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.


1