We have already completed binary addition, binary subtraction and binary multiplication. In this lecture we will learn binary division. We want to divide 101010 by 110 and before this binary division we will first revise division of decimal numbers. We want to divide 128 by 8. So this 128 is our dividend and this 8 is our divisor. We will start our division from the left most digit and in this case we have 1 as the left most digit.
1 is smaller than 8 so quotient is 0. 8 times 0 is 0 and for remainder I will subtract this 0 from 1. 1 minus 0 is 1. This 2 will come down and we have 12. 12 is greater than 8 so quotient is 1. Now we will again. Subtract this 8 from 12 to get the remainder. Remainder is 4. Now this 8 will come down.
So we have 48 and 8 times 6 will give us 48 exact division and remainder is 0. So this is our answer the quotient of this division and we have to find quotient in every division. Now we will move to binary division and follow the same steps. We will start our division from the left most bit, the most significant bit. 1 is smaller than 110. 110 is the divisor and 1 is the left most bit of the dividend.
So we have 0 as the quotient. 110, 0 times will give 0 and remainder is 1. Now the 0 will come and we have one zero still one zero is smaller than one one zero so again quotient is zero and remainder is one zero this one will come down and we have one zero one one zero one is again smaller than one one zero one zero one is five one one zero is six so six is greater than five and again we have quotient equal to zero So remainder is 101 and this 0 will now come down and we have 1010. Now 1010 is greater than 110 and we only have two possibilities. We can multiply 110 by 0 or by 1. So we will multiply 110 by 1 this time and we have 110. To find out remainder we will perform subtraction.
0 minus 0 is 0, 1 minus 1 is 0, 0 minus 1 we have to take the borrow and we have 2 here, 2 minus 1 is 1 and as we have taken the borrow from this position we have 0. So we have 1 0 0 as the remainder and this 1 will now come down and we have 1 0 0 1. So 1 0 0 1 is again greater than 1 1 0. So we will multiply 110 by 1 and again we will perform the subtraction to find out the remainder. 1 minus 0 is 1. 0 minus 1 we have to take the borrow. We have 2 here and when I give borrow to this position I will have 1 and here I have 2. So 2 minus 1 is 1. 1 minus 1 is 0 and here I have 0 after giving borrow. So I have 0. So 11 is the remainder and now this 0, the last bit, the least significant bit will come.
110 we have. 110 is equal to 110. So we have 1 as the multiplier and after subtraction we have 0. So we have complete division and the quotient is 111. You can neglect the 0s because it will make no change and we have 111 as our Now we can definitely cross check our answer 101010 is 42. You can easily find out the decimal equivalent of this binary number. Weight of this position is one, then we have 248 16 and 32. We have 32, we don't have 16, we have 8, we don't have 4, we have 2 and we don't have 1. So 32 plus 8 plus 2 is equal to 42. So dividend is 42 and divisor is 6. 42 divided by 6 is definitely 7. 7 times 6 is 42 and 1 1 1 is 7. So our answer is correct.
And now we will move to homework problem. I have two homework problems. In first homework problem, you have to divide 1 0 0 1 by 1 1. And in second homework problem, you have to divide by 1, 1, 1. So these are the two homework problems and if you have any doubt regarding any part of this presentation you may ask in the comment section.
This is all for this presentation. See you in the next one.