Question

show the relation

R={(a,b):(a-b) is divisible by 5} in the set of integers is an equivalence relation​

Answer 1

$\large\underline{\sf{Solution-}}$

It is given that

⟼ R = {(a, b): (a – b) is divisible by 5, where a, b ∈ Z}

➢ In order to show that R is an equivalence relation. We have to first show that R is Reflexive, Symmetric and Transitive.

Reflexive :- Let a ∈ S, then (a, a) ∈ R

Symmetric :- If (a, b) ∈ R, then (b, a) ∈ R

Transitive :- If (a, b) ∈ R, (b, c) ∈ R, then (a, c) ∈ R.

Now,

Consider,

$\red{\rm :\longmapsto\:Reflexive}$

Let assume a ∈ Z,

Now,

We know, 0 is divisible by 5.

It implies, a - a is divisible by 5.

⇛ (a, a) ∈ R

Therefore, R is reflexive.

Consider,

$\red{\rm :\longmapsto\:Symmetric}$

Let assume a, b ∈ Z such that (a, b) ∈ R

⇛ a – b is divisible by 5.

⇛ b – a is also divisible by 5

⇛ (b, a) ∈ R

Therefore, R is symmetric.

Consider,

$\red{\rm :\longmapsto\:Transitive}$

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

As (a, b) ∈ R,

⇛ a – b is divisible by 5.

⇛ a – b = 5m -----(1) where m is integer

Also, (b, c) ∈ R

⇛ b – c is divisible by 5

⇛ b – c = 5n ------(2) where n is integer

➢ On adding equation (1) and (2), we get

⟼ a – c + b - c = 5 (m + n)

⇛ a - c = 5 (m + n)

⇛ a – c is divisible by 5.

⇛ (a, c) ∈ R

Therefore, R is transitive.

Since,

• R is reflexive, symmetric and transitive.

Therefore,

R is an equivalence relation.