Answer :

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 n.

Now,

__R is Reflexive if (a,a)__ __∈__ __R__ __∀__ __a__ __∈__ __Z__

aRa ⇒ (a-a) is divisible by n.

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

⇒ a-a is divisible by n

⇒ (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 n

⇒ (a-b) = nz for some z ∈ Z

⇒ -(b-a) = nz

⇒ b-a = n(-z) [∵ z ∈ Z ⇒ -z ∈ Z ]

⇒ (b-a) is divisible by n

⇒ (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 n

⇒ a-b = nz_{1} for some z_{1}∈ Z

(b,c) ∈ R ⇒ (b-c) is divisible by n

⇒ b-c = nz_{2} for some z_{2}∈ Z

Now,

a-b = nz_{1} and b-c = nz_{2}

⇒ (a-b) + (b-c) = nz_{1} + nz_{2}

⇒ a-c = n(z_{1} + z_{2} ) = nz_{3} where z_{1} + z_{2} = z_{3}

⇒ a-c = nz_{3} [∵ z_{1},z_{2} ∈ Z ⇒ z_{3}∈ Z]

⇒ (a-c) is divisible by n.

⇒ (a, c) ∈ R

Thus, R is transitive on Z.

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

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