Question

Show that the relation R on the set Z of integers, given by R = {(a,b):2divides (a - b)} is an equivalence relation. 

Answer 1

The relation R on Z is given by R = {(a,b) :2divides a - b}.
We observe the following properties of relation R.
Refelxivity : For any a ∈ Z

a - a = 0 = 0 × 2
⇒ 2 divides a - a
⇒  (a, a) ∈  R
So, R is a reflexive relation on Z.

Symmetry: Let a,b ∈ Z be such that
(a,b) ∈  R
⇒ 2 divides a - b
⇒ a - b = 2λ for some  λ ∈ Z
⇒ b -  a = 2(- λ ),where - λ ∈ Z

⇒ 2 divides b -  a 

⇒ (b, a) ∈ R

Thus, (a,b) ∈ R  ⇒ (b, a) ∈ R. So, R is a symmetric relation on Z.
Transitivity: Let a,b, c ∈ Z be such that (a,b) ∈ R and (b, c) ∈ R. Then,

(a,b) ∈ R ⇒ 2 divides a  - b ⇒ a  - b =  2λ for some λ ∈ Z
and (b, c) ∈ R ⇒ 2 divides b - c ⇒ b - c  = 2 μ for some μ ∈ Z
a - b + b - c = 2( λ + μ  )
2 divides a - c
⇒ (a, c) ∈ R
Thus, (a,b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R.
So, R is a transitive relation on Z.
Since R is symmetric and transitive
reflexive therefore an equivalence relation
Hence, R is a transitive relation on Z.