In order to show R is an equivalence relation we need to show R is Reflexive, Symmetric and Transitive.

Given that, ∀ a, b ∈A, R = {(a, b) : |a – b| is even}.

Now,

R is Reflexive if (a,a) ∈ R ∀ a ∈ A

For any a ∈ A, we have

|a-a| = 0, which is even.

⇒ (a,a) ∈ R

Thus, R is reflexive.

R is Symmetric if (a,b) ∈ R ⇒ (b,a) ∈ R ∀ a,b ∈ A

(a,b) ∈ R

⇒ |a-b| is even.

⇒ |b-a| is even.

⇒ (b,a) ∈ R

Thus, R is symmetric .

R is Transitive if (a,b) ∈ R and (b,c) ∈ R ⇒ (a,c) ∈ R ∀ a,b,c ∈ A

Let (a,b) ∈ R and (b,c) ∈ R ∀ a, b,c ∈ A

⇒ |a – b| is even and |b – c| is even

⇒ (a and b both are even or both odd) and (b and c both are even or both odd)

Now two cases arise:

Case 1 : when b is even

Let (a,b) ∈ R and (b,c) ∈ R

⇒ |a – b| is even and |b – c| is even

⇒ a is even and c is even [∵ b is even]

⇒ |a – c| is even [∵ difference of any two even natural numbers is even]

⇒ (a, c) ∈ R

Case 2 : when b is odd

Let (a,b) ∈ R and (b,c) ∈ R

⇒ |a – b| is even and |b – c| is even

⇒ a is odd and c is odd [∵ b is odd]

⇒ |a – c| is even [∵ difference of any two odd

natural numbers is even]

⇒ (a, c) ∈ R

Thus, R is transitive on Z.

Since R is reflexive, symmetric and transitive it is an equivalence relation on Z.