Let R = {(a, b) : a, b ϵ Z and (a – b) is even}.
Then, show that R is an equivalence relation on Z.
(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.