A = {1, 2, 3, 4, 5} 

R = {(a, b) : |a – b| is even}

R = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (1, 3), (3, 1), (1, 5), (5, 1), (2, 4), (4, 2), (3, 5), (5, 3)} 

Reflexive: As |a – a| = 0 (an even number) 

(a, a) ∈ R So R is reflexive. 

Symmetric: (a, b) ∈ R 

|a – b| is even = |-(b – a)| is even = |b – a| is even = (b, a) ∈ R

So R is symmetric. 

Transitive: (a, b) ∈ R, (b, c) ∈ R 

⇒ |a – b| is even and |b – c| is even 

⇒ |a – b| is even and (b – c) is even 

= (a – b + b – c) is even = (a – c) is even = |a – c| is even = (a, c) ∈ R 

So R is transitive. 

R is reflexive, symmetric and transitive, so R is an equivalence relation. as 

|1 – 3|, |3 – 1|, |3 – 5|, |5 – 3| and |1 – 5|, |5 – 1| are even numbers. 

Therefore all elements of {1, 3, 5} are related to each other. 

Also |2 – 4|, |4 – 2| is even. So all elements of {2, 4} are related to each other. |1 – 2|, |2 – 1|; |1 – 4|, |4 – 1|; |3 – 2|, |2 – 3|; |3 – 4|, |4 – 3|; |5 – 2|, |2 – 5|; |5 – 4|, |4 – 5| are not even. 

So no element of {1, 3, 5} is related to any element of {2, 4}.