Question -
Answer -
Given (a, b) ∈ R ⇔ a − b is divisible by n is a relation R defined on Z.
To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.
Let us check these properties on R.
Reflexivity:
Let a ∈ N
Here, a − a = 0 = 0 × n
⇒ a − a is divisible by n
⇒ (a, a) ∈ R
⇒ (a, a) ∈ R for all a ∈ Z
So, R is reflexive on Z.
Symmetry:
Let (a, b) ∈ R
Here, a − b is divisible by n
⇒ a − b = n p for some p ∈ Z
⇒ b − a = n (−p)
⇒ b − a is divisible by n [ p ∈ Z⇒ − p ∈ Z]
⇒ (b, a) ∈ R
So, R is symmetric on Z.
Transitivity:
Let (a, b) and (b, c) ∈ R
Here, a − b is divisible by n and b − c is divisible by n.
⇒ a − b= n p for some p ∈ Z
And b−c = n q for some q ∈ Z
a – b + b − c = n p + n q
⇒ a − c = n (p + q)
⇒ (a, c) ∈ R for all a, c ∈ Z
So, R is transitive on Z.
Therefore R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on Z.