 Hello and welcome to the session. In this session, we discussed the following question that says, let z be the set of all integers and r be the relation on z defined as r is equal to the set containing the elements a being such that a and b belongs to z and a minus b is divisible by 5. So, we have to prove that r is an equivalence relation. So, we have given a relation r defined as this, we have to prove that r is an equivalence relation. First of all, we have a relation r in a set a is first is reflexive if a a belongs to r for every a belongs to the set a. Then the relation r is said to be symmetric if we have a1 a2 belongs to r implies that a2 a1 also belongs to for all a1 a2 belongs to the set a. The relation r transitive belongs to a2 a3 belongs to implies that a1 a3 also belongs to r for all a1 a2 a3 belongs to the set. The next is a relation r in a set a to be reflexive symmetric transitive. This is the key idea that we use in this question. Let's proceed with the solution now. We have given a relation r defined as a set a, b such that a and b belongs to z which is the set of integers and sb is divisible equivalence relation. As we know from the key idea that a relation r in a set a is said to be an equivalence relation if that relation is reflexive symmetric and transitive. And we already know the conditions for reflexive symmetric and transitive. So, let us see if this relation r defined as this is reflexive symmetric and transitive or not. For the check whether r is reflexive or not, r in a set a is said to be reflexive if a a belongs to the relation r for every a belongs to the set a. Now for any to the set of integers that we have is equal to 0 or you can say 0 into 5. So, this means that a minus a divisible by 5 the conditions of the relation r as it belongs to that and a minus a is also divisible by 5. So, this means that a a belongs to the relation and therefore we say that the relation r is reflexive. Now next we see if the relation r is symmetric or not. A relation r in a set a is said to be symmetric if we have a1 a2 belongs to r implies that a2 a1 also belongs to r where a1 and a2 are the elements of the set a. For this we take a and b belongs to the set of integers z and we assume that a b belongs to the relation r. So, this means that a minus b is divisible by 5. That is we can say that a minus b is equal to 5 p where this p belongs to the set of integers z. This means that b minus is equal to 5 into minus p. This shows that b minus a is divisible by a is divisible by 5. So, this means that b a belongs to and therefore we can say that the relation r is symmetric. Now next we have to check if the relation r is considered and see which belongs to the set z of integers and a relation r is said to be transitive if a1 a2 belongs to r, a2 a3 belongs to r implies that a1 a3 belongs to r where these a1 a2 a3 are the elements of the set a. So, if we take a b belongs to r this means that a minus b divisible by which shows that a minus b is equal to 5 p where this p is the element of the set z. Now again consider bc belongs to r. We have that b minus c is divisible by 5 that is b minus c is equal to 5 q where this q is an element of the set of integers z. So, now we have a minus b equal to 5 p and b minus c equal to 5 q then a minus b plus b minus c is equal to 5 p plus 5 q or you can say that a minus b plus b minus c is equal to 5 into p plus q the whole. Now b and minus b cancels so we have a minus c is equal to 5 into p plus q the whole which shows divisible by belongs to but as we have a minus c is divisible by 5. So, this shows that ac belongs to thus we have shown that if we have a b belongs to r bc belongs to this implies that ac belongs to r and c belongs to the set of integers z. This shows that the relation r is transitive. Now finally we have that the relation r is reflexive symmetric transitive therefore we can say that r is a relation. So, hence proved to be it is the session hope you understood the solution of this question.