Given as R = {(a, b): a – b is divisible by 3; a, b ∈ Z} is a relation
To prove equivalence relation, the given relation should be reflexive, symmetric and transitive.
We have to check these properties on R.
Reflexivity:
Let a be an arbitrary element of R.
Then, a – a = 0 = 0 × 3
⇒ a − a is divisible by 3
⇒ (a, a) ∈ R for all a ∈ Z
Therefore, R is reflexive on Z.
Symmetry:
Let (a, b) ∈ R
⇒ a − b is divisible by 3
⇒ a − b = 3p for some p ∈ Z
⇒ b − a = 3(−p)
Here, −p ∈ Z
⇒ b − a is divisible by 3
⇒ (b, a) ∈ R for all a, b ∈ Z
So, R is symmetric on Z.
Transitivity:
Let (a, b) and (b, c) ∈ R
⇒ a − b and b − c are divisible by 3
⇒ a – b = 3p for some p ∈ Z
And b − c = 3q for some q ∈ Z
Adding two equations above, we get
a − b + b – c = 3p + 3q
⇒ a − c = 3(p + q)
Here, p + q ∈ Z
⇒ a − c is divisible by 3
⇒ (a, c) ∈ R for all a, c ∈ Z
So, R is transitive on Z.
∴ R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on Z.
