Question

Show that the relation R in the set A=(a, b) :(a-b) is even) is an equivalence relation

Answer 1

Answer:

(i) Since a−a=0 and 0 is an even integer ⇒(a,a)∈R

∴R is reflexive

(ii) If (a−b) is even, then (b−a) is also even. Then, if (a,b)∈R⇒(b,a)∈R

∴ The relation is symmetric

(iii) If (a,b)∈R,(b,c)∈R, then a−b and b−c are even

Sum of two even integers is even

So, (a−b+b−c)=(a−c) is even

∴ If (a,b)∈R,(b,c)∈R implies (a,c)∈R

∴ R is transitive

Since R is reflexive, symmetric and transitive, it is an equivalence relation.

Answer 2

Answer:

a−a=0 and 0 is an even integer ⇒(a,a)∈R

∴R is reflexive

(ii) If (a−b) is even, then (b−a) is also even. Then, if (a,b)∈R⇒(b,a)∈R

∴ The relation is symmetric

(iii) If (a,b)∈R,(b,c)∈R, then a−b and b−c are even

Sum of two even integers is even

So, (a−b+b−c)=(a−c) is even

∴ If (a,b)∈R,(b,c)∈R implies (a,c)∈R

∴ R is transitive