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

Given that, ∀ a, b ∈Z, aRb if and only if a – b is divisible by 5.

Now,

R is Reflexive if (a,a) ∈ R ∀ a ∈ Z

aRa ⇒ (a-a) is divisible by 5.

a-a = 0 = 0 × 5 [since 0 is multiple of 5 it is divisible by 5]

⇒ a-a is divisible by 5

⇒ (a,a) ∈ R

Thus, R is reflexive on Z.

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

(a,b) ∈ R ⇒ (a-b) is divisible by 5

⇒ (a-b) = 5z for some z ∈ Z

⇒ -(b-a) = 5z

⇒ b-a = 5(-z) [∵ z ∈ Z ⇒ -z ∈ Z ]

⇒ (b-a) is divisible by 5

⇒ (b,a) ∈ R

Thus, R is symmetric on Z.

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

(a,b) ∈ R ⇒ (a-b) is divisible by 5

⇒ a-b = 5z₁ for some z₁∈ Z

(b,c) ∈ R ⇒ (b-c) is divisible by 5

⇒ b-c = 5z₂ for some z₂∈ Z

Now,

a-b = 5z₁ and b-c = 5z₂

⇒ (a-b) + (b-c) = 5z₁ + 5z₂

⇒ a-c = 5(z₁ + z₂ ) = 5z₃ where z₁ + z₂ = z₃

⇒ a-c = 5z₃ [∵ z₁,z₂ ∈ Z ⇒ z₃∈ Z]

⇒ (a-c) is divisible by 5.

⇒ (a, c) ∈ R

Thus, R is transitive on Z.

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