Let n be a fixed positive integer.

Solution:

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

Now, for

aRa ⇒ (a – a) is divisible by n, which is true for any integer a as ‘0’ is divisible by n.

Thus, R is reflective.

Now, aRb

So, (a – b) is divisible by n.

⇒ – (b – a) is divisible by n.

⇒ (b – a) is divisible by n

⇒ bRa

Thus, R is symmetric.

Let aRb and bRc

Then, (a – b) is divisible by n and (b – c) is divisible by n.

So, (a – b) + (b – c) is divisible by n.

⇒ (a – c) is divisible by n.

⇒ aRc

Thus, R is transitive.

So, R is an equivalence relation.