems of numbers we've got the natural numbers then we've got the integers where we add negative numbers or we had negative versions of the natural numbers then we have the rationals where we have quotients of integers these are good for counting these are good for doing arithmetic when we want to have negative values like my bank account these are good if we want to measure things if we want to do mathematics we have to go one step further to the rails by the way this was in in many ways it was a historical progression uh not quite because z sort of got in in a different way but going from the natural numbers to the integers to the rational numbers to the real numbers that was largely a historical progression and it was done in order to have greater exp expressivity when when arithmetic began about 10 000 years ago it was exclusively the natural numbers and it was introduced to provide a monetary system in some area in in what was known as the fertile crescent region uh present-day iraq essentially and then as uh as bankers came along and they wanted to keep track of people's negative bank accounts and been a little bit flippant but that was pretty close to what happened uh negatives were introduced uh rationals were introduced to measure things in in the world and then the real numbers are required to do mathematics and to get the real numbers from the rational numbers what mathematicians in the late 19th century did was find a way to fill in the holes there were these holes in the rational line and the rails were constructed by filling in the holes okay fill in the holes in the rational line well i'm going to give you a few more details about how this was done but i should mention that there were some mind-blowing elements to this there were some really interesting surprises in store for the mathematicians that made this step one was the fact that however you count it there are more holes in the rational line than there are rational numbers now there are infinitely many rational numbers nevertheless the infinitude of the holes in the rational line vastly outweighs on an infinite scale the infinitude of the rational numbers themselves it was a super infinity of holes that meant that when those holes were filled in the number system that was obtained the real numbers contained incomparably more numbers than the rational numbers they're both infinite but mathematicians had to develop systems for counting infinite collections in order to cope with this and it turned out that the right that the real numbers as a set has infinitely many more numbers than the rational numbers in an infinite sense now making that precise is definitely beyond the scope of of this course but i will be able to get into the beginnings of the considerations that led to those surprising conclusions okay well that's all coming up within the next few minutes well before i give you any details of how the real numbers were constructed i need to do a little bit of preliminary work i need to introduce or reintroduce if you've seen it before the notion of intervals of the real line let a b be real numbers with a less than b the open interval a b written with parentheses this way is a set x in r such that a less than x less than b so a simple diagram would be this we've got a point a we've got a point b this is now the real line and the open interval a b is a set of all the numbers strictly bigger than a and strictly less than b but the the interval excludes these two so it's an interval of the line it's a segment of the line but it excludes the two end points the closed interval a b written with square brackets like this is the set x in r a less than or equal to x less than or equal to b so the diagram for this would be we've got a then we've got b this interval actually includes the two endpoints so the distinction between these two is that a and b are elements of the closed interval but a and b are not elements of the open interval now this may seem like splitting hairs after all this is a segment of the real line so there are infinitely many numbers in here in fact as i just indicated a moment ago on that previous page the infinitude of the real numbers in there is mind-bogglingly bigger than the infinity would have the rational numbers in there and the rational numbers in there is already an infinite set so we've got infinitely many numbers in in here infinitely many points in this line and yet i'm splitting hairs between these two points at the end that sounds like splitting hairs but take it from me this is a big big big difference or distinction the distinction between these two turns out to be huge and it's closely bound up with the reasons why the rationals are inadequate to do with mathematics and the rails are actually good for doing mathematics we're going to use this notation in order to talk about what had to be done to go from the rational numbers to the real numbers i'm not going to do the construction it's way too deep for a course like this it's first and second year university level mathematics and it takes most of us a long time to understand it but i will open the door to such a study there are some variations on the on the notation let me give you those i'm not doing any mathematics here i'm just giving you some notation okay one of them is what's known as a half open or half closed intervals so we have half open intervals or half closed they're the same thing really such as a b where that's closed and that's open square brackets parenthesis that's a set of all x in r such that a less than or equal to x less than b and the other way would be a b this way is a set of all x in r such that a less than x less than or equal to b and this one is called left closed right open and this one is called left open right closed so the word closed means you include the end point and open means you exclude the end point so if it's left closed the left endpoint is included if it's right open the right endpoint is excluded and one more bit of notation we sometimes include intervals that stretch all the way to infinity and we write things like negative infinity a is the set of all x in our well such that x is straight left in a it's everything to the left of a or i might write something like uh negative infinity a with a closed bracket here with a with a square bracket to denote a closed part of the interval that would be the set of all x in r such that x less than or equal to a um similarly i could have a going out to plus infinity would be the set of all x in r so it should x strictly greater than a or i could have closed a infinity which is a set of all x in r so should x greater than or equal to a and with one final remark i'm done with this summary of notation we don't have something together with infinity closed and we don't have infinity closed together with something you can't have infinity next to a square permanent to a square bracket because infinity is not a real number so there's no possibility of this guy whatever it is being an element of the interval notice that when i defined these there was no mention of infinity i just said all those x's to the left of a deter dit okay with this notation available we can now move ahead and look at how we go from the rational numbers to the real numbers we can look at what goes wrong with the rational numbers in a deep precise sense and i can give you some indication of how we would go about rectifying the the deficiencies of the rational line okay coming up next where the key property that the real numbers have that the rationals don't have is what's known as a completeness property in a nutshell the completeness property is what makes the real numbers great for doing mathematics and the absence of which makes the rationals inadequate for doing mathematics so it was a formulation of a system of numbers that satisfied the completeness property and indeed a formulation of the completeness property itself that was one of the crowning glories of late 19th century mathematics that set up the 20th century mathematics upon which most of modern science and technology depends so this is a big deal folks given the set a of reals a number b such that for all a and a a is less than or equal to b is said to be an upper bound of a we say b is at least upper bound of a if in addition for any upper bound c of a we have b less than or equal to c well duh what else would a least upper bound be it's the least one it's an upper bound and it's the least one why am i making such a big deal of this because it was by making a big deal of this that mathematicians were able to formulate this and it was by formulating this that mathematicians were able to construct the real number system which meant that after two thousand years of effort mathematicians finally had a system of numbers that was adequate for doing modern science physics technology et cetera et cetera et cetera in mathematics it's often the case that something that seems trivial turns out to be fundamental and and have enormous consequences and this is one of those moments okay so i've been very precise about the definition because mathematicians found that it was only about being precise that they were able to figure out how to proceed the notation we use for least upper bounds is this least upper bound of a so the notation for the least upper bound of a set here is lub a let me make a note we can make the same definitions for n for z and for q the completeness property of the real number system says that every non-empty set of rails that has an upper bound has at least upper bound and let me stress that that least upper bound is in the set of real numbers this one simple elegant statement is the key to the real number system and to most of modern mathematics before i go any further you really ought to take a look at assignment 10.1 i see you should at least look at it ideally you should try to do it or at least try to do as many of the exercises as you can you need to be familiar with the background material about upper bounds and least upper bounds before we progress we're about to meet material that most beginners beginners at university-level mathematics find incredibly difficult you know to go back to my favorite example about riding a bike it's one of those transitions in life where it seems impossibly difficult until you get the hang of it and then it suddenly seems remarkably straightforward and you wonder why it took you so long so it's not that we're facing something terribly difficult it's just that we need to go through some kind of a shift going from thinking it's impossible to thinking well okay that seems straightforward it's one of those difficult transitions that when you look back actually with retrospect don't seem too difficult at all we're facing one of those and i strongly recommend that you take a look at that assignment before you go any further otherwise you might very rapidly find yourself lost in the next few minutes of lecture okay um having said that i'm going to take a pause now and then i'll come back in the second video connected with lecture 10. and i'll see you then welcome back as i say in the media where the correct answers number two and number three five is not the maximum element of this interval in fact five is not an element of this interval at all this remember is a set of all x in the rails such that zero strictly less than x strictly less than five so five itself is not a member of there so not only is it not the maximum element it's not an element at all seven is actually not a member of this interval for the same reason as five wasn't an element of this interval but nevertheless seven is still the least upper bound there is no smaller upper bound of that interval than seven so it is a at least upper bound so at least upper bounds are not the same as maximums and in this case zero is a member of that set that interval is defined remember as a set of rails so it's a zero less than or equal to x less than or equal to one and in this case the end point zero and one are elements of the interval so zero is in there and it's clearly the minimum element of that interval where the answer is the first one is correct this is what it means to say that the rational line is dense between any two rationals you can find a third one the second one is actually true but it doesn't express density it doesn't express density because whether or not there was an irrational number between two rationals is sort of irrelevant uh it's it's with the question is about the rational line being dense uh this is actually making a statement about the real line so it's true but actually irrelevant to the notion of density of the rational line so that one's not true i mean that one's true but it's not the answer to the question and this one um that's actually expressing the notion of completeness now if by at least upper bound we mean least upper bound in the rationals in the rational line q then that would say that the rational line is complete which is false if however we interpreted this to mean every set of rationals that is bounded above has at least upper bound in the real numbers then that would be an instance of the completeness of the real line and this issue about the existence of at least upper bounds is what distinguishes the rails from the rationals and it's what makes the rails a very powerful system for doing uh advanced mathematics and calculus in particular and and demonstrating the the fact that the rationals is not complete is is what demonstrates the the the impoverished nature of the rationals in terms of mathematics and doing things like calculus okay how did you get on what i want to do now is uh well actually really what i want to do is introduce the beginnings of the subject known as real analysis now this isn't real analysis as opposed to fake analysis real here is is essentially short for real numbers or for the real number system if you like it's the analysis of the real numbers and i'm going to begin with a theorem the rational line is not complete now if you've done that assignment assignment 10.1 that i asked you to do you should be familiar with what that means but let me remind you in case you decided to play play risky and go ahead without doing that assignment well let me just remind you that completeness means if a subset of rails has an upper bound then it has at least upper bound in the set of rails that is that was completeness as applied to rails but as i mentioned at the time these notions also apply to any set so in terms of the rationals completeness would mean if a is a set of rationals having an upper bound then it has a least upper bound in the rationals what this theorem says is that this property does not hold for the rational numbers remember r stands for real numbers here but if i replaced r by q and talked about the rational numbers then this property would not hold it does however hold for the real numbers that is the completeness property for the real line we won't be able to prove that but i'll be able to indicate how it's possible to construct the rails in order to make it possible to prove that okay here's the proof of the theorem let a be the set of all rationals r such that r is non-negative and r squared is less than two now already you can probably sense what's going on this is going to hinge around that property that the square root of two is irrational so let me draw a picture here's zero here's two a is going to be a set well everything is going to be greater than or equal to zero and a is going to go up to some point less than two well a only contains rationals less than two whose square is less than two so those rationals themselves are less than two so it's going to be something like this and we all know that sis lurking in there somewhere is the square root of two i should stress that throughout this argument the argument i'm about i'm about to give we're talking purely about the rationals so i'm not going to be talking about any reals this is why i sort of put this down here somewhat faintly this is to help guide our intuition this is to sort of motivate what's going to go on but the entire argument i give is going to be in terms of rational numbers not real numbers i deliberately did not write our less than the square root of 2 because there is no such thing as a square root of 2 in the rationals i'm using sets of rationals in this argument it's an argument about the rational numbers not about the real numbers okay well a is bounded above for example two is an upper bound we only need to find one and two will do just fine i will show that a has no least upper bound that would mean that a is a set of rationals which has an upper bound but nor least upper bound and hence the rational line is not complete because completeness would say that any set of rationals with an upper bound in the rationals has at least upper bound in the rationals well how would i show that there's no least upper bound well let x and q be any upper bound of a and sure there's a smaller one again let me stress smaller one in the rationals remember we use the letter q to denote rationals because q stands for quotient and rational numbers are numbers of the quotients of integers we can't use the letter r for rational because r is used for real numbers unfortunate i know but there we are well since we're talking about the rationals that upper bound x is going to be of the form p over q where p and q are introduced in fact they can be natural numbers because this set a is is positive individual he's non-negative number is it this set is non-negative rationals it's everything's to the right of the origin so everything's positive so i don't have to worry about negative numbers here so these two integers can be chosen positive and i want to show that there's a smaller upper bound well let's suppose x squared is less than two it's either less than two or it's greater than or equal to two it's one of the two let's just see what happens if x is less than two in that case looking at this equation two q squared is bigger than p squared now as n gets larger n squared divided by two n plus one increases without bound so we can pick an n n so large that n squared over 2n plus 1 is bigger than p squared divided by 2q squared minus p squared now you might not see where i'm going with this but hopefully you can believe everything i've said okay we're assuming x squared is less than two we'll actually in a moment we'll arrive at a contradiction so the conclusion i'm going to get out of this is that x squared is in fact not less than 2 but this is where we're starting if x squared is less than 2 then because of that definition 2q squared is greater than p squared okay so 2q squared minus p squared is positive that means this number is a positive number and what i'm saying is that because we've got an n squared here and a linear term involving n here the bigger n gets this gets increasingly large it gets as large as you want it to be so i can pick an end big enough so that this number is bigger than that one and if you rearrange that you'll find that 2 n squared q squared is bigger than n plus 1 squared p squared okay i'll leave you to do the algebra for getting from there to there hence n plus 1 over n squared times p squared over q squared is less than two just rearranging that taking those terms to the other side now we'll let y be n plus 1 over n times p over q notice that y is a rational number it's a quotient of integers and y squared is less than two because this says that y squared is less than 2. by the way this by now you should have begin to smell why i i i i started looking at this term i was trying to get this number y remember i started with an x as an upper bound and i wanted to show that there's a smaller one and i'm going to work towards that um and now i've got to i've introduced this y okay so y is in q and y squared less than 2. so y is an element of that set a but wait a minute y is equal to a number slightly bigger than one times x so that means that y is actually bigger than x so the number y that i've constructed is in the set a and yet is bigger than x well that's a contradiction since x here's an upper bound of a that supposition must be false so x squared has to be greater than or equal to two okay so what i've done is i've taken an upper bound of a i'm going to show there's a smaller one and as a first step towards doing that i've shown by contradiction that that upper bound has to have its square greater than or equal to two now i'm going to go ahead using this extra information to show that there's a smaller upper bound and hence there's no chance of any x being at least up about let me recap where we've got to we let a be the set of all rationals that are non-zero and for which r squared is less than two let x be an upper bound of a and we had x in the form p over q where p and q are integers okay so we we have that and our goal is to show that a has an upper bound smaller than x hence there cannot be a least upper bound which would show that the rationals are not complete and we just showed that x squared is greater than or equal to 2. hence since the square root of 2 is irrational x squared is strictly bigger than 2. x is irrational x squared can't be equal to two so it's strictly greater than two thus since x equals p over q p squared is bigger than two q squared i'm going to use this fact to find an upper bound of a smaller than x to do that i'm going to pick n an integer so large that the following is true n squared divided by 2n plus 1 is bigger than 2q squared over p squared minus 2q squared i e rearranging that p squared n squared greater than two q squared times m plus one squared so you just rearrange this to a little bit of algebra and you get this ie p squared over q squared times n over n plus one squared is greater than two again you just rearrange that and do a little bit of algebra to get that let y be n over n plus 1 times p over q then y is an element of q y is a rational number it's a quotient of integers so it's in q and moreover y squared is bigger than 2. moreover since n divided by n plus 1 is less than 1 y is less than x because p over q is x and y is just this guy times x so it sits less than x but for any a and a a squared is less than 2 is less than y squared so a is less than y hence y is an upper bound of a which is smaller than x thus a does not have a least upper bound and this proves the theorem i guess my mathematics is better than my handwriting this proves the theorem final remark the construction of r from q can be done in several different ways but in all cases the aim is to prevent an argument like the above going through for r and with that you're the very gateway to modern real analysis for our final topic in this course i'd like to say a little bit about real number sequences these are connected with one of the ways of constructing the real numbers from the rationals and they also give us a a technique or a concept for doing an awful lot of work in real analysis to put in another way sequences of real numbers are a big deal in modern real analysis which means they're a big deal in calculus and anything that's a big deal in calculus is a big deal in science and engineering and technology so whichever way you cut it sequences are a big deal well what is a sequence well in everyday terms it's a list one two three let's put some commas in here of numbers so we have a number a number a number going on to infinity the way we normally express this and try to capture this uh this infinite extent here is by writing it a n where n goes from 1 to infinity and this is what's called an infinite sequence if you look in textbooks you'll find a more formal definition of the sequence is a function from the set of natural numbers into the real numbers but for the purposes of one of the kind of things i want to talk about here it's enough to think of it simply as an infinite list of real numbers for example the sequence of natural numbers 1 2 3 and so forth that's an infinite sequence in terms of our notation i would just write that as n for n goes from one to infinity or i could have um the sequence that consists simply of an infinite sequence of sevens seven going on forever that would be expressed in this way or i could have the following sequence 3 1 4 1 5 9. etc enumerating the decimal digits of pi well there's no simple formula like this to capture this one i have to use some expression like this well let me give you another example i could have the sequence consisting of negative 1 to the n plus 1 from n equals 1 to infinity that's the sequence that consists of plus 1 negative 1 plus 1 negative 1 plus 1 negative 1 etc that's an example of what's known as an alternating sequence meaning that the sign alternates as you go through the sequence okay so that's what sequences are just infinite lists of numbers now let's look at the following example look at the sequence consisting of the numbers one over n from n equals one to infinity okay that's the sequence one a half a third a quarter and so on and the thing to notice about this is that the numbers get closer and closer to zero in fact they get arbitrarily close to zero or this one one plus one over two to the n from n equals one to infinity that consists of the numbers one and a half one and a quarter one and an eighth one and the 16th etc and these numbers get arbitrarily close to 1 and going back to this example here the sequence 3 3.1 3.14 3.141 3.141 3.14159 3.141592 [Music] 3.1415926 [Music] that sequence gets arbitrarily close to pi that's as far as i know the decimal expansion by the way so these sequences have the property that as you go along them they get arbitrarily close to a fixed number zero in the first case one in the second case and pi in the third case and there's a there's a general property here that we're going to capture in the by way of a definition if the numbers in the sequence n n from one to infinity gets arbitrarily closer to some fixed number a we say that sequence tends to the limit a and write n arrows a as n hours infinity an alternative notation is we sometimes write it this way limit as n goes to infinity of a sub n equals a not all sequences tend to a limit look at this one for example this alternating sequence plus one negative one plus one negative one that doesn't approach any particular number it bounces back and forth between plus one and negative one this one in a trivial sense doesn't tend to a limit this one tends to the limit seven it doesn't just tend to win it never gets away from it this one doesn't tend to eliminate at all these numbers get bigger and bigger and bigger we would sometimes say that the sequence tends to infinity but that's beyond the scope of the limited amount i want to talk about sequences in in this course the point is some sequences don't tend to limit other sequences do tend to limit okay let's move on well so far everything's been very intuitive let's get a little bit more formal now i've got a sequence a n and from one to infinity the intuitive explanation or the intuitive description i gave of this that the sequence or the members of the sequence tend to a limit as n tends to infinity that sort of corresponds fairly roughly to the fact that the absolute value of a n minus a becomes arbitrarily close to zero boy my writing really is a problem isn't it close now let me give you the formal definition and the formal definition involves being precise about what that means a n tends to a as n tends to infinity if and only if for all real numbers epsilon greater than zero there is a natural number n such that for all m greater than or equal to n absolute value of am minus a is less than epsilon and now perhaps for the first time you see why we spent so much effort understanding quantifiers and being particular about the order in which quantifiers appeared this definition is absolutely crucial in real analysis it's absolutely crucial to our definition and our understanding of the real numbers and that means it's absolutely crucial to calculus and hence it's absolutely crucial to physics science engineering technology etc etc etc this is the real stuff folks okay let's try and understand how that captures this intuition the geometric intuition we have i think it's fairly clear that you're going along a sequence and the numbers get closer and closer and closer to some fixed number how does this somewhat complicated looking expression capture that in a precise formal way well let's peel away this part for a minute by the way it's it's traditional in mathematics to use epsilon in this context and at the back of your mind you should think that epsilon is not just positive but it's very small and positive like a tenth or a hundredth or a millionth or a zillionth or whatever um as we'll see when i as you'll see when i go through this it's it's when epsilon becomes small but remains positive that this thing kicks into power okay so let's forget that a bit let's look at the other part exists in n such a for all m greater than or equal to n absolute value a minus a is less than epsilon and i've avoided mentioning explicitly the set n now notice that in the first place i didn't mention that m was also a natural number because the context makes that clear n is natural number and we're talking about all m's greater than or equal to n and when we discussed quantifies earlier on i i pointed out that mathematicians do this kind of simplification all the time so it makes it easy to read that's all okay what does this part mean so we've been given an epsilon we'll assume we've given some positive number epsilon and at the back of your mind think of it as a small positive number what this means is that from some point onwards so from some point n onwards all the members of the sequence are within a distance epsilon of a so all the numbers in the sequence a n n equals one to infinity are within a distance of epsilon from a you still with me now let's see what this thing means when we when we have the whole expression this says for any epsilon greater than zero this thing holds so for any epsilon zero there is a point such that from that point onwards all the numbers are within a distance epsilon for me as i mentioned a moment ago the intuition here is that we can take epsilon greater than zero as small as we want so let me draw a picture now we've got some number a and there's a sequence bouncing around here somewhere you know maybe it's coming in from the left from the right or maybe it's bouncing around but there were these numbers a1 a2 scattered along this line here now we're given an epsilon greater than 0. so let's swift was it's down here so here's let's say negative here's a minus epsilon and here is a plus epsilon okay i've just got a distance epsilon to the left and a distance epsilon to the right and the arguments i'm about to give will hold for any epsilon at least that's what this says so i'm given an epsilon and i look at this interval and what the rest of this formula says is that from some point on all the elements of the sequence are in here if i take a smaller epsilon let's do it say here and here let's call it epsilon prime so that would be a minus x one prime and here would be a plus epsilon prime so i'll take a smaller epsilon still positive so the formula will hold for the epsilon prime there will be some point from which all the elements of the sequence beyond that point are within this region and then i could take an even smaller one and an even smaller one notice that the n depends on the epsilon remember that example about the american melanoma foundation wasn't a problem for them getting the quantifiers the wrong way around if we cut the quantifiers the wrong way around here we'd be in big trouble this wouldn't work the point is given an epsilon you could find an n for each epsilon you may have to go further out in the sequence until you come to a point where all the elements are within that distance but what this says is that you can always by going sufficiently far out in the sequence reach a stage where all of the numbers in the sequence from then onwards are within a given distance of a and you can do that for smaller and smaller and smaller a and that's capturing this intuition and it does it beautifully and elegantly and with enormous power okay let me give you a couple of examples first example let's look at the sequence 1 over n n goes from 1 to infinity now we know that 1 over n turns to 0 as n tends to infinity let's prove this rigorously in terms of the definition of limits that i just gave okay so i'm going to prove that fact what i have to show is that for all epsilon greater than zero all real numbers epsilon greater than zero there's an n in the natural numbers such that for all natural numbers m greater than or equal to n absolute value one over m minus zero is less than epsilon well let's simplify that for all real numbers x longer than zero there is an n and let me just drop that explicit mention of the natural numbers for all m greater than or equal to n that just says absolute value of 1 over m less than epsilon which actually just means 1 over m less than epsilon because m is a positive integer okay so i have to verify this in order to prove that okay so how do i verify that well let epsilon greater than zero be given another way of saying it would be to say let's epsilon greater than zero will be arbitrary different way of saying the same thing essentially at least in this circumstance what i need to do is find an n such that for all m greater than or equal to n one over m is less than epsilon well pick any n such that n is bigger than one over epsilon epsilon if epsilon's a very small number this would be a very large number so i might have to pick a large head but pick such an n now intuitively you know that since the natural numbers go on forever as it were getting bigger and bigger that's always possible um that actually uses a principle of mathematics it archimedean property so named after archimedes see assignments 10.2 for a little bit more about that so pick an n big enough so that n greater than one over epsilon then if m greater than or equal to n one over m is less than or equal to 1 over n is less than epsilon and we're done notice that the choice of n depending on epsilon given an epsilon i picked my n in terms of the epsilon please note american melanoma foundation at least when you apply quantifiers in mathematics it's a big deal if you get them in the wrong order the n depends upon the epsilon the smaller the epsilon is the bigger the n has to be intuitively for this sequence the smaller the tolerance you impose on the number being close to zero the further out in the sequence you have to go the bigger the n is before you're within that tolerance so quantify all the matters the choice of n depended on epsilon different epsilon different n okay let me give you one more example and let me look at this sequence n over n plus 1 from n equals 1 to infinity that's the sequence a half two-thirds three-quarters four-fifths and so on and let me prove in terms of the definition that n over n plus one tends to one as n tends to infinity now intuitively it obviously does these numbers get closer and closer and closer to one but to prove that what i have to show is the following for all epsilon greater than zero there is a natural number n such that for all natural numbers m greater than or equal to n absolute value of m over m plus one minus one is less than epsilon okay i'm going to prove it the same way as before i'm going to assume i'm given an epsilon then i'm going to find an n dependent on epsilon that makes this thing true let epsilon greater than zero will be given we need to find an n such that for all m greater than or equal to n m over m plus 1 minus 1 in absolute value is less than epsilon well i'm going to pick n so large that n is bigger than one over epsilon that's actually the same choices in the previous example it just works out that way because i'm using very simple examples in more complicated examples the choice is perhaps not quite so simple as straightforward as this okay but let's see why that one works in this case then for any m greater than or equal to n i've got the following m over m plus 1 minus 1 in absolute value equals minus 1 over m plus 1 in absolute value just work that out as a single fraction which is just 1 over n plus 1 which is less than 1 over m which is less than or equal to one over n because m is greater than or equal to n which is less than epsilon and we're done okay you should now be in a position to attempt the questions on assignment 10.2 bear in mind that the two examples i've given were particularly simple ones where the choice of the n was very simple you won't always be so lucky you might have to work a little bit harder to find the n but my focus here as it's been throughout the course wasn't so much on the details and on the procedures and if you graduated from high school i assume you can do mathematical procedures you can do the algebra the focus is on the reasoning in particular in the context of sequences the key thing is how given an epsilon you have to choose an n that depends upon it it's this quantifier switch that's really crucial for all epsilon there is an n if you go on to further studies in in mathematics and particularly real analysis you're going to see variance of this kind of definition not just for the limits of sequences but for the definition of continuity and uh and then when you come to look at the formal definitions of integration and so forth so you're going to come across this kind of a definition a lot uh coming up with this formulation was one of the jewels in the crown of late 19th century mathematics absolutely brilliant piece of work and it gave rise to pretty well all of modern mathematics at least all of modern mathematics is to do with with real numbers okay well good luck with assignment 10.2 well you made it all you have to do now is complete the final assignment and you should be well set for taking college level mathematics courses or for simply reasoning more effectively in your job or in your everyday life since you got this far i'm sure you'll complete these last steps so let me offer you my congratulations now these massive online courses are still hugely experimental so simply sticking with us and finishing the course is an accomplishment all by itself future generations of students will surely have it easier in some ways what they won't find easier however is the material itself learning facts is relatively easy learning to think a whole new way is extremely difficult however well you think you've mastered the material you should feel pleased with completing the entire process under your own motivation as a lifelong teacher much of my motivation and reward has been getting to know each generation of students and working with them helping them overcome difficulties and then seeing them succeed with a mooc that regular close contact's absent and i miss it on the other hand being able to reach tens of thousands of students as opposed to 25 brings its own reward to complete the basic course all you have to do now is work on assignment 10 and submit problem set 8. if you are doing the extended course the next two weeks are devoted to the test flight process see the website for details okay well we're at the end of the course now so in keeping with my basic philosophy behind the course i'm only going to answer one or two of the questions from from assignment 10. since there's no more lectures to come no more material to come in this course there's nothing that's going to depend upon on anything now you'll only need this stuff for future courses or for using it in the real world so i'm going to leave most of the questions on assignment 10 for you to do in your own time and to uh to resolve them with colleagues and so forth um but i'll just do one or two uh just to sort of set the ball warning and uh and show a good faith if you like okay um so this one um i'll do assignment 10.1 number one the intersection of two intervals is again an interval it's sort of easy if you think in terms of diagrams i mean what are the possibilities you could have two intervals they could sort of overlap or one could be inside one of another they could overlap the other way around um they could be disjoint completely separate uh and i guess this isn't it they could have overlapped one word they could overlap another way uh one could be completely inside another one or they could be completely disjoint um but let's just do it in a symbolic fashion let's just let's uh be well i'll take the case for open intervals an analogous argument will work for closed intervals so let's say they be a b and let c be the interval c d then by definition a intersection c is the set of all x such that a is less than x is less than b intersected with the set of all x this is x in the reals such that um c is less than x is less than d already and because of the way conjunction works that's the set of x such that the maximum of a and c is less than x is less than the minimum of uh of b and d okay i think i'll use the right letters there yeah that's correct okay which is an interval um it's in fact the interval open interval maximum of a and c to the minimum of b and d and that's an interval it may be empty you know if it's disjoint it's it's empty uh the empty set is an interval it's the uh it's still a set of uh numbers between two points okay so um that i think is uh we'll deal with that one and uh let me just say similarly for closed intervals and for half half open intervals half open half closed okay so um that's really it and in the case of uh unions it's false for unions for example if i take the open interval zero one and i form the union with the open interval three four then that's not an interval okay and that's that one okay which one shall i do next let me oh do shall i do next um [Music] oh let's do question 5 in 10.1 as a question five asks us to to um to verify this alternative definition of least upper bound okay so let's see first of all a just says that uh b is an upper bound okay so let me write that a says b is an upper bound so the issue is does condition b say that it's at least one okay well let's see what's what that amounts to b is at least upper bound if and only if no c less than b is an upper bound okay that's what the word least means right there isn't a smaller one so that's the original concept there's nothing small just a little a little bound and upper bound okay well that's true if and only if for any c less than b c is not an upper bound that's just another way of saying the same thing if and only if for any c less than b if it's not an upper bound that means there is an a in a such that it's not the case that a is less than or equal to c okay you can find an a for which c is not bigger than it is not an upper bound if and only if and i'm really laying this one on with incredible detail over here for any c less than b there is an a in a such that a if it's not less than or equal to c is bigger than c now arguably i could have deleted one of these lines i'm really been very pedantic about writing everything down but in my experience both when i was a student and from teaching this kind of material for many years even though these are very simple ideas uh it's very confusing at first dealing with the different variations of definitions of greatest low bounds least upper bounds there's something about the way the human brain works that even though we recognize on one level that this has got to be really really trivial uh it causes problems um so it's just something to do with the way the brain works that makes this a challenge at least for most of us okay well that's that one um let's see what we're gonna do next well i think i'll do just one more and i'll pick one of those limit questions in number four okay this is from assignment 10.2 and i'll pick this one show that the limit as of n over n plus 1 all squared tends to 1 as n goes to infinity and these all follow a standard pattern you start by with a given epsilon greater than zero and what we need to do is we need to find an integer n such that for any natural number n bigger than or equal to n it is the case that the absolute value of n over n plus one all squared minus one is less than epsilon from some point on in the sequence the difference between the the that term and one is less than epsilon okay so now let's just pull this apart and see what this really wants us to find we need to find an integer n such that n greater than or equal to n implies let me write everything over n plus 1 all squared and i've got n squared minus n squared minus two n minus one that is less than epsilon i e i need to find an n such says n bigger than or equal to n implies the n squares disappear i've got a 2n plus 1 over n plus 1 squared less than epsilon okay so the n squared's disappear i'm left with minus two n minus one over n plus one squared in an absolute value and taking the absolute value and so when i do take the absolute value the minus two n minus one just becomes two n plus one and the n plus one squared is positive of course so that just reduces to that okay now um we can see what to do what we need to do remember we're looking for a big n so let's pick n so big that n plus one squared over two n plus one is bigger than one over epsilon we can always do that epsilon is given it's a it's a small number we'll assume i mean these really they are we use a symbol epsilon to sort of emphasize the fact that uh that in reality epsilon's very small that's not a logical restriction it's just our intuitions behind what the proof's doing okay so this is going to be a very large number but in this term in in this this quotient the numerator is squared and the denominator is linear in n so eventually the numerator dominates the denominator so we can make this as big as we want and we can certainly make it bigger than one over epsilon and when we do that we find of course that if n is big or equal to n then 2n plus 1 over n plus 1 squared is certainly less than or equal to big 2 big n plus 1 over big n plus 1 squared which is less than epsilon because we picked big end to make that happen okay and that's it okay we sort of we just worked backwards we looked at the the goal and we worked backwards and and indeed we can we can always find an n with this property because the numerator grows faster than the denominator without without bound squared over over linear and so we get this expression this is very typical of of limit verifications uh i wouldn't say they were all like this but the vast majority of them follow this pattern um you work backwards and then you just uh ask yourself where you have to look how big you have to how far you have to go out in order to make something happen and you you very often end up saying well pick something so big that it's bigger than one over epsilon and very very common uh for limits that involve quotients okay well um that's um i think that's all i'm going to do for as i said i'm not going to do many on question on assignment 10. i'll leave it with those uh there are plenty more to to work with and i hope you'll enjoy uh working through those i remember when i let this stuff for the first time as a student uh i actually enjoyed it i thought it was a lot of fun first of all it's dealing with infinity in a vigorous way and that's that's pretty cool anyway right infinity dealing with infinity you know holding infinity in the palm of your hand that's that's really pretty cool um and uh and it's just an intellectual challenge it's a fun game to play okay so uh enjoy it uh have fun with these things and i'll uh i hope you do okay in the in the final exam okay bye for now you