Question

on the set z of all integers,consider the relation r={(a,b):(a-b) is divisible by 3}.show that r is an equivalence relation on z​.

Answer 1

Given:

Relation r={(a,b):(a-b) is divisible by 3}

Defined on set of all integers Z.

To Prove:

r is an equivalence relation on z.

Solution:

First of all, let us learn what is a equivalence relation. An equivalence relation is a relation which is all three:

1. Reflexive

2. Symmetric and

3. Transitive

Reflexive relation is one which has:

$a\ r\ a \forall a\in z$

i.e. (a,a) will be in relation r for all a in z.

The relation's values for (a,a) = a - a to be divisible by 3.

a - a = 0, 0 is always divisible by 3 so it is reflexive.

Symmetric relation is one which has:

$a\ r\ b \Rightarrow b\ r\ a\forall \{a,b\}\in z$

i.e. $if\ \{(a,b)\} \in r \Rightarrow \{(b,a)\} \in r \forall \{a,b\} \in z.$

The relation's values for (a,b) = a - b to be divisible by 3.

Let a - b = 3x ..... (1)

Then (b,a) $\Rightarrow$ b - a

From (1) b - a = -3x which is again divisible by 3.

So, (b,a) is also an element in r.

It is Symmetric.

Transitive relation is one in which:

$if\ \{(a,b),(b,c)\} \in r \Rightarrow \{(a,c)\} \in r \forall \{a,b,c\} \in z.$

If (a,b) is in r then a-b is divisible by 3

Let a-b = 3p ..... (2)

If (b,c) is in r then b-c is divisible by 3

Let b-c = 3q ..... (3)

Let us add (2) and (3):

a - c = 3p + 3q = 3(p+q)

Hence, a-c is divisible by 3.

$\therefore$ {(a,c)} $\in$ r

It is transitive.

The relation is all three i.e. reflexive, symmetric and transitive.

$\therefore$ the relation is equivalence relation.