 Hello friends, so welcome again to this session on real numbers. We are dealing with Euclid's algorithm, Euclid's division algorithm. We have seen how to find out the GCT of two positive integers. We also understood the geometrical interpretation of Euclid's division algorithm. We understood the meaning of coprimes. We also learned the process of finding GCT using Euclid's division algorithm. In this session, we are going to prove a theorem and the theorem is that if A and B are positive integers such that A equals BQ plus R, which you would have seen that it is resembling our Euclid's division lemma, then every common divisor of A and B is a common divisor of B and R and vice versa. Little complicated, isn't it? But don't worry, we'll try and break this up into simpler terms and smaller parts so that we could understand it properly. So what does the theorem say? It says that if A and B are positive integers, so no problem in understanding that these are two positive integers such that A equals BQ plus R. So you have already seen that in our Euclid's division lemma, A is equal to BQ plus R where A and B are two positive integers and Q and R or rather A and B are integers also. That's not a problem but we are in this chapter, we are going to take into consideration only positive integers. So A and B are positive integers and for every pair of A and B, we get unique pair of Q and R such that A equals BQ plus R and R was less than, sorry, R was greater than equal to 0 and less than B. That's what we have learned so far. So this was our Euclid's division lemma. So hence, what does it say? So it says that A and B are positive integers such that A equals BQ plus R, if you have ascertained this, then every common divisor of A and B, that is some divisor which is dividing A as well as B will also be a common divisor of B and R. So let's say if just for an example, if 5 divides A as well as 5 divides B, that means 5 is a common divisor of A and B, if that is true and if this holds, then 5 will also divide B and R together. That means 5 will be common divisor of B and R as well. So let us take an example and understand what does it mean. So for example, it says that let us take our common example which we have been using so far, 42 and 30, that is 42 is A and 30 is B, then I can express 42S, 42 equals 30 into 1 plus 12. That's what we said. And then we know that if there is a common divisor of 42 and 30, for example, let's say 2 divides 42 as well as 2 divides 30. So 2 is a common divisor of 42 and 30. So if that is so, then 2 will also be a common divisor of 30 as well as 12. So clearly if you see 2 divides both 30 and 12. Another example, let's say 4. Now 4 is not 3, let's say 3. So 3 divides 42 as well as 3 divides 30. So 3 is a common divisor of 42 and 30. So if you see 3 also is a common divisor of 30 as well as 12. So remainder and the divisor will have the same common divisor which the divisor and the dividend are having. So that's what this particular theorem says. Let us take another example. Let me take 25 or rather 250 and let's say 60, 250 and 60. These are two integers. So this is example number 2. This was example number 1, isn't it? Now if you see 250 and 60. So what is a common divisor for both? Again, if 2 is a common divider clearly, so 2 divides 250 and 2 also divides 60. Now let us express in Euclid's division lemma form. So 250 is clearly 60 into 4 plus 10. So if 2 divides 250 and 2 divides 60, so 2 will divide 60 as well as 2 will divide 10. So this is holding true so far. Now another example, 5. 5 is a common divisor to both. So 5 divides 250. That is a. 5 divides 60 also. That is b. So you'll see 5 will divide 60 as well as the remainder which is true. So everywhere we are seeing that this particular theorem is holding true, right? Now the question arises, how do we prove it? So let's prove this theorem. Okay friends, so now let us try and prove this theorem. So we saw two examples and now it's time to prove this theorem. So how do we prove it? So it says that if a and b are positive integers such that a is equal to bq plus r and every common divisor of a and b is a common divisor of b and r and yc versus r. So we'll have to prove that. If let us say c divides a, c is an integer which divides a and c also divides b, then c will divide r as well. c will divide r as well. That's what we have to prove. Okay? So how do we prove? So this is how the proof should look like. So let us say, let us say since there's a common divisor, let's say c is a c is a common divisor, common, common divisor, divisor of a and b. So by divisibility's definition we know a will be equal to k1c and b will be equal to k2c where k1 and k2 are positive, positive integers. Yeah? If c divides a, then there must be integer k1 such that ck1 is a and if c divides b, then there must be another integer k2 such that ck2 is b. Now, so from here we see from and we also know that a is equal to bq plus r, isn't it? So this implies I can say r will be equal to a minus b times q. Now from these two expressions, I can write a as k1c and I can write b as k2c times q. This implies c, I can take common and this is k1 minus k2c, k2q. Sorry. Yeah? What do we see? We see that c, c times an integer. Is it an integer? Yes, this one is an integer. Why is it an integer? Because k1, k2q, all are integers. So you know that operations on integers will give you integers. So multiplication and subtraction operation will give you an integer. So we can say that c also divides r because r was c times integer. So c divides r. Hence, we could prove that yes, when c divides a and c divides b, then c will also divide r. That what did we prove? We proved that if c divides a and c divides b and a is equal to bq plus r, then c divides, then c also divides r. Now we will have to prove the vice versa part of it. Right? So what is it? Now we have to prove that if there is a common divisor of what? b and r, b and r, then it would be the common divisor of a and b as well if a can be expressed as bq plus r. So that's what it says now. So let us prove this. So what can we say again? So if let's say c divides b and c divides r. Let us say this is the given case. That means I can write by divisibility's definition. b is equal to again, let's say some k3 times c and r can be expressed as k4 times c where k3 and k4 are integers. It can or positive integers are positive integers. So that means in set theory notation I can say k3 belongs to z plus and k4 also belongs to z plus. Correct? Now this is given. Now also what is given is a equals bq plus r. Correct? So I can write from these two expressions. I can write b was k3cq plus k4c. That means I can take c common and I can write k3q plus k4. Again if you see, again if you see b is this part. This is an integer. Why? Because k3q and k4 are all integers and there is a multiplication and addition operation happening on integers. So hence it is an integer. So we can say c divides a. So c was already dividing b and c was dividing r. Now c is also dividing a. That means the vice versa of the converse or the real theorem, the original theorem is also true. This is what was the objective of this theorem. Now this particular theorem if you see becomes the premise of the Euclidean division algorithm to find GCD. So thanks for watching this video. Hope it was useful to you. We will be coming back with the next video soon. Thanks.