Answer :

(i) Reflexivity: Let a є Z, a - a = 0 є Z which is also even.

Thus, (a, a) є R for all a є Z. Hence, it is reflexive

(ii) Symmetry: Let (a, b) є R

(a, b) є R è a - b is even

-(b - a) is even

(b - a) is even

(b, a) є R

Thus, it is symmetric

(iii) Transitivity: Let (a, b) є R and (b, c) є R

Then, (a – b) is even and (b – c) is even.

[(a - b) + (b - c)] is even

(a - c) is even.

Thus (a, c) є R.

Hence, it is transitive.

Since, the given relation possesses the properties of reflexivity, symmetry and transitivity, it is an equivalence relation.

Rate this question :

Find <a name="_HlRS Aggarwal - Mathematics

Let A be the set RS Aggarwal - Mathematics

If R is a binary RS Aggarwal - Mathematics

Let f : R <span lRS Aggarwal - Mathematics

Let R = (x, y) : RS Aggarwal - Mathematics

Mark the correct RD Sharma - Mathematics

If *R* is a RD Sharma - Mathematics

What is an equivaRS Aggarwal - Mathematics

Let R = {(a, b) :RS Aggarwal - Mathematics

Mark the correct RD Sharma - Mathematics