Show that the relation R on the set Z of integers, given by R = {(a, b

1.	Show that the relation R on the set Z of integers, given by R = {(a, b) : 2 divides a – b}, is an equivalence relation.
Answer» We have, R = {(a, b) : a – b is divisible by 2; a, b ∈ Z} To prove : R is an equivalence relation Proof : To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive. Reflexivity : For Reflexivity, we need to prove that- (a, a) ∈ R Let a ∈ Z ⇒ a – a = 0 ⇒ a – a is divisible by 2 ⇒ (a, a) ∈ R ⇒ R is reflexive Symmetric : For Symmetric, we need to prove that- If (a, b) ∈ R, then (b, a) ∈ R Let a, b ∈ Z and (a, b) ∈ R ⇒ a – b is divisible by 2 ⇒ a – b = 2p For some p ∈ Z ⇒ b – a = 2 × (–p) ⇒ b – a ∈ R ⇒ R is symmetric Transitive : : For Transitivity, we need to prove that- If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R Let a, b, c ∈ Z and such that (a, b) ∈ R and (b, c) ∈ R ⇒ a – b = 2p(say) and b – c = 2q(say) , For some p, q ∈ Z ⇒ a – c = 2 (p + q) ⇒ a – c is divisible by 2 ⇒ (a, c) ∈ R ⇒ R is transitive Now, since R is symmetric, reflexive as well as transitive- ⇒ R is an equivalence relation.

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

Answer»

We have,

R = {(a, b) : a – b is divisible by 2; a, b ∈ Z}

To prove : R is an equivalence relation

Proof :

To prove that relation is equivalence, we need to prove that it is reflexive, symmetric and transitive.

Reflexivity : For Reflexivity, we need to prove that-

(a, a) ∈ R

Let a ∈ Z

⇒ a – a = 0

⇒ a – a is divisible by 2

⇒ (a, a) ∈ R

⇒ R is reflexive

Symmetric : For Symmetric, we need to prove that-

If (a, b) ∈ R, then (b, a) ∈ R

Let a, b ∈ Z and (a, b) ∈ R

⇒ a – b is divisible by 2

⇒ a – b = 2p For some p ∈ Z

⇒ b – a = 2 × (–p)

⇒ b – a ∈ R

⇒ R is symmetric

Transitive : : For Transitivity, we need to prove that-

If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R

Let a, b, c ∈ Z and such that (a, b) ∈ R and (b, c) ∈ R

⇒ a – b = 2p(say) and b – c = 2q(say) , For some p, q ∈ Z

⇒ a – c = 2 (p + q)

⇒ a – c is divisible by 2

⇒ (a, c) ∈ R

⇒ R is transitive

Now, since R is symmetric, reflexive as well as transitive-

⇒ R is an equivalence relation.