Understanding Rational Numbers Operations

Hello everyone. In the previous lecture we did a sort of a review survey of fractions, mixed numbers, rational numbers, and decimals, and we also talked briefly about irrational numbers. real numbers. Now we want to start looking at how to do the four basic arithmetic operations with rational numbers and of course the four basic arithmetic operations are addition, subtraction, multiplication, and division. So today in Lecture 5, which covers, we're still looking at material in Section 1.3 in the book, today we look at multiplication and division of rational numbers. So to discuss multiplication and division of rational numbers, in the first part of the lecture we are going to review and survey how multiplication and division is done for fractions, for ordinary fractions. how multiplication divisional fractions is taught to children. So the first bullet, it says, we have three familiar interpretations of fraction multiplication. I don't know if this is focused. Let me try to focus it. Okay, I think that's a little bit better. Okay, so it says we have three familiar interpretations of fraction multiplication. Number one, multiplication of a fraction by a positive integer n can be thought of as repeated addition of the fraction. where the fraction is repeated n times. So then that's an interpretation where a fraction is multiplied by a positive integer n. Number two, it says multiplication of a positive integer n by a fraction, such as 2 fifths, be thought of as two-fifths of an identical object. And number three, it says multiplication of a fraction such as two-thirds by another fraction such as four-fifths can be thought of as two-thirds of four-fifths of one whole object. So these three interpretations are taught to children in elementary school and we will now review them by looking at specific examples. So if you look at example 1 and 2 together, example 1, it says compute 2 fifths times 3, which is the same as 2 fifths times 3 over 1. So remember, the integer 3 is the same as the rational number 3 over 1. We saw that in the previous lecture. So we want to compute this multiplication, which is the same as this multiplication, by interpreting this multiplication as a repeated addition. as the fraction 2 fifths repeated 3 times, whereas over here, 3 times 2 fifths, which is the same as 3 over 1 times 2 fifths, we'll interpret this multiplication as 2 fifths of 3 identical rectangular objects that are given here. Now, we could have interpreted this one as 2 fifths of 3 objects, and we could have interpreted this one as 2 fifths repeated 3 times. That would have been fine as well, but we just need to be consistent with how we interpret things from the beginning. So the way I have stated the interpretations here, I'm just being consistent with that. So now let's try example 1. So this multiplication, I'll write that first here. We have 2 over 5. times 3 over 1 that is the same as 2 over 5 times 3 so our interpretation is that this is repeated addition so the fraction 2 over 5 is repeated three times Okay, so that's interpreting the multiplication as repeated addition. Now, can we interpret this repeated addition? Yes, we can, because remember, what is the interpretation of the fraction 2 over 5? It means you take an object, cut it up into 5 equal parts, and then... take two of those parts. So here you have two parts out of five equal parts. You're adding to that two parts out of five equal parts. So overall here you have four parts out of five equal parts and then you're adding two more parts out of five equal parts. So overall you have six parts. out of 5 equal parts and can we write 6 parts out of 5 equal parts as a fraction? Yes, it's written like that. So this, with our usual interpretations of fractions, this addition expression is equal to that fraction. So there we go, we have managed Do this multiplication by interpreting it as repeated addition. Now how about example two? So this time we want to interpret that as two-fifths of three identical objects. So I want to somehow take two-fifths of these three objects. So I can be clever about this. I can say to myself, okay, I'll first focus on the first object and take two-fifths of that. Then I'll take two-fifths of the middle object. Then I'll take two-fifths of the third object. and then I can combine all to get 2 fifths of all three objects. So for the first object we saw this same object in the previous lecture it's roughly 5 centimeters long so we will look at 1 centimeter intervals and now use those to divide up the object into five equal parts Okay, so I've divided the first of the three objects into five equal parts, and I'll take two-fifths of that object. So I can take any two of these five equal pieces. So just to keep things simple, we'll take the first piece and the second piece. Now I will cut up the middle object into five equal parts. Okay, and now take any two of these five equal parts, doesn't matter which two, so let's take this one and this one and finally cut up the third object into five equal parts. Okay, so, and now take two of those five equal parts. So now I have 2 fifths of this first object, 2 fifths of the middle object, and 2 fifths of the third object. So if I take 2 fifths of all three objects, how many of these 5 equal parts do I get in total? I get 1, 2, 3, 4, 5, 6 out of 5 equal parts. So this multiplication, 3 over 1 times... 2 over 5, so again since 3 over 1 is the same as 3, this multiplication is the same as 3 times 2 over 5, that is 2 fifths of these three identical objects, and 2 fifths of all three objects gives 1, 2, 3, 4, 5, 6 fifths. And that's example two. And notice something interesting. So this multiplication was interpreted as repeated addition, whereas this multiplication was interpreted as two-fifths of three objects. but with these two different interpretations the numerical answer is still the same the numerical answer to the multiplication is still 6 over 5 in both cases and what is the relationship the 6 and the 5 to the numerators and denominators of the fractions that we multiplied. Well notice that 2 times 3 gives you the 6 here, here the 3 times 2 gives you the 6, the 5 times 1 gives you the 5, the 1 times 5 gives you the 5. So that's an interesting observation isn't it? It's as though when we multiplied these fractions we separately multiplied the numerators to get the numerator of the answer and we multiplied the denominators to get the to get the denominator of the answer. So that's an interesting observation. But of course, all this meant something. In example one, the multiplication meant repeated addition. And in example two, the multiplication meant two-fifths of three objects. Okay, so now we have dealt with multiplication of a fraction by a positive integer from either direction. And here the positive integer three was on the... On the right, here the positive integer 3 was on the left. Now, going into page 2, we are going to look at the case where we multiply a fraction by another fraction. So, in example 3, we will compute this multiplication, 2 thirds times 4 fifths. And in example 4, we will look at it in the reverse order. We'll have 4 fifths times 2 thirds. And how are we going to interpret these multiplications? Well, the same interpretations that children are taught. So here in example 3, it says compute 2 thirds times 4 fifths by interpreting this multiplication as 2 thirds of 4 fifths of this rectangular object. In example 4, we will compute 4 fifths times 2 thirds by interpreting this multiplication as 2 thirds of 4 fifths of this rectangular object. Interpreting this multiplication is 4 fifths of 2 thirds of the same rectangular object, so we'll do it both We'll do both and then see What we get and how they compare? Okay, so this first one, we interpret 2 3rds times 4 5ths as 2 3rds of 4 5ths of the rectangular object. So what I'm going to do is I'm going to figure out first 4 5ths. of the rectangular object. So four fifths means you divide the object into five equal parts and take four of them. So let's see how long is this object. If I use my ruler and centimeters, this object is roughly 15 centimeters long, roughly. So to divide it into fifths, into five equal parts, each part will have length 3 centimeters, right? Because 15 centimeters divided by 5 is 3 centimeters. So I'll have, let's see, from 0 to 3, then 3 to 6. then six to nine nine to twelve and then twelve to fifteen so that is dividing the object into five equal parts Okay, so I've taken my original rectangular object and I've divided it up into five equal parts. And now I want to take four-fifths, right? So four-fifths means four out of five equal parts. So I can take any four of these parts, but just to keep things simple, I'll take this, this, this, and this. So in this diagram, that is going to represent four-fifths. Okay, so, so far I've taken 4 fifths of the original rectangular object, and now I want to take 2 thirds of 4 fifths, right? So 4 fifths is starting here and ending here. So let me make this a little bit thicker. So that's the boundary of 4 fifths of the object. And I'll also make these a little bit thicker. Okay, so from here to here, that's four-fifths of the original object. And now I want to take two-thirds of this four-fifths, right? How can I take two-thirds of this much of the original object? Well, again, I can be clever about this. I can say, all right, instead of focusing on all that, let's focus on this and take two-thirds of that, then take two-thirds of this part, take two-thirds of this part, and then take two-thirds of this part, and in that way, I'll have two-thirds of this. part from here to here. So let's divide up this one-fifth into two-thirds right or into thirds and then we'll take two of them. So this one is roughly three centimeters long so to divide that into thirds we'll have one centimeter units so let me do it like this kind of lightly Okay, so that's dividing this bit into thirds, right? One third, one third, one third, one third of that. And then I'm going to take two thirds, so I'm going to take two of those. So this one and this one. Then I'll do the same thing over here. Okay, so this one fifth has been divided into one third, and I'll take two of those. Then do the same thing over here and over here. Okay, so take two-thirds here and two-thirds here. Okay, so let's summarize what we have done so far. This was our original object, right, that complete rectangle. We divided it first into fifths. So this here is one-fifth, this here is one-fifth. fifth, one fifth, one fifth, and this is one fifth, and then we took four fifths, so we took one, two, three, four of those fifths, but we want two-thirds of four fifths, so to figure out two-thirds of this much, we were clever about it. it we took two-thirds of this one-fifth took two-thirds of this one-fifth took two-thirds of this one-fifth and two-thirds of this one-fifth and now the shaded shaded bits represent two-thirds of four-fifths of the original of the original rectangular object now how much of the original rectangular object are the shaded portions right the shaded rectangles we have one two three four four, five, six, seven, eight of them, but it's eight out of how many equal parts, right? So we know there are three equal parts here, three equal parts here, three equal parts here, three equal parts here. We can also have three equal parts of this bit that has been left out so far. Let's do that. Okay, so notice that this, this, this, right? These little mini rectangles, whether shaded or unshaded, they're all the same, they all have the same area, right? So one of these is... what fractions, so we have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 of them. So one of these is 1 15th, right? One part out of five equal parts. So two thirds of four fifths. of the entire rectangle, that's what this multiplication means, so 2 3rds times 4 5th, that is represented by the areas of the shaded rectangles, and how many of those do we have? We have 1, 2, 3, 4, 5, 6, 7, 8 of them, so that's 8 out of 15 equal parts. So that is the answer to this multiplication with this interpretation. So do we notice something interesting in the numerical answer? Well, the 8 is 2 times 4. it's as though we multiply the numerators and the 15 is 3 times 5 so we are multiplying the denominators. Okay so that was example 3 and now example 4 we do the same thing but we now reverse the order. We have 4 fifths times 2 thirds. So we'll interpret this multiplication as 4 fifths of 2 thirds. So this one was 2 thirds of 4 fifths. This one is 4 fifths of 2 thirds. So let's see if there is a difference. Okay, so we want 4 fifths of 2 thirds. So first step is to take 2 thirds of this object. So we'll divide it up into thirds. The object as we observed already is roughly 15 centimeters long. So if we divide it into 3 equal parts, each part will be roughly 5 centimeters long. So let's do that. So 5 centimeters. five centimeters and then ten okay so there we go these are a bit thick Okay, so I have taken my object, I've divided it into three equal parts, and then I take two-thirds, so I take two of those three equal parts. So I can take this one and this one, or this one and this one, or this one and this one. So again, for convenience, let's take this one and this one. So that represents two-thirds of the original object. And now I want 4 fifths of 2 thirds, so I want 4 fifths of this. So to be clever about it, I will first divide this bit into 5 equal parts and take 4 fifths of that. And then I will divide this part into... 5 equal parts and take 4 of those. So how do I divide this bit into 5 equal parts? Well this bit is roughly 5 centimeters long so we just cut off 1 centimeter length Okay, so we have five equal parts here and we take four of those right four fifths So if you take four of those so take any four of these I'll take this one this one And then do the same thing for this one so divide that into five equal parts Okay Okay, now take four of these. So take this one, this one, this one, this one. Okay, so the shaded rectangles represent four-fifths of two-thirds of the original object. Again, how much is one of these shaded mini-rectangles? Well, this has been divided into five equal parts, this has been divided into five equal parts. Let's do the same thing on this part that has been left out. Okay, so each one of these mini rectangles, whether shaded or unshaded, how many of these do we have? We have 5 here, 5 here, 5 here, so we have 15 altogether. So each one of these mini rectangles represents 1 15th. one part out of 15 equal parts. And now, 4 fifths of 2 thirds of the original object is represented by these shaded rectangles. And how many of them do we have? We have 1, 2, 3, 4, 5, 6, 7, 8. so that means this multiplication 4 over 5 times 2 over 3 that is 8 out of 15 equal parts so we get 8 over 15 Okay, so in example 4 and in example 3, we did have slightly different interpretations. Right here we had 2 thirds of 4 fifths, but here we have 4 fifths of 2 thirds. But numerically, the final answer is the same, 8 over 15, 8 over 15. And numerically, again, all we seem to be doing is that we're multiplying the numerators together and we're multiplying the denominators together. So 4 times 2 gives you the 8 and 5 times 3 gives you the 15. So the actual... arithmetic is it seems to be simple you just multiply the numerators together you multiply the denominators together and and the arithmetic is the same in both cases although the interpretations are somewhat different here and going back to page one again the arithmetic was the same you multiplied the numerators to get the numerator you multiplied the denominators to get the denominator but the interpretations were different Here we had repeated addition, here we had two-fifths of three objects. Okay, so the bottom line is that the arithmetic of multiplication of fractions seems to be simple, but the interpretation may be different in individual situations. And hundreds of such situations literally, or probably thousands, have been observed, and the arithmetic is always the same. Thank you. multiply the numerators to get the numerator of your answer, and you multiply the denominators to get the denominators of your answer when you're trying to do an interpret multiplication. So with that observation, we can now extend our idea of multiplication to rational numbers. So remember, regular fractions, we have a positive numerator or zero in the numerator, and we have a positive denominator. With rational numbers, we can have negative numerators and negative denominators. So how should we define multiplication for rational numbers? We'll do it the same way as for regular fractions. So this is what's mentioned at the bottom of page 2. It says examples 1 through 4 suggest that to multiply two fractions, we need to multiply the numerators and multiply the denominators. Because of similar observations in other situations, interpretations in science, and the necessity of consistency of various arithmetic properties, rational number multiplication is defined as follows. For rational numbers a over b and c over d, we have a over b times, the dot here means times, a over b times c over d is a times c, so you multiply the numerators, and b times d, you multiply the denominators. So hopefully this definition... now makes sense, you understand why rational number multiplication is defined that way, because at the very least it's consistent with the examples of regular fraction multiplication that we have seen so far. So with that out of the way let's try page 3, example 5. So example 5, it says simplify, and we have several parts. In part A, we're multiplying two fractions, right? We're multiplying two rational numbers. So how do we do that? Well, how is multiplication defined? So here we're not bothering interpreting this in a physical situation. We just want to do the arithmetic of multiplication. So the way multiplication is defined is you... That's a pretty ugly equal sign. It's meant to be an equal sign. So you multiply the numerators together, so you get 10 times 12. And you multiply the denominators together, 21 times 55. So we're doing brute computation here. We're not bothering. with interpreting this in a physical situation. Okay, so as far as multiplication of rational numbers is concerned, we are done, but you are not allowed to leave your answer like that. As is usual in mathematics, you have to simplify your answer. So this resulting rational number, this resulting fraction, you have to simplify. And how do you do that? Well, as we saw in the previous lecture, you need to cancel common factors. So... So we need to factor these integers a bit more. So the 10, that's 2 times 5. You can write 2 times 5 or 5 times 2. The 12, we have choices. You have 2 times 6 or you have 3 times 4. Which one would be better here? So here you have to be a little bit alert. Notice that the 21, that's 3 times 7, right? So for the 12, it would be better to do 3 times 4, so the 3s can cancel. So for the 12... we don't want to do 2 times 6, although if you do there's no harm, eventually things will work out, but if you can save yourself a little bit of time by being alert that's a good thing. So because 21 is 3 times 7, the 12 we will factor as 3 times 4. So 21 that's 3 times 7, and then 55 that's 5 times 11. And now, as you saw in the previous lecture, we can cancel common factors. You don't have to line them up. If you want to, you can, but let's not waste time. So can the 2 cancel anything at the bottom? No. The 5 can cancel this 5. and the 3 can cancel this 3. So those cancel, and then what are we left with? We are left with 2 times 4 at the top, and we are left with 7 times 11 at the bottom. So without interpreting anything, just doing brute arithmetic, the rational number of fraction 10 over 21 times the rational number of fraction 12 over 55 is the rational number of fraction 8 over 77. So that's our answer. Now, you will be expected to do these kind of multiplications by hand. You can always put them in your calculator to check your final answer, but it is important that you understand and are able to do this kind of multiplication by hand. And the other thing to point out... here is that if you wish you can skip this step because you know you write 10 times 12 here 21 times 55 here but eventually you have to break up those numbers anyway so if you wish you could skip this step and go directly from here to here so skip if you wish but you have to be able to go directly from here to here or some other kind of breaking whatever you feel like and eventually show that everything simplifies or reduces to 8 over 77. Okay in part b we have this negative this opposite of 15 over 16 we have that rational number multiplying 24 over 66. So how do we handle this negative sign? Pretend you are seeing this for the first time. How do we handle this? Well, we learned in the previous lecture that The negative sign, which is in front of the rational number, and now it means the opposite of 15 over 16, the negative sign could go with the 15, the numerator, or it could go with the 16. So you can tag negative sign to the 15 or the 16, so it's your choice. Let me tag negative to the 15, the negative sign to the 15. So we have this rational number multiplying 24 over 66. Okay, so this multiplication is this multiplication. Now, how do we multiply those? Well, the definition is you multiply the numerators and the denominators separately. So we have negative 15 times 16, sorry, times 24 at the top, and then we have 16 times 66 at the bottom. Okay. Now, What can we do next? The negative 15, we can break that up as negative 3 times 5 or 5 times negative 3. That's integer multiplication that we already saw several lectures ago. So let's write that as negative 3 times 5. We have the 24. We could break that up as 3 times 8. 4 times 6. The 16, you know, we could do 4 times 4, or 2 times 2 times 2 times 2, right? The 6 to 6 is 6 times 11. So, since the 6 to 6 has 6 in it, that's one possibility. How about for the 24, we break that up as 4 times 6. And the 16 will do 4 times 4 because we have a 4 here. The 66 is 6 times 11. So this is one of several possible choices, right? If you made a different choice, that's perfectly fine. But given this choice, let's see. what cancels now. This 4 can cancel this 4 or this 4, one of them, right? Let's cancel this one. And this 6 can cancel this 6. So what are we left with at the end? At the top, we're left with negative 3 times 5, which is negative 15. that's integer multiplication, and at the bottom we're left with 4 times 11 which is 44. We could leave our final answer like that or we could make the negative go with the denominator 44, so we could write our answer as 15 over negative 44, or we could leave this negative sign in front of the fraction, and that's what usually people do. So I will leave the negative in front of the fraction, and I have the answer negative. 15 over 44 which is the opposite of 15 over 44 So that's our final answer. And now a couple of observations. So what was the effect of this negative sign? It's as though it didn't move at all, right? Here it was in front, here it is in front. So in the future, when you have a single negative sign like this in the front, you don't need to tag it along with the numerator or the denominator, just leave it in the front. That's what we observe in this example. And so you can skip this step if you wish. You can skip this step if you wish. So I'll just indicate that you can skip that if you wish. I'll put a question mark. It's up to you. You don't have to skip, but you can if you want. You can skip this. So skip question mark. But you should be able to do this by hand, so you should be able to show this work. And again, you don't have to factor the way I did. You could have other choices. Eventually, things will simplify. So definitely you have to include. this step or something equivalent to that step and then have the final answer. Okay, so let's try part C. Part C, we're multiplying three rational numbers together. Okay, now the lesson we learned from Part B is that if there is a single negative sign somewhere, you can just leave that in the front. So here we have a single negative sign. If you really want, you can tag it along with the numerator 18 or the denominator 7, but eventually it will come out in front anyway. So leave that negative sign in the front. Okay, so that negative sign will stay in the front. And then we multiply the numerators, 3. So if we were to multiply these two fractions, we would multiply the 3 and the 18, right? And whatever result is, we'll multiply with that fraction. So whatever the numerator is, we would have to multiply with the 28. we might as well multiply all the three numerators in one go. Similarly, if we were just multiplying this rational number with this rational number, we'll multiply the denominators to get 4 times 7. And then whatever the answer to that is, when we multiply that with that, that denominator has to multiply the 9, so the 4 times 7 has to multiply the 9. So again, we might as well do it all in one go. So that's that, and as in part B and A, if you wish to skip this step, if you are comfortable, you can skip this step. So I'll write it like this, skip with a question mark. It's up to you. You can skip that step or not. But the next step you definitely should not skip, or some version of the next step. So I have the negative sign. I have a 3 here, and here the 9 would be 3 times 3, so it's good. I have the 3 there. It can cancel. I have an 18 here, right? Now, I have a 9 here, so the 9 is... Well, since I have the 9 here, maybe I should write the 18 as 2 times 9, right? Then in that way I can actually cancel the entire 9. So let me do that. That's one possible choice. So the 18, I'll write that as 2 times 9. And then the 28... Should I do 4 times 6, or should I do, sorry, 4 times 7, or should I do 2 times 14? Well, I have a 4 here, so the 28 I can do 4 times 7. And that's actually good because then that 7 will cancel this 7 down here. Okay, so I won't break up the 9. This 9 can cancel this 9. This 4 can cancel this 4. And this 7 can cancel this 7. So at the end of the day, what am I left with? I have the negative in front. In the numerator, I am left with 3 times 2, which is 6. What am I left with in the denominator? Here some students get confused. They... they say, okay, in the denominator there's nothing left, so I get a zero. But that's not correct, because originally in the denominator I had 4 times 7 times 9, right? And the 4 times 7 times 9 can be thought of as 4 times 7 times 9 times 1. The 1 is always there, we just don't write it. So it's not true when some students say that since everything got cancelled, since the 4, the 7, and the 9 got cancelled, there is zero left. It's not true that there is zero left, because if there was zero here, then the original denominator... denominators would have been 0, right? And that's not the case. So what we have left in the denominator is actually 1, we just don't write it. So it's 1 left in the denominator, so we get negative 6 over 1. But remember, what is negative 6 over 1? It's the opposite of 6 over 1, and 6 over 1 is just the integer 1. So the final answer is negative 6. And then finally part D. Here we are multiplying two terminating decimals, right? So those are rational numbers. They can be written as fractions. Now, when you do these four Homer problems, just use your calculator. So just use your calculator. Okay, so when you do homopropylism webassign, just do this multiplication in your calculator and you have your answer. But for the purposes of this lecture, I want to illustrate how you could do this by hand if necessary. So we have negative this number times this number. So as we saw over here, the negative can stay out in front. So we have negative here. And then what's 6.89? You should remember from elementary school, 6.89 is the fraction 689 over 100, right? Because you have the number 689 for that integer, the decimal. point is here after the 9 and when you divide by 100 the decimal point moves two places to the left and you get 6.89 so that's something you should remember from elementary school and then you're multiplying that by 0.0003 35. How do you write that as a fraction? You have 35. For 35, the decimal point is here after the 5, and the decimal point has to move how many places to the left to get this decimal? It's 1, 2, 3, 4, 5. So you have to divide by 1 and 5 zeros. 1, 2, 3, 4, 5. So 35 divided by 100,000 is 0.00035. So this is something from elementary school that you are expected to remember. And now we're just multiplying fractions, aren't we? So now we're in the same situation as here or here. So we multiply the... We'll leave the negative out in front. And we multiply... the numerators, so we get 689 times 35, and we multiply the denominators. Now this is easy, because here we have one and two zeros, here we have one and five zeros, so when you multiply those you get one and seven zeros. So again that's from elementary school, one, two, three, four, five, six, seven, so that's ten million. And now we need to do 689 times 35. Can we do that by hand? Yeah, it's a little bit easiest, but in principle we can, right? Let me quickly do it here. 6, 8, 9, 35. This is something you learned in elementary school, right? So 5 times 9, 45. So 5. carry a 4, and then 5 times 8 is 40, and then plus the 4 gives you 44, and then you carry a 4. 5 times 6 is 30, carry a 4, you get 34. So that's the elementary school multiplication algorithm. And then 3 times 9 is 27, so write a 7, carry a 2. 3 times 8 is 24, plus the 2 gives you 26. So write 6 and carry a 2. 3 times 6 is 18, plus the 2 gives you 20. And now you add these rows, so 5, 7 plus 4 is 11, carry a 1, 6 plus 4 is 10, plus the 1 is 11, carry a 1, 3, and 1 is 4, and 2. So you get 24,115. Okay, so nothing is mysterious here, it's just elementary school stuff. So we get minus... 24,115 over 10 million. And now one last bit of elementary school stuff. What is that? So we have the negative in front, so that negative means opposite of that rational number. And how do you write that rational number as a decimal? Well, you have 1 and 7 zeros, so the decimal point for this integer is after the 5. you have to move that 7 places, right? So let's see, let me write 24115 here. So the decimal point here, you have to move 7 places to the left. So 1, 2, 3, 4, 5, 6, and then 7. So the decimal point ends up here. So we end up with negative 0.0024115. So that's our answer. Okay, so in principle when you have to multiply these two repeating decimals you can convert them to fractions pretty easily. That's elementary school mathematics and then just carry out the multiplication like we did in parts A and B and you have your rational number answer which can be reconverted to a decimal using elementary school mathematics. Okay, so there's no mystery here but as I said when you do these homework problems please don't do that that's just too tedious right so when you see problems like this in WebAssign just use your calculator. Okay, so we're done with looking at multiplication of rational numbers, both the brute calculation and also how to interpret them when we're dealing with regular fractions. Now we want to talk about how to divide a rational number by another rational number. So to get into division of rational numbers, we'll again use the ordinary fractions as our guide. How did we do division of fractions when we were children, right? How is division of fractions taught to children? So let's review that, let's survey that. So at the top of page 4 of your notes... The bullet here says, a familiar interpretation of the fraction division, A over B divided by C over D, is as the answer to the question, how many C over D of a unit can we fit inside A over B of a unit? Now, when you read this, something should stir in your memory, should be vaguely familiar from your childhood in elementary school. On the other hand, even in English, this kind of phrase is a bit of a mouthful, isn't it? It just feels weird because ordinary people don't talk like that. At least I hope you don't talk like that, right? So to make this a little bit less weird, let's look at a concrete situation. So let's look at this example. It says, for example, the... division 9 over 11 divided by 4 over 7 can be the answer to the following question. Suppose bucket A holds 4 sevenths of a gallon. So this is a concrete situation, right? This is not weird. Bucket A, let's say, when it's full, it holds. holds 4 sevenths of a gallon. So it's not a very big bucket, but that's fine. And bucket B holds 9 elevenths of a gallon. Okay, so bucket B is a little bit bigger than bucket A. The question is how many bucketfuls of A can be poured into bucket B before B overflows? So bucket B is a little bit bigger than bucket A. Imagine you fill up bucket A with water or whatever or sand, and then you pour it into bucket B. and fill bucket A up again and pour it into bucket B. And keep doing it until bucket B reaches its limit, until it begins to overflow. So this is a perfectly concrete situation. number of times that you can do this, the number of times bucket A can be poured into bucket B, so how many bucketfuls of A can be poured into bucket B, that number can be interpreted to be the result of this division problem, right? And this is similar to the situation with integers, right? Imagine if bucket A holds two gallons and bucket B holds ten gallons. So again think about that, bucket A holds two gallons, bucket B holds ten gallons, you can ask how many bucketfuls of A can be poured into bucket B. right and the answer would be 10 divided by 2 so that would be a division problem for integers here we have a similar division problem but involving fractions okay so this rather awkward English phrasing has hopefully been more clear using a specific example such as this one now pretend this is the first time in your life that you're seeing this I'm sure you you already know the answer but pretend that you are seeing this for the first time as a child so you are perhaps totally confused how on earth do I carry out this division well to help make sense of it to someone who is seeing this for the first time we have example six example six says compute this division 9 over 11 divided by 4 over 7 using the following three steps so we have three steps on page four step one step two step three to figure out this division and the idea is that each step should be clear each step should be perfect to clear And then at the end of step three, we'll see what the overall process is. At the moment, especially if someone is seeing this for the first time, this is a little bit complicated, right? It's not perhaps entirely clear immediately what the answer is. In the case of integers, right, if bucket A holds 2 gallons and bucket B holds 10 gallons, that's a much simpler situation. You do 10 divided by 2 and you get 5. So bucket A, you can fill up bucket A 5 times and pour into bucket B before bucket B overflows. So in the case... of integers the situation is simple. In the case of these fractions it's not as clear, certainly not to me. So let's follow these three steps and again fingers crossed each step is going to be clear. So step one it says compute the simpler division 1 divided by 1 over 7. So forget this horrible 9 over 11 and this horrible 4 over 7. Let's look at the simpler situation. We're going to divide the integer 1 by the fraction 1 over 7 and how hard we're going to do this computation. We're going to compute this by interpreting it as the answer to the following question. How many one-seventh gallons can fit inside one gallon? That should make perfect sense to you, even if you are somebody who is seeing this for the first time, right? We have one full gallon. One over seven can be one-seventh of a gallon. And this division can be interpreted to mean how many one-seventh gallons can you pour into one gallon, right? So imagine a bucket holds one gallon so let's say this here is one gallon so one gallon and then you have 1 7th of a gallon so you have divided up into seven equal parts so let's see I have one two three four 6, 7. Okay, so pretend that was one full gallon, and then I've divided it into seven equal parts. So just this much is one-seventh gallon, right? So how many of these one-seventh gallons can fit inside one gallon? The answer is clearly seven, right? So the answer to step one is as follows. One divided by one-seventh should be seven. So maybe you want to pause the video and think about this for a few moments. My hope is that the answer to step one is clear to everybody. How many want seven gallons, let's say, fit inside one gallon? The answer is seven. Now, step two, we jazz things up a bit. This time we will compute 1 divided by 4 sevenths. So here we are dividing 1 by 1 seventh. That was a very simple situation. Now we have jazzed things up a bit. Now we want to divide 1 divided by... by 4 7ths. So it says compute that and why 4 7ths? Well 4 7ths appear in our original problem so we are we're getting to this problem bit by bit in using several hopefully easy steps. So compute 1... divide by 4 sevenths by interpreting it as the answer to the following question, how many 4 seventh gallons can fit inside 1 gallon? Okay, so here in step 1, how many 1 seventh gallons can fit inside 1 gallon? That was pretty straightforward. Now we have a slightly more complicated situation, how many 4 seventh gallons can fit inside 1 gallon? Now, to figure out this answer, I want to make the following observations, which should... Well, I hope this is relatively straightforward for you. So observe this. So this is a bulleted point. Observe the following. 4 sevenths is 4 times as large... as one seventh. So think about that. One seventh is one part out of seven equal parts. Four sevenths is four parts out of seven equal parts. So four sevenths is four times as large as one seventh, right? Going back to this diagram, from here to here is one gallon. that bit is 1 seventh of a gallon, and then if you take four of those, this, this, this, and this, that's 4 sevenths, and then 4 sevenths is, you know, 4 times as large as 1 seventh. So, again, maybe pause the video and think about this for a second, but I hope this observation is clear to you. So if that observation is clear to you, look back to the answer for step 1. How many 1 7ths can you fit inside 1? The answer is 7. Since 4 7ths is 4 times as large as 1 7th, you are going to be able to fit 4 times less 4 7ths inside 1, right? Again, 4 7ths is 4 times as large as 1 7th. So if you can finish 7 1... if you can fit 7 1 7ths inside 1, how many 4 7ths can you fit inside 1? It will be 4 times as... 4 times less. So let me write that down. So the answer to step 2... The answer to step 2 is going to be 1 fourth as much as the answer in step 1. So once again, perhaps you want to pause the video and think about this for a few moments. This number 4 7, this fraction is 4 times as large as 1 7th. You can fit 7 1 7ths inside 1. gallon, right, you won't be able to fit as many 4 sevenths because 4 sevenths is larger than 1 seventh. And 4 sevenths is how much larger is 4 times as large. So the answer to step 2 will be 1 fourth as much as the answer in step 1. So hopefully that makes sense. So this implies, the arrow means implies, 1 divided by 4 sevenths So we're interpreting that to mean how many 4 7s can you fit inside one gallon, right? How many 4 7 gallons can you fit inside one gallon? So that would be 1 4th of the answer to part A. So that would be 1 4th of 7. And then how did we interpret 1 4th of 7? Back in page 1. You know, 2 fifths of 3 was interpreted to be 3 times 2 fifths. So by the same analogy, 1 fourth of 7 would be 7 times 1 fifth. So 1 fourth of 7, by staying consistent with our interpretation, that's 7 times 1 fourth, just like in page 1. And then how do we do 7 times 1 fourth? Well, 7 is the same as 7 over 1, so we get 7 over 1 times 1 fourth. And now how do you multiply the fractions? You just multiply the numerators. So 7 times 1 is 7 and 1 times 4 is 4. So the answer to step 2 is 7 over 4. Now, do you notice something interesting about step 1 and step 2? In step 1, we were trying to figure out 1 divided by 1 seventh. The answer was 7, so it's as though the fraction 1 over 7 got converted into 7, right? In step 2, we're trying to figure out 1 divided by 4 sevenths. Well, what is the answer? The answer is 7 over 4. That's 4 over 7 turned upside down, right? So here the 1 over 7 got turned upside down to 7 over 1, which is just 7. Here the 4 over 7 got turned upside down to 7 over 4. That's interesting, huh? We had some interpretation going on, but numerically, what we're doing numerically seems pretty straightforward. We're just flipping fractions. Okay, now let's continue. We're almost done. So step 3, we want to finally figure out the answer to this division problem. That was the problem that is asked in example 6, right? 9 over 11 divided by 4 over 7. So in step 3, we want to figure this out. And how are we going to compute this? We're going to compute this by interpreting it as the answer to the following question. How many 4 seventh gallons can fit inside 9 elevenths of a gallon? Okay, so imagine this is 9 elevenths of a gallon. The question is how many 4 seventh gallons can you fit inside there? And this interpretation here is precisely the one that we had up here when we had bucket A and bucket B. So how can we handle step three? Well, again, the... The answer here is going to be the following, and I'll write this down for your benefit. So, in step 2, how many 4 7's can you fit inside 1? It's this many. Now, how many 4 7's can you fit inside here? here you're fitting 4 sevenths into one whole. Here you're fitting 4 sevenths into not one whole but 9 elevenths of a whole. So the answer is going to be 9 elevenths of the answer in step 2. So once again, maybe pause the video and think about that for a second. In step 2, you're trying to figure out how many 4 7ths can you fit inside one hole. But in step 3, you're trying to figure out how many 4 7ths can you fit inside 9 11ths of a hole. So the answer would be 9 11ths of whatever the answer is in step 2. In step 2 you have fitted 4 7th gallons into 1 whole gallon. Now you don't have a 1 whole gallon, you have 9 11ths. of a gallon. So however many 4 sevenths you could fit inside one whole gallon, if you take 9 11ths of that answer in step two, you now get the answer to step three. So we have 9 11ths of the answer in step two. How do we write that? Well, we saw this back on page two, right? On page two, we saw that 2 thirds of 4 fifths is 2 thirds times 4 fifths. So using that same interpretation, 9 11ths of the answer in step 2 would be 9 11ths times the answer in step 2, which is 7 fourths. And what is that? Well, you just multiplied the numerators, so 9 times 7 is 63, and 11 times 4 is 44. Are there any common factors? Well, the common factors of... The factors of 63 are 9 and 7, so 3 and 7, basically. And 3 and 7 are not factors of 44, so there are no common factors. So that's the answer to step 3. So that's the answer to the problem. But overall, what is happening? Overall, let's write this down. So overall, we were trying to figure out 9 over 11 divided by 4 over 7. That's what we were trying to figure out. 9 over 11 divided by 4 over 7. So in step 3, we observed that... 9 over 11 divided by 4 over 7. Ignore this 4, that's the page number. So 9 over 11 divided by 4 over 7. What is that? That is equal to this. 9 over 11 times 7 over 4. And that turns out to be 63 over 44. So notice what has happened. 9 over 11 divided by 4 over 7 in this concrete example is the same as 9 over 11 times... this fraction 4 over 7 flipped over, right? And then that's how we get the answer. So we will use this as our guide to define the division of rational numbers. So in this particular concrete example using buckets and whatnot we see that 9 over 11 divided by 4 over 7 is the same as 9 over 11 times 4 over 7 flipped over and again literally thousands of such examples have been observed in various areas of science and mathematics and engineering and so we use this as our guide to define division of rational numbers so moving on to the last page page five Top of page 5. First bullet, it says the reciprocal of a non-zero rational number a over b is the number b over a. So here we have a non-zero rational number, which means a can't be 0, right? b cannot be 0 anyway. Denominator is not allowed. zero anyway but since a over B is nonzero that means a cannot be zero so the reciprocal of a over B is just B over a is just a over B flipped over and now the second bullet says example 6 which we just finished in the pre previous page. Example 6 suggests that dividing by a fraction is equivalent to multiplying by the reciprocal of that fraction. So that's what we observed. And then it says because of similar observations in other situations, interpretations of science and the necessity of consistency of various arithmetic properties, rational number division is defined as follows. For rational numbers a over b and c over d, where c is not equal to zero, we have a over b divided by c over d. It's defined to be over b over c over d. times the reciprocal of C over D, which is D over C. So division by a fraction is defined to be multiplication by the reciprocal of the fraction. And that's what you need to remember. So with that, let's do our final example of the lecture. So part A, we are dividing the rational number of fraction 3 tenths by the rational number of fraction negative 15 over 14. Again, that means the opposite of negative 15 over 40 means the opposite of 15 over 40. Okay, so how do you do this? Now, this kind of fraction division you are expected to be able to do by hand. So you can put this in your calculator to check your final answer, but you should be able to do this by hand. Now, how is division defined? Division is defined to be multiplication by the reciprocal. So this is defined to be the fraction or rational number 3 tenths times the reciprocal of this. Now, what is the ratio of the fraction to the reciprocal? the reciprocal of this negative rational number? Do we just flip? Well let's think about this. What does the negative sign in front mean? The negative sign in front means you could put the negative with the 15 the numerator or with the denominator 14, right? Either is fine. We saw this in a previous lecture. So now don't write this down, but imagine this. Imagine the negative goes with the 15. So you have negative 15 over 14. If you flip that, you will get 14 over negative 15, right? You will get 14 over negative 15, but then the negative can come to the front again. So the overall effect is this. If you take the reciprocal of this, you just get negative 14 over 15. so it's as though the negative doesn't change position at all. And let me say this one more time. So you can imagine the negative going with the 15 or with the 14. If you put the negative with the 15 up here, you have negative 15 over 14. When you take the reciprocal of that, you'll get 14 over negative 15, but then the negative with the 15 in the denominator now can again come to the front. So it's as though the negative doesn't change position at all, and you're just flipping the 15. and the 14. So we have obeyed the definition here and written the division problem as a multiplication problem. So you should show this step when you do this by hand. And now what happens to the negative sign in this multiplication problem? Well as we saw before with rational number multiplication you can put that out in front. And you multiply the numerators, so you have 3 times 14. And you multiply the denominators, so you have 10 times 15. And this step you can skip if you wish. So I'll put skip and then question mark. It's up to you. You can skip that step. And then in the next step you cancel common factors. So this next step you should not skip. So we have the 3. 14 is 2 times 7. The 10 here is 2 times 5. The 15 is 3 times 5. So now what cancels? This 2 can cancel this 2. The 3 can cancel the 3. And that's it. Right. So we're left with. Oops sorry you couldn't see that. We're left with 7 at the top and 5 times 7, so 25 at the bottom. So the result of this division, the fraction or rational number 3 tenths divided by the rational number negative 15 over 14, the result of that division is... division is negative 7 over 25. So we are not interpreting this in a physical situation here like example 6 on the previous page. We are just doing the brutal computation of division, nothing more. and the answer turns out to be negative 7 over 25. And once again, you are expected to be able to do this by hand, and you can skip this step if you wish, but you should not skip this step where you show the cancellation of common factors, and then you get to the final answer in simplified form. Okay, that's part A. And then part B, here you're dealing with decimals. So as with back in example, example 5 part d, if you look back to example 5 part d where you're multiplying two decimals, I said you can just use your calculator when you do those homework problems in WebAssign. So same here. When you see a division problem involving decimals, just use your calculator. So unlike something like this where you are expected to be able to do this by hand, so you must learn that, this one you don't have to do by hand. So just put them in your calculator and you'll get your answer and then you'll enter those answers for the HOMA problem. consumer have assigned but for this lecture let's just do this by hand so you understand the process so we have negative 0.27 that's the opposite of 0.27 can I write 0.27 as a fraction yeah because it's a terminating decimal so 0.27 is 27 over 100 this is again a elementary school arithmetic that you are supposed to know so the decimal point after seven moves two places when you divide by 100 to get 0.27 when you're dividing that by 1.32 1.32 is 132 over 100. And now division by rational number is defined to be multiplication by the reciprocal. So we have negative 27 over 100 times the reciprocal of 132 over 100, which is 100 over 132. Good. Next, the negative sign can stay in the front. We multiply the numerator, so we get 27 times 100. And we multiply the denominator, so 100 times 132. And this step, you know, you could skip. And now what? The hundreds cancel, right? They're common. Then we have negative 27. You can factor that as 3 times 9, or 3 times 3 times 3. How about the 132? If you divide 132... So 132 by 3, what do you get? 3 goes into 13 four times. Then 3 times 4 is 12, and then 1 left, and then 3 goes into 12 four times. So 132 is 3 times 44. Now the 3s cancel, and you're left with negative 9 over 44. So when you divide this by this, the answer as a fraction is negative 9 over 44. If you want to get the answer as a decimal, you can just put negative 9 over 44 in your calculator, and you do not get a terminating decimal. Well, you do get a terminating decimal. Well, sorry, I'm saying the wrong thing. I said the correct thing the first time. You do not get a terminating decimal, you get a repeating decimal. So if you put this in your calculator, you will get the following. Negative 0.20. And it turns out that you have 4, 5, and then 4, 5, 4, 5, and so on. So that turns out to be 0.2045 repeating. Now, when you do the Hummer problems, as I said, you just put this in your calculator. And if you put that division problem in your calculator, you will get... Get 0.204545 and so on. The problem with that is the calculator only gives you a finite number of decimal places. So this is something we mentioned before in a previous lecture. So how can you be 100% sure that 4, 5 repeats forever? Maybe after a million decimal places, the 4, 5 changes. to something else like 6, 7, 9, right? So how can you be sure from your calculator that this 4, 5 really does repeat forever? And the answer is you cannot be sure from a calculator because the calculator will only give you a finite number of decimal places. so to be absolutely 100% sure unfortunately you will have to do this by hand and see that okay we really do get a fraction and fractions are either repeating decimals or terminating decimals so only doing something like this proves to you that yes this 4, 5 repeats forever but again as I said when you do the Homer problems in WebAssign don't worry about things like that if the answer to the problem from your calculator looks like it's being repeated it's unlikely that WebAssign is trying to cheat you you So you can just assume that yes, like the 4-5 really is repeated and you could put a bar over the 4-5 in your answer or write 4-5, 4-5, 4-5, dot, dot, dot. But you will have to see what the instrument does. to the problem say in WebAssign. Okay so with that we are finally done with lecture 5 and now you can do the homework for lecture 5 in WebAssign so good luck with that

So to discuss multiplication and division of rational numbers, in the first part of the lecture we are going to review and survey how multiplication and division is done for fractions, for ordinary fractions. how multiplication divisional fractions is taught to children. So the first bullet, it says, we have three familiar interpretations of fraction multiplication.

I don't know if this is focused. Let me try to focus it. Okay, I think that's a little bit better. Okay, so it says we have three familiar interpretations of fraction multiplication.

Number one, multiplication of a fraction by a positive integer n can be thought of as repeated addition of the fraction. where the fraction is repeated n times. So then that's an interpretation where a fraction is multiplied by a positive integer n.

Number two, it says multiplication of a positive integer n by a fraction, such as 2 fifths, be thought of as two-fifths of an identical object. And number three, it says multiplication of a fraction such as two-thirds by another fraction such as four-fifths can be thought of as two-thirds of four-fifths of one whole object. So these three interpretations are taught to children in elementary school and we will now review them by looking at specific examples. So if you look at example 1 and 2 together, example 1, it says compute 2 fifths times 3, which is the same as 2 fifths times 3 over 1. So remember, the integer 3 is the same as the rational number 3 over 1. We saw that in the previous lecture. So we want to compute this multiplication, which is the same as this multiplication, by interpreting this multiplication as a repeated addition.

as the fraction 2 fifths repeated 3 times, whereas over here, 3 times 2 fifths, which is the same as 3 over 1 times 2 fifths, we'll interpret this multiplication as 2 fifths of 3 identical rectangular objects that are given here. Now, we could have interpreted this one as 2 fifths of 3 objects, and we could have interpreted this one as 2 fifths repeated 3 times. That would have been fine as well, but we just need to be consistent with how we interpret things from the beginning. So the way I have stated the interpretations here, I'm just being consistent with that. So now let's try example 1. So this multiplication, I'll write that first here.

We have 2 over 5. times 3 over 1 that is the same as 2 over 5 times 3 so our interpretation is that this is repeated addition so the fraction 2 over 5 is repeated three times Okay, so that's interpreting the multiplication as repeated addition. Now, can we interpret this repeated addition? Yes, we can, because remember, what is the interpretation of the fraction 2 over 5?

It means you take an object, cut it up into 5 equal parts, and then... take two of those parts. So here you have two parts out of five equal parts.

You're adding to that two parts out of five equal parts. So overall here you have four parts out of five equal parts and then you're adding two more parts out of five equal parts. So overall you have six parts. out of 5 equal parts and can we write 6 parts out of 5 equal parts as a fraction? Yes, it's written like that.

So this, with our usual interpretations of fractions, this addition expression is equal to that fraction. So there we go, we have managed Do this multiplication by interpreting it as repeated addition. Now how about example two?

So this time we want to interpret that as two-fifths of three identical objects. So I want to somehow take two-fifths of these three objects. So I can be clever about this. I can say to myself, okay, I'll first focus on the first object and take two-fifths of that. Then I'll take two-fifths of the middle object.

Then I'll take two-fifths of the third object. and then I can combine all to get 2 fifths of all three objects. So for the first object we saw this same object in the previous lecture it's roughly 5 centimeters long so we will look at 1 centimeter intervals and now use those to divide up the object into five equal parts Okay, so I've divided the first of the three objects into five equal parts, and I'll take two-fifths of that object.

So I can take any two of these five equal pieces. So just to keep things simple, we'll take the first piece and the second piece. Now I will cut up the middle object into five equal parts. Okay, and now take any two of these five equal parts, doesn't matter which two, so let's take this one and this one and finally cut up the third object into five equal parts. Okay, so, and now take two of those five equal parts.

So now I have 2 fifths of this first object, 2 fifths of the middle object, and 2 fifths of the third object. So if I take 2 fifths of all three objects, how many of these 5 equal parts do I get in total? I get 1, 2, 3, 4, 5, 6 out of 5 equal parts.

So this multiplication, 3 over 1 times... 2 over 5, so again since 3 over 1 is the same as 3, this multiplication is the same as 3 times 2 over 5, that is 2 fifths of these three identical objects, and 2 fifths of all three objects gives 1, 2, 3, 4, 5, 6 fifths. And that's example two.

And notice something interesting. So this multiplication was interpreted as repeated addition, whereas this multiplication was interpreted as two-fifths of three objects. but with these two different interpretations the numerical answer is still the same the numerical answer to the multiplication is still 6 over 5 in both cases and what is the relationship the 6 and the 5 to the numerators and denominators of the fractions that we multiplied. Well notice that 2 times 3 gives you the 6 here, here the 3 times 2 gives you the 6, the 5 times 1 gives you the 5, the 1 times 5 gives you the 5. So that's an interesting observation isn't it? It's as though when we multiplied these fractions we separately multiplied the numerators to get the numerator of the answer and we multiplied the denominators to get the to get the denominator of the answer.

So that's an interesting observation. But of course, all this meant something. In example one, the multiplication meant repeated addition. And in example two, the multiplication meant two-fifths of three objects. Okay, so now we have dealt with multiplication of a fraction by a positive integer from either direction.

And here the positive integer three was on the... On the right, here the positive integer 3 was on the left. Now, going into page 2, we are going to look at the case where we multiply a fraction by another fraction. So, in example 3, we will compute this multiplication, 2 thirds times 4 fifths. And in example 4, we will look at it in the reverse order.

We'll have 4 fifths times 2 thirds. And how are we going to interpret these multiplications? Well, the same interpretations that children are taught.

So here in example 3, it says compute 2 thirds times 4 fifths by interpreting this multiplication as 2 thirds of 4 fifths of this rectangular object. In example 4, we will compute 4 fifths times 2 thirds by interpreting this multiplication as 2 thirds of 4 fifths of this rectangular object. Interpreting this multiplication is 4 fifths of 2 thirds of the same rectangular object, so we'll do it both We'll do both and then see What we get and how they compare?

Okay, so this first one, we interpret 2 3rds times 4 5ths as 2 3rds of 4 5ths of the rectangular object. So what I'm going to do is I'm going to figure out first 4 5ths. of the rectangular object.

So four fifths means you divide the object into five equal parts and take four of them. So let's see how long is this object. If I use my ruler and centimeters, this object is roughly 15 centimeters long, roughly. So to divide it into fifths, into five equal parts, each part will have length 3 centimeters, right? Because 15 centimeters divided by 5 is 3 centimeters.

So I'll have, let's see, from 0 to 3, then 3 to 6. then six to nine nine to twelve and then twelve to fifteen so that is dividing the object into five equal parts Okay, so I've taken my original rectangular object and I've divided it up into five equal parts. And now I want to take four-fifths, right? So four-fifths means four out of five equal parts.

So I can take any four of these parts, but just to keep things simple, I'll take this, this, this, and this. So in this diagram, that is going to represent four-fifths. Okay, so, so far I've taken 4 fifths of the original rectangular object, and now I want to take 2 thirds of 4 fifths, right? So 4 fifths is starting here and ending here.

So let me make this a little bit thicker. So that's the boundary of 4 fifths of the object. And I'll also make these a little bit thicker.

Okay, so from here to here, that's four-fifths of the original object. And now I want to take two-thirds of this four-fifths, right? How can I take two-thirds of this much of the original object?

Well, again, I can be clever about this. I can say, all right, instead of focusing on all that, let's focus on this and take two-thirds of that, then take two-thirds of this part, take two-thirds of this part, and then take two-thirds of this part, and in that way, I'll have two-thirds of this. part from here to here. So let's divide up this one-fifth into two-thirds right or into thirds and then we'll take two of them. So this one is roughly three centimeters long so to divide that into thirds we'll have one centimeter units so let me do it like this kind of lightly Okay, so that's dividing this bit into thirds, right?

One third, one third, one third, one third of that. And then I'm going to take two thirds, so I'm going to take two of those. So this one and this one.

Then I'll do the same thing over here. Okay, so this one fifth has been divided into one third, and I'll take two of those. Then do the same thing over here and over here. Okay, so take two-thirds here and two-thirds here.

Okay, so let's summarize what we have done so far. This was our original object, right, that complete rectangle. We divided it first into fifths. So this here is one-fifth, this here is one-fifth. fifth, one fifth, one fifth, and this is one fifth, and then we took four fifths, so we took one, two, three, four of those fifths, but we want two-thirds of four fifths, so to figure out two-thirds of this much, we were clever about it.

it we took two-thirds of this one-fifth took two-thirds of this one-fifth took two-thirds of this one-fifth and two-thirds of this one-fifth and now the shaded shaded bits represent two-thirds of four-fifths of the original of the original rectangular object now how much of the original rectangular object are the shaded portions right the shaded rectangles we have one two three four four, five, six, seven, eight of them, but it's eight out of how many equal parts, right? So we know there are three equal parts here, three equal parts here, three equal parts here, three equal parts here. We can also have three equal parts of this bit that has been left out so far.

Let's do that. Okay, so notice that this, this, this, right? These little mini rectangles, whether shaded or unshaded, they're all the same, they all have the same area, right?

So one of these is... what fractions, so we have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 of them. So one of these is 1 15th, right? One part out of five equal parts.

So two thirds of four fifths. of the entire rectangle, that's what this multiplication means, so 2 3rds times 4 5th, that is represented by the areas of the shaded rectangles, and how many of those do we have? We have 1, 2, 3, 4, 5, 6, 7, 8 of them, so that's 8 out of 15 equal parts.

So that is the answer to this multiplication with this interpretation. So do we notice something interesting in the numerical answer? Well, the 8 is 2 times 4. it's as though we multiply the numerators and the 15 is 3 times 5 so we are multiplying the denominators.

Okay so that was example 3 and now example 4 we do the same thing but we now reverse the order. We have 4 fifths times 2 thirds. So we'll interpret this multiplication as 4 fifths of 2 thirds. So this one was 2 thirds of 4 fifths.

This one is 4 fifths of 2 thirds. So let's see if there is a difference. Okay, so we want 4 fifths of 2 thirds. So first step is to take 2 thirds of this object.

So we'll divide it up into thirds. The object as we observed already is roughly 15 centimeters long. So if we divide it into 3 equal parts, each part will be roughly 5 centimeters long. So let's do that.

So 5 centimeters. five centimeters and then ten okay so there we go these are a bit thick Okay, so I have taken my object, I've divided it into three equal parts, and then I take two-thirds, so I take two of those three equal parts. So I can take this one and this one, or this one and this one, or this one and this one.

So again, for convenience, let's take this one and this one. So that represents two-thirds of the original object. And now I want 4 fifths of 2 thirds, so I want 4 fifths of this.

So to be clever about it, I will first divide this bit into 5 equal parts and take 4 fifths of that. And then I will divide this part into... 5 equal parts and take 4 of those.

So how do I divide this bit into 5 equal parts? Well this bit is roughly 5 centimeters long so we just cut off 1 centimeter length Okay, so we have five equal parts here and we take four of those right four fifths So if you take four of those so take any four of these I'll take this one this one And then do the same thing for this one so divide that into five equal parts Okay Okay, now take four of these. So take this one, this one, this one, this one.

Okay, so the shaded rectangles represent four-fifths of two-thirds of the original object. Again, how much is one of these shaded mini-rectangles? Well, this has been divided into five equal parts, this has been divided into five equal parts.

Let's do the same thing on this part that has been left out. Okay, so each one of these mini rectangles, whether shaded or unshaded, how many of these do we have? We have 5 here, 5 here, 5 here, so we have 15 altogether.

So each one of these mini rectangles represents 1 15th. one part out of 15 equal parts. And now, 4 fifths of 2 thirds of the original object is represented by these shaded rectangles. And how many of them do we have?

We have 1, 2, 3, 4, 5, 6, 7, 8. so that means this multiplication 4 over 5 times 2 over 3 that is 8 out of 15 equal parts so we get 8 over 15 Okay, so in example 4 and in example 3, we did have slightly different interpretations. Right here we had 2 thirds of 4 fifths, but here we have 4 fifths of 2 thirds. But numerically, the final answer is the same, 8 over 15, 8 over 15. And numerically, again, all we seem to be doing is that we're multiplying the numerators together and we're multiplying the denominators together. So 4 times 2 gives you the 8 and 5 times 3 gives you the 15. So the actual... arithmetic is it seems to be simple you just multiply the numerators together you multiply the denominators together and and the arithmetic is the same in both cases although the interpretations are somewhat different here and going back to page one again the arithmetic was the same you multiplied the numerators to get the numerator you multiplied the denominators to get the denominator but the interpretations were different Here we had repeated addition, here we had two-fifths of three objects.

Okay, so the bottom line is that the arithmetic of multiplication of fractions seems to be simple, but the interpretation may be different in individual situations. And hundreds of such situations literally, or probably thousands, have been observed, and the arithmetic is always the same. Thank you. multiply the numerators to get the numerator of your answer, and you multiply the denominators to get the denominators of your answer when you're trying to do an interpret multiplication.

So with that observation, we can now extend our idea of multiplication to rational numbers. So remember, regular fractions, we have a positive numerator or zero in the numerator, and we have a positive denominator. With rational numbers, we can have negative numerators and negative denominators. So how should we define multiplication for rational numbers? We'll do it the same way as for regular fractions.

So this is what's mentioned at the bottom of page 2. It says examples 1 through 4 suggest that to multiply two fractions, we need to multiply the numerators and multiply the denominators. Because of similar observations in other situations, interpretations in science, and the necessity of consistency of various arithmetic properties, rational number multiplication is defined as follows. For rational numbers a over b and c over d, we have a over b times, the dot here means times, a over b times c over d is a times c, so you multiply the numerators, and b times d, you multiply the denominators. So hopefully this definition... now makes sense, you understand why rational number multiplication is defined that way, because at the very least it's consistent with the examples of regular fraction multiplication that we have seen so far.

So with that out of the way let's try page 3, example 5. So example 5, it says simplify, and we have several parts. In part A, we're multiplying two fractions, right? We're multiplying two rational numbers.

So how do we do that? Well, how is multiplication defined? So here we're not bothering interpreting this in a physical situation. We just want to do the arithmetic of multiplication.

So the way multiplication is defined is you... That's a pretty ugly equal sign. It's meant to be an equal sign.

So you multiply the numerators together, so you get 10 times 12. And you multiply the denominators together, 21 times 55. So we're doing brute computation here. We're not bothering. with interpreting this in a physical situation.

Okay, so as far as multiplication of rational numbers is concerned, we are done, but you are not allowed to leave your answer like that. As is usual in mathematics, you have to simplify your answer. So this resulting rational number, this resulting fraction, you have to simplify.

And how do you do that? Well, as we saw in the previous lecture, you need to cancel common factors. So... So we need to factor these integers a bit more. So the 10, that's 2 times 5. You can write 2 times 5 or 5 times 2. The 12, we have choices.

You have 2 times 6 or you have 3 times 4. Which one would be better here? So here you have to be a little bit alert. Notice that the 21, that's 3 times 7, right? So for the 12, it would be better to do 3 times 4, so the 3s can cancel.

So for the 12... we don't want to do 2 times 6, although if you do there's no harm, eventually things will work out, but if you can save yourself a little bit of time by being alert that's a good thing. So because 21 is 3 times 7, the 12 we will factor as 3 times 4. So 21 that's 3 times 7, and then 55 that's 5 times 11. And now, as you saw in the previous lecture, we can cancel common factors. You don't have to line them up. If you want to, you can, but let's not waste time.

So can the 2 cancel anything at the bottom? No. The 5 can cancel this 5. and the 3 can cancel this 3. So those cancel, and then what are we left with?

We are left with 2 times 4 at the top, and we are left with 7 times 11 at the bottom. So without interpreting anything, just doing brute arithmetic, the rational number of fraction 10 over 21 times the rational number of fraction 12 over 55 is the rational number of fraction 8 over 77. So that's our answer. Now, you will be expected to do these kind of multiplications by hand. You can always put them in your calculator to check your final answer, but it is important that you understand and are able to do this kind of multiplication by hand. And the other thing to point out...

here is that if you wish you can skip this step because you know you write 10 times 12 here 21 times 55 here but eventually you have to break up those numbers anyway so if you wish you could skip this step and go directly from here to here so skip if you wish but you have to be able to go directly from here to here or some other kind of breaking whatever you feel like and eventually show that everything simplifies or reduces to 8 over 77. Okay in part b we have this negative this opposite of 15 over 16 we have that rational number multiplying 24 over 66. So how do we handle this negative sign? Pretend you are seeing this for the first time. How do we handle this? Well, we learned in the previous lecture that The negative sign, which is in front of the rational number, and now it means the opposite of 15 over 16, the negative sign could go with the 15, the numerator, or it could go with the 16. So you can tag negative sign to the 15 or the 16, so it's your choice.

Let me tag negative to the 15, the negative sign to the 15. So we have this rational number multiplying 24 over 66. Okay, so this multiplication is this multiplication. Now, how do we multiply those? Well, the definition is you multiply the numerators and the denominators separately.

So we have negative 15 times 16, sorry, times 24 at the top, and then we have 16 times 66 at the bottom. Okay. Now, What can we do next? The negative 15, we can break that up as negative 3 times 5 or 5 times negative 3. That's integer multiplication that we already saw several lectures ago. So let's write that as negative 3 times 5. We have the 24. We could break that up as 3 times 8. 4 times 6. The 16, you know, we could do 4 times 4, or 2 times 2 times 2 times 2, right?

The 6 to 6 is 6 times 11. So, since the 6 to 6 has 6 in it, that's one possibility. How about for the 24, we break that up as 4 times 6. And the 16 will do 4 times 4 because we have a 4 here. The 66 is 6 times 11. So this is one of several possible choices, right?

If you made a different choice, that's perfectly fine. But given this choice, let's see. what cancels now. This 4 can cancel this 4 or this 4, one of them, right?

Let's cancel this one. And this 6 can cancel this 6. So what are we left with at the end? At the top, we're left with negative 3 times 5, which is negative 15. that's integer multiplication, and at the bottom we're left with 4 times 11 which is 44. We could leave our final answer like that or we could make the negative go with the denominator 44, so we could write our answer as 15 over negative 44, or we could leave this negative sign in front of the fraction, and that's what usually people do.

So I will leave the negative in front of the fraction, and I have the answer negative. 15 over 44 which is the opposite of 15 over 44 So that's our final answer. And now a couple of observations.

So what was the effect of this negative sign? It's as though it didn't move at all, right? Here it was in front, here it is in front.

So in the future, when you have a single negative sign like this in the front, you don't need to tag it along with the numerator or the denominator, just leave it in the front. That's what we observe in this example. And so you can skip this step if you wish. You can skip this step if you wish.

So I'll just indicate that you can skip that if you wish. I'll put a question mark. It's up to you.

You don't have to skip, but you can if you want. You can skip this. So skip question mark. But you should be able to do this by hand, so you should be able to show this work. And again, you don't have to factor the way I did.

You could have other choices. Eventually, things will simplify. So definitely you have to include.

this step or something equivalent to that step and then have the final answer. Okay, so let's try part C. Part C, we're multiplying three rational numbers together.

Okay, now the lesson we learned from Part B is that if there is a single negative sign somewhere, you can just leave that in the front. So here we have a single negative sign. If you really want, you can tag it along with the numerator 18 or the denominator 7, but eventually it will come out in front anyway.

So leave that negative sign in the front. Okay, so that negative sign will stay in the front. And then we multiply the numerators, 3. So if we were to multiply these two fractions, we would multiply the 3 and the 18, right?

And whatever result is, we'll multiply with that fraction. So whatever the numerator is, we would have to multiply with the 28. we might as well multiply all the three numerators in one go. Similarly, if we were just multiplying this rational number with this rational number, we'll multiply the denominators to get 4 times 7. And then whatever the answer to that is, when we multiply that with that, that denominator has to multiply the 9, so the 4 times 7 has to multiply the 9. So again, we might as well do it all in one go. So that's that, and as in part B and A, if you wish to skip this step, if you are comfortable, you can skip this step. So I'll write it like this, skip with a question mark.

It's up to you. You can skip that step or not. But the next step you definitely should not skip, or some version of the next step.

So I have the negative sign. I have a 3 here, and here the 9 would be 3 times 3, so it's good. I have the 3 there. It can cancel. I have an 18 here, right?

Now, I have a 9 here, so the 9 is... Well, since I have the 9 here, maybe I should write the 18 as 2 times 9, right? Then in that way I can actually cancel the entire 9. So let me do that. That's one possible choice.

So the 18, I'll write that as 2 times 9. And then the 28... Should I do 4 times 6, or should I do, sorry, 4 times 7, or should I do 2 times 14? Well, I have a 4 here, so the 28 I can do 4 times 7. And that's actually good because then that 7 will cancel this 7 down here.

Okay, so I won't break up the 9. This 9 can cancel this 9. This 4 can cancel this 4. And this 7 can cancel this 7. So at the end of the day, what am I left with? I have the negative in front. In the numerator, I am left with 3 times 2, which is 6. What am I left with in the denominator?

Here some students get confused. They... they say, okay, in the denominator there's nothing left, so I get a zero. But that's not correct, because originally in the denominator I had 4 times 7 times 9, right?

And the 4 times 7 times 9 can be thought of as 4 times 7 times 9 times 1. The 1 is always there, we just don't write it. So it's not true when some students say that since everything got cancelled, since the 4, the 7, and the 9 got cancelled, there is zero left. It's not true that there is zero left, because if there was zero here, then the original denominator...

denominators would have been 0, right? And that's not the case. So what we have left in the denominator is actually 1, we just don't write it. So it's 1 left in the denominator, so we get negative 6 over 1. But remember, what is negative 6 over 1?

It's the opposite of 6 over 1, and 6 over 1 is just the integer 1. So the final answer is negative 6. And then finally part D. Here we are multiplying two terminating decimals, right? So those are rational numbers.

They can be written as fractions. Now, when you do these four Homer problems, just use your calculator. So just use your calculator. Okay, so when you do homopropylism webassign, just do this multiplication in your calculator and you have your answer.

But for the purposes of this lecture, I want to illustrate how you could do this by hand if necessary. So we have negative this number times this number. So as we saw over here, the negative can stay out in front.

So we have negative here. And then what's 6.89? You should remember from elementary school, 6.89 is the fraction 689 over 100, right?

Because you have the number 689 for that integer, the decimal. point is here after the 9 and when you divide by 100 the decimal point moves two places to the left and you get 6.89 so that's something you should remember from elementary school and then you're multiplying that by 0.0003 35. How do you write that as a fraction? You have 35. For 35, the decimal point is here after the 5, and the decimal point has to move how many places to the left to get this decimal? It's 1, 2, 3, 4, 5. So you have to divide by 1 and 5 zeros. 1, 2, 3, 4, 5. So 35 divided by 100,000 is 0.00035.

So this is something from elementary school that you are expected to remember. And now we're just multiplying fractions, aren't we? So now we're in the same situation as here or here.

So we multiply the... We'll leave the negative out in front. And we multiply... the numerators, so we get 689 times 35, and we multiply the denominators.

Now this is easy, because here we have one and two zeros, here we have one and five zeros, so when you multiply those you get one and seven zeros. So again that's from elementary school, one, two, three, four, five, six, seven, so that's ten million. And now we need to do 689 times 35. Can we do that by hand? Yeah, it's a little bit easiest, but in principle we can, right?

Let me quickly do it here. 6, 8, 9, 35. This is something you learned in elementary school, right? So 5 times 9, 45. So 5. carry a 4, and then 5 times 8 is 40, and then plus the 4 gives you 44, and then you carry a 4. 5 times 6 is 30, carry a 4, you get 34. So that's the elementary school multiplication algorithm.

And then 3 times 9 is 27, so write a 7, carry a 2. 3 times 8 is 24, plus the 2 gives you 26. So write 6 and carry a 2. 3 times 6 is 18, plus the 2 gives you 20. And now you add these rows, so 5, 7 plus 4 is 11, carry a 1, 6 plus 4 is 10, plus the 1 is 11, carry a 1, 3, and 1 is 4, and 2. So you get 24,115. Okay, so nothing is mysterious here, it's just elementary school stuff. So we get minus...

24,115 over 10 million. And now one last bit of elementary school stuff. What is that?

So we have the negative in front, so that negative means opposite of that rational number. And how do you write that rational number as a decimal? Well, you have 1 and 7 zeros, so the decimal point for this integer is after the 5. you have to move that 7 places, right?

So let's see, let me write 24115 here. So the decimal point here, you have to move 7 places to the left. So 1, 2, 3, 4, 5, 6, and then 7. So the decimal point ends up here. So we end up with negative 0.0024115. So that's our answer.

Okay, so in principle when you have to multiply these two repeating decimals you can convert them to fractions pretty easily. That's elementary school mathematics and then just carry out the multiplication like we did in parts A and B and you have your rational number answer which can be reconverted to a decimal using elementary school mathematics. Okay, so there's no mystery here but as I said when you do these homework problems please don't do that that's just too tedious right so when you see problems like this in WebAssign just use your calculator. Okay, so we're done with looking at multiplication of rational numbers, both the brute calculation and also how to interpret them when we're dealing with regular fractions.

Now we want to talk about how to divide a rational number by another rational number. So to get into division of rational numbers, we'll again use the ordinary fractions as our guide. How did we do division of fractions when we were children, right? How is division of fractions taught to children? So let's review that, let's survey that.

So at the top of page 4 of your notes... The bullet here says, a familiar interpretation of the fraction division, A over B divided by C over D, is as the answer to the question, how many C over D of a unit can we fit inside A over B of a unit? Now, when you read this, something should stir in your memory, should be vaguely familiar from your childhood in elementary school. On the other hand, even in English, this kind of phrase is a bit of a mouthful, isn't it?

It just feels weird because ordinary people don't talk like that. At least I hope you don't talk like that, right? So to make this a little bit less weird, let's look at a concrete situation.

So let's look at this example. It says, for example, the... division 9 over 11 divided by 4 over 7 can be the answer to the following question. Suppose bucket A holds 4 sevenths of a gallon. So this is a concrete situation, right?

This is not weird. Bucket A, let's say, when it's full, it holds. holds 4 sevenths of a gallon. So it's not a very big bucket, but that's fine.

And bucket B holds 9 elevenths of a gallon. Okay, so bucket B is a little bit bigger than bucket A. The question is how many bucketfuls of A can be poured into bucket B before B overflows?

So bucket B is a little bit bigger than bucket A. Imagine you fill up bucket A with water or whatever or sand, and then you pour it into bucket B. and fill bucket A up again and pour it into bucket B.

And keep doing it until bucket B reaches its limit, until it begins to overflow. So this is a perfectly concrete situation. number of times that you can do this, the number of times bucket A can be poured into bucket B, so how many bucketfuls of A can be poured into bucket B, that number can be interpreted to be the result of this division problem, right? And this is similar to the situation with integers, right? Imagine if bucket A holds two gallons and bucket B holds ten gallons.

So again think about that, bucket A holds two gallons, bucket B holds ten gallons, you can ask how many bucketfuls of A can be poured into bucket B. right and the answer would be 10 divided by 2 so that would be a division problem for integers here we have a similar division problem but involving fractions okay so this rather awkward English phrasing has hopefully been more clear using a specific example such as this one now pretend this is the first time in your life that you're seeing this I'm sure you you already know the answer but pretend that you are seeing this for the first time as a child so you are perhaps totally confused how on earth do I carry out this division well to help make sense of it to someone who is seeing this for the first time we have example six example six says compute this division 9 over 11 divided by 4 over 7 using the following three steps so we have three steps on page four step one step two step three to figure out this division and the idea is that each step should be clear each step should be perfect to clear And then at the end of step three, we'll see what the overall process is. At the moment, especially if someone is seeing this for the first time, this is a little bit complicated, right? It's not perhaps entirely clear immediately what the answer is. In the case of integers, right, if bucket A holds 2 gallons and bucket B holds 10 gallons, that's a much simpler situation.

You do 10 divided by 2 and you get 5. So bucket A, you can fill up bucket A 5 times and pour into bucket B before bucket B overflows. So in the case... of integers the situation is simple.

In the case of these fractions it's not as clear, certainly not to me. So let's follow these three steps and again fingers crossed each step is going to be clear. So step one it says compute the simpler division 1 divided by 1 over 7. So forget this horrible 9 over 11 and this horrible 4 over 7. Let's look at the simpler situation. We're going to divide the integer 1 by the fraction 1 over 7 and how hard we're going to do this computation. We're going to compute this by interpreting it as the answer to the following question.

How many one-seventh gallons can fit inside one gallon? That should make perfect sense to you, even if you are somebody who is seeing this for the first time, right? We have one full gallon. One over seven can be one-seventh of a gallon.

And this division can be interpreted to mean how many one-seventh gallons can you pour into one gallon, right? So imagine a bucket holds one gallon so let's say this here is one gallon so one gallon and then you have 1 7th of a gallon so you have divided up into seven equal parts so let's see I have one two three four 6, 7. Okay, so pretend that was one full gallon, and then I've divided it into seven equal parts. So just this much is one-seventh gallon, right?

So how many of these one-seventh gallons can fit inside one gallon? The answer is clearly seven, right? So the answer to step one is as follows. One divided by one-seventh should be seven.

So maybe you want to pause the video and think about this for a few moments. My hope is that the answer to step one is clear to everybody. How many want seven gallons, let's say, fit inside one gallon? The answer is seven.

Now, step two, we jazz things up a bit. This time we will compute 1 divided by 4 sevenths. So here we are dividing 1 by 1 seventh. That was a very simple situation. Now we have jazzed things up a bit.

Now we want to divide 1 divided by... by 4 7ths. So it says compute that and why 4 7ths?

Well 4 7ths appear in our original problem so we are we're getting to this problem bit by bit in using several hopefully easy steps. So compute 1... divide by 4 sevenths by interpreting it as the answer to the following question, how many 4 seventh gallons can fit inside 1 gallon?

Okay, so here in step 1, how many 1 seventh gallons can fit inside 1 gallon? That was pretty straightforward. Now we have a slightly more complicated situation, how many 4 seventh gallons can fit inside 1 gallon?

Now, to figure out this answer, I want to make the following observations, which should... Well, I hope this is relatively straightforward for you. So observe this.

So this is a bulleted point. Observe the following. 4 sevenths is 4 times as large...

as one seventh. So think about that. One seventh is one part out of seven equal parts.

Four sevenths is four parts out of seven equal parts. So four sevenths is four times as large as one seventh, right? Going back to this diagram, from here to here is one gallon. that bit is 1 seventh of a gallon, and then if you take four of those, this, this, this, and this, that's 4 sevenths, and then 4 sevenths is, you know, 4 times as large as 1 seventh.

So, again, maybe pause the video and think about this for a second, but I hope this observation is clear to you. So if that observation is clear to you, look back to the answer for step 1. How many 1 7ths can you fit inside 1? The answer is 7. Since 4 7ths is 4 times as large as 1 7th, you are going to be able to fit 4 times less 4 7ths inside 1, right?

Again, 4 7ths is 4 times as large as 1 7th. So if you can finish 7 1... if you can fit 7 1 7ths inside 1, how many 4 7ths can you fit inside 1?

It will be 4 times as... 4 times less. So let me write that down.

So the answer to step 2... The answer to step 2 is going to be 1 fourth as much as the answer in step 1. So once again, perhaps you want to pause the video and think about this for a few moments. This number 4 7, this fraction is 4 times as large as 1 7th. You can fit 7 1 7ths inside 1. gallon, right, you won't be able to fit as many 4 sevenths because 4 sevenths is larger than 1 seventh. And 4 sevenths is how much larger is 4 times as large.

So the answer to step 2 will be 1 fourth as much as the answer in step 1. So hopefully that makes sense. So this implies, the arrow means implies, 1 divided by 4 sevenths So we're interpreting that to mean how many 4 7s can you fit inside one gallon, right? How many 4 7 gallons can you fit inside one gallon?

So that would be 1 4th of the answer to part A. So that would be 1 4th of 7. And then how did we interpret 1 4th of 7? Back in page 1. You know, 2 fifths of 3 was interpreted to be 3 times 2 fifths. So by the same analogy, 1 fourth of 7 would be 7 times 1 fifth.

So 1 fourth of 7, by staying consistent with our interpretation, that's 7 times 1 fourth, just like in page 1. And then how do we do 7 times 1 fourth? Well, 7 is the same as 7 over 1, so we get 7 over 1 times 1 fourth. And now how do you multiply the fractions?

You just multiply the numerators. So 7 times 1 is 7 and 1 times 4 is 4. So the answer to step 2 is 7 over 4. Now, do you notice something interesting about step 1 and step 2? In step 1, we were trying to figure out 1 divided by 1 seventh.

The answer was 7, so it's as though the fraction 1 over 7 got converted into 7, right? In step 2, we're trying to figure out 1 divided by 4 sevenths. Well, what is the answer? The answer is 7 over 4. That's 4 over 7 turned upside down, right? So here the 1 over 7 got turned upside down to 7 over 1, which is just 7. Here the 4 over 7 got turned upside down to 7 over 4. That's interesting, huh?

We had some interpretation going on, but numerically, what we're doing numerically seems pretty straightforward. We're just flipping fractions. Okay, now let's continue.

We're almost done. So step 3, we want to finally figure out the answer to this division problem. That was the problem that is asked in example 6, right? 9 over 11 divided by 4 over 7. So in step 3, we want to figure this out.

And how are we going to compute this? We're going to compute this by interpreting it as the answer to the following question. How many 4 seventh gallons can fit inside 9 elevenths of a gallon?

Okay, so imagine this is 9 elevenths of a gallon. The question is how many 4 seventh gallons can you fit inside there? And this interpretation here is precisely the one that we had up here when we had bucket A and bucket B. So how can we handle step three?

Well, again, the... The answer here is going to be the following, and I'll write this down for your benefit. So, in step 2, how many 4 7's can you fit inside 1? It's this many. Now, how many 4 7's can you fit inside here?

here you're fitting 4 sevenths into one whole. Here you're fitting 4 sevenths into not one whole but 9 elevenths of a whole. So the answer is going to be 9 elevenths of the answer in step 2. So once again, maybe pause the video and think about that for a second. In step 2, you're trying to figure out how many 4 7ths can you fit inside one hole.

But in step 3, you're trying to figure out how many 4 7ths can you fit inside 9 11ths of a hole. So the answer would be 9 11ths of whatever the answer is in step 2. In step 2 you have fitted 4 7th gallons into 1 whole gallon. Now you don't have a 1 whole gallon, you have 9 11ths.

of a gallon. So however many 4 sevenths you could fit inside one whole gallon, if you take 9 11ths of that answer in step two, you now get the answer to step three. So we have 9 11ths of the answer in step two.

How do we write that? Well, we saw this back on page two, right? On page two, we saw that 2 thirds of 4 fifths is 2 thirds times 4 fifths.

So using that same interpretation, 9 11ths of the answer in step 2 would be 9 11ths times the answer in step 2, which is 7 fourths. And what is that? Well, you just multiplied the numerators, so 9 times 7 is 63, and 11 times 4 is 44. Are there any common factors? Well, the common factors of... The factors of 63 are 9 and 7, so 3 and 7, basically.

And 3 and 7 are not factors of 44, so there are no common factors. So that's the answer to step 3. So that's the answer to the problem. But overall, what is happening?

Overall, let's write this down. So overall, we were trying to figure out 9 over 11 divided by 4 over 7. That's what we were trying to figure out. 9 over 11 divided by 4 over 7. So in step 3, we observed that... 9 over 11 divided by 4 over 7. Ignore this 4, that's the page number. So 9 over 11 divided by 4 over 7. What is that?

That is equal to this. 9 over 11 times 7 over 4. And that turns out to be 63 over 44. So notice what has happened. 9 over 11 divided by 4 over 7 in this concrete example is the same as 9 over 11 times... this fraction 4 over 7 flipped over, right? And then that's how we get the answer.

So we will use this as our guide to define the division of rational numbers. So in this particular concrete example using buckets and whatnot we see that 9 over 11 divided by 4 over 7 is the same as 9 over 11 times 4 over 7 flipped over and again literally thousands of such examples have been observed in various areas of science and mathematics and engineering and so we use this as our guide to define division of rational numbers so moving on to the last page page five Top of page 5. First bullet, it says the reciprocal of a non-zero rational number a over b is the number b over a. So here we have a non-zero rational number, which means a can't be 0, right? b cannot be 0 anyway.

Denominator is not allowed. zero anyway but since a over B is nonzero that means a cannot be zero so the reciprocal of a over B is just B over a is just a over B flipped over and now the second bullet says example 6 which we just finished in the pre previous page. Example 6 suggests that dividing by a fraction is equivalent to multiplying by the reciprocal of that fraction.

So that's what we observed. And then it says because of similar observations in other situations, interpretations of science and the necessity of consistency of various arithmetic properties, rational number division is defined as follows. For rational numbers a over b and c over d, where c is not equal to zero, we have a over b divided by c over d.

It's defined to be over b over c over d. times the reciprocal of C over D, which is D over C. So division by a fraction is defined to be multiplication by the reciprocal of the fraction. And that's what you need to remember. So with that, let's do our final example of the lecture.

So part A, we are dividing the rational number of fraction 3 tenths by the rational number of fraction negative 15 over 14. Again, that means the opposite of negative 15 over 40 means the opposite of 15 over 40. Okay, so how do you do this? Now, this kind of fraction division you are expected to be able to do by hand. So you can put this in your calculator to check your final answer, but you should be able to do this by hand. Now, how is division defined? Division is defined to be multiplication by the reciprocal.

So this is defined to be the fraction or rational number 3 tenths times the reciprocal of this. Now, what is the ratio of the fraction to the reciprocal? the reciprocal of this negative rational number? Do we just flip? Well let's think about this.

What does the negative sign in front mean? The negative sign in front means you could put the negative with the 15 the numerator or with the denominator 14, right? Either is fine. We saw this in a previous lecture.

So now don't write this down, but imagine this. Imagine the negative goes with the 15. So you have negative 15 over 14. If you flip that, you will get 14 over negative 15, right? You will get 14 over negative 15, but then the negative can come to the front again. So the overall effect is this. If you take the reciprocal of this, you just get negative 14 over 15. so it's as though the negative doesn't change position at all.

And let me say this one more time. So you can imagine the negative going with the 15 or with the 14. If you put the negative with the 15 up here, you have negative 15 over 14. When you take the reciprocal of that, you'll get 14 over negative 15, but then the negative with the 15 in the denominator now can again come to the front. So it's as though the negative doesn't change position at all, and you're just flipping the 15. and the 14. So we have obeyed the definition here and written the division problem as a multiplication problem.

So you should show this step when you do this by hand. And now what happens to the negative sign in this multiplication problem? Well as we saw before with rational number multiplication you can put that out in front.

And you multiply the numerators, so you have 3 times 14. And you multiply the denominators, so you have 10 times 15. And this step you can skip if you wish. So I'll put skip and then question mark. It's up to you.

You can skip that step. And then in the next step you cancel common factors. So this next step you should not skip. So we have the 3. 14 is 2 times 7. The 10 here is 2 times 5. The 15 is 3 times 5. So now what cancels?

This 2 can cancel this 2. The 3 can cancel the 3. And that's it. Right. So we're left with. Oops sorry you couldn't see that.

We're left with 7 at the top and 5 times 7, so 25 at the bottom. So the result of this division, the fraction or rational number 3 tenths divided by the rational number negative 15 over 14, the result of that division is... division is negative 7 over 25. So we are not interpreting this in a physical situation here like example 6 on the previous page.

We are just doing the brutal computation of division, nothing more. and the answer turns out to be negative 7 over 25. And once again, you are expected to be able to do this by hand, and you can skip this step if you wish, but you should not skip this step where you show the cancellation of common factors, and then you get to the final answer in simplified form. Okay, that's part A.

And then part B, here you're dealing with decimals. So as with back in example, example 5 part d, if you look back to example 5 part d where you're multiplying two decimals, I said you can just use your calculator when you do those homework problems in WebAssign. So same here.

When you see a division problem involving decimals, just use your calculator. So unlike something like this where you are expected to be able to do this by hand, so you must learn that, this one you don't have to do by hand. So just put them in your calculator and you'll get your answer and then you'll enter those answers for the HOMA problem.

consumer have assigned but for this lecture let's just do this by hand so you understand the process so we have negative 0.27 that's the opposite of 0.27 can I write 0.27 as a fraction yeah because it's a terminating decimal so 0.27 is 27 over 100 this is again a elementary school arithmetic that you are supposed to know so the decimal point after seven moves two places when you divide by 100 to get 0.27 when you're dividing that by 1.32 1.32 is 132 over 100. And now division by rational number is defined to be multiplication by the reciprocal. So we have negative 27 over 100 times the reciprocal of 132 over 100, which is 100 over 132. Good. Next, the negative sign can stay in the front. We multiply the numerator, so we get 27 times 100. And we multiply the denominator, so 100 times 132. And this step, you know, you could skip. And now what?

The hundreds cancel, right? They're common. Then we have negative 27. You can factor that as 3 times 9, or 3 times 3 times 3. How about the 132? If you divide 132...

So 132 by 3, what do you get? 3 goes into 13 four times. Then 3 times 4 is 12, and then 1 left, and then 3 goes into 12 four times.

So 132 is 3 times 44. Now the 3s cancel, and you're left with negative 9 over 44. So when you divide this by this, the answer as a fraction is negative 9 over 44. If you want to get the answer as a decimal, you can just put negative 9 over 44 in your calculator, and you do not get a terminating decimal. Well, you do get a terminating decimal. Well, sorry, I'm saying the wrong thing.

I said the correct thing the first time. You do not get a terminating decimal, you get a repeating decimal. So if you put this in your calculator, you will get the following.

Negative 0.20. And it turns out that you have 4, 5, and then 4, 5, 4, 5, and so on. So that turns out to be 0.2045 repeating.

Now, when you do the Hummer problems, as I said, you just put this in your calculator. And if you put that division problem in your calculator, you will get... Get 0.204545 and so on. The problem with that is the calculator only gives you a finite number of decimal places.

So this is something we mentioned before in a previous lecture. So how can you be 100% sure that 4, 5 repeats forever? Maybe after a million decimal places, the 4, 5 changes. to something else like 6, 7, 9, right? So how can you be sure from your calculator that this 4, 5 really does repeat forever?

And the answer is you cannot be sure from a calculator because the calculator will only give you a finite number of decimal places. so to be absolutely 100% sure unfortunately you will have to do this by hand and see that okay we really do get a fraction and fractions are either repeating decimals or terminating decimals so only doing something like this proves to you that yes this 4, 5 repeats forever but again as I said when you do the Homer problems in WebAssign don't worry about things like that if the answer to the problem from your calculator looks like it's being repeated it's unlikely that WebAssign is trying to cheat you you So you can just assume that yes, like the 4-5 really is repeated and you could put a bar over the 4-5 in your answer or write 4-5, 4-5, 4-5, dot, dot, dot. But you will have to see what the instrument does.

to the problem say in WebAssign. Okay so with that we are finally done with lecture 5 and now you can do the homework for lecture 5 in WebAssign so good luck with that

Transcript for:Understanding Rational Numbers Operations

Transcript for:
Understanding Rational Numbers Operations