Question

Q23. If R is a relation on set of integers defined as R = {(x, y)/(x - y) is divisible by 5} show that R is an equivalence relation.​

Answer 1

Consider any a,b,c∈Z.

Since a−a=0=3.0⇒(a−a) is divisible by 3.

⇒(a,a)∈R⇒ is reflexive.

Let (a,b)∈R⇒(a−b) is divisible by 3.

⇒a−b=3q for some q∈Z⇒b−a=3(−q)

⇒(b−a) is divisible by 3 (∵q∈Z⇒−q∈Z⇒−q∈Z)

Thus, (a,b)∈R⇒(b,a)∈R⇒R is symmetric.

Let (a,b)∈R and (b,c)∈R

⇒(a−b) is divisible by 3 and (b−c) is divisible by 3

⇒a−b=3q and b−c=3q′ for some q,q′∈Z

⇒(a−b)+(b−c)=3(q+q′)⇒a−c=3(q+q′)

⇒(a−c) is divisible by 3 (∵q.q′∈Z⇒q+q′∈Z)

⇒(a,c)∈R

Thus, (a,b)∈R and (b,c)∈R⇒(a,c)∈R⇒R is transitive.

Therefore, the relation R is reflexive, symmetric and transitive, and hence it is an equivalence relation.

Answer 2

★ Concept :-

Here the concept of Relations has been used. We see that we are given a relation on the set of integers. To begin our answer with, we must understand the set on which the relation is there. Then for proving the relationship as equivalence, Here we shall prove the answer using theoretical method and example method. we need to prove it reflexive, symmetric and then transitive. If the condition is proved, then the relationship is equivalence.

Let's do it !!

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★ Solution :-

Given,

» R = {(x, y) : (x - y) is divisible by 5}

» Set = Integer (Z)

» Set of Integers = Z

We know that, for a Relation to be equivalence, it should be reflexive, symmetric and transitive.

• For Reflexive ::

Here since relation is based on Integers. Then, (x, y) ∈ Z

For the relation,

→ R = {(x, y) : (x - y) is divisible by 5}

to be reflexive, (x, x) ∈ R.

>> Theoretical Prove ::

According to relation,

→ (x - y) ÷ 5

Let y be x . So,

→ (x - x) ÷ 5

→ 0 ÷ 5 = 0

Clearly, (x - x) ÷ 5 = 0 (0 is also an integer)

This means, (x, x) ∈ R .

>> Prove with Example ::

Let x = 5 . Then,

→ (x - x) ÷ 5

→ (5 - 5) ÷ 5 = 0

So, 0 is also an integer.

Thus, (x, x) ∈ R

Hence,

• R is Reflexive Relation .

--------------------------------------------------

• For Symmetric ::

We already know that, (x, y) ∈ Z

For a relation,

→ R = {(x, y) : (x - y) is divisible by 5}

to be symmetric, (y, x) ∈ R .

>> Theoretical Prove ::

We know that,

→ (x - y) / 5 = p (where p ∈ Z , p is the quotient of division)

→ (x - y) = 5p ...(i)

Now, let,

→ (y - x) / 5 = p' (where p' is the quotient of division)

→ (y - x) = 5p' ...(ii)

Here we see that,

||→ (x - y) = -(y - x)

From equation (i) and (ii), we get

||→ p = -p'

Since, p ∈ Z then p' ∈ Z . (Obviously Z is set of integers)

This also proves that, (y, x) ∈ R.

• Prove with Example ::

Let x = 8 and y = 3 .

Here x, y ∈ Z.

Then,

→ (x - y) / 5 = (8 - 3) / 5 = 1 ∈ Z

→ (y - x) / 5 = (3 - 8) / 5 = -1 ∈ Z

Clearly, (y, x) ∈ R

Hence,

• R is a Symmetric Relation .

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• For Transitive ::

We already know that (x, y) ∈ Z . Let (y, z) ∈ Z .

Then, applying this in the relation, we get

→ (x - y) / 5 = p

→ (y - z) / 5 = p'

Adding these both equations, we get

(since denominators of fractions at L.H.S. are equal)

→ [(x - y) + (y - z) / 5] = p + p'

→ (x - y) + (y - z) = 5(p + p')

→ x - y + y - z = 5(p + p')

→ (x - z) = 5(p + p')

→ (x - z) / 5 = (p + p')

Here we see that (x - z) is divisible by 5 giving result as (p + p') . This means that (p + p') is the quotient when (x - z) is divided by 5. Since a, z ∈ Z then (p + p') ∈ Z .

(When integers are divided by integers, the result is integer)

So, (x, z) ∈ R .

>> Prove with Example ::

Let x = 8, y = 3 and z = -2

Here, x, y, z ∈ Z

Then,

→ (x - y)/5 = (8 - 3)/5 = 1 (1 ∈ Z)

→ (y - z)/5 = (3 - (-2))/5 = 1 (1 ∈ Z)

→ (x - z)/5 = (8 - (-2))/5 = 2 (2 ∈ Z)

Here, (1 + 1) = 2

This means (x, z) ∈ R.

Hence,

• R is a Transitive Relation .

Since, R is reflexive, symmetric and transitive, this means that R is a reflexive relation.

Thus,

~~ R is a Reflexive Relation.

Hence proved :D