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

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

• 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.