Question -
Answer -
Given relation R on Z defined by (a, b) ∈ R ⇔ a − b is divisible by 5
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 be an arbitrary element of R. Then,
⇒ a − a = 0 = 0 × 5
⇒ a − a is divisible by 5
⇒ (a, a) ∈ R for all a ∈ Z
So, R is reflexive on Z.
Symmetry:
Let (a, b) ∈ R
⇒ a − b is divisible by 5
⇒ a − b = 5p for some p ∈ Z
⇒ b − a = 5 (−p)
Here, −p ∈ Z [Since p ∈ Z]
⇒ b − a is divisible by 5
⇒ (b, a) ∈ R for all a, b ∈ Z
So, R is symmetric on Z.
Transitivity:
Let (a, b) and (b, c) ∈ R
⇒ a − b is divisible by 5
⇒ a − b = 5p for some Z
Also, b − c is divisible by 5
⇒ b − c = 5q for some Z
Adding the above two equations, we get
a −b + b − c = 5p + 5q
⇒ a − c = 5 ( p + q )
⇒ a − c is divisible by 5
Here, p + q ∈ Z
⇒ (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.