What is transitive relation on set?

Let A be a set in which the relation R defined.

R is said to be transitive, if

(a, b) ∈ R and (b, a) ∈ R ⇒ (a, c) ∈ R,

That is aRb and bRc ⇒ aRc where a, b, c ∈ A.

The relation is said to be non-transitive, if

(a, b) ∈ R and (b, c) ∈ R do not imply (a, c ) ∈ R.

For example, in the set A of natural numbers if the relation R be defined by ‘x less than y’ then

a

Hence this relation is transitive.

Solved

example of transitive relation on set:

**1.** Let k be given fixed positive integer.

Let

R = {(a, a) : a, b ∈ Z and (a – b) is divisible by k}.

Show

that R is transitive relation.

**Solution:**

Given

R = {(a, b) : a, b ∈ Z, and (a – b) is divisible by k}.

Let

(a, b) ∈ R and (b, c) ∈ R. Then

(a, b) ∈ R and (b, c) ∈ R

⇒ (a

– b) is divisible by k and (b – c) is divisible by k.

⇒ {(a

– b) + (b – c)} is divisible by k.

⇒ (a – c) is divisible by k.

⇒ (a, c) ∈ R.

Therefore,

(a, b) ∈ R and (b, c) ∈ R ⇒ (a,

c) ∈ R.

So,

R is transitive relation.

**2.** A relation ρ on the set N is given by “ρ = {(a, b) ∈ N × N : a is divisor of b}”. Examine

whether ρ is transitive or not transitive

relation on set N.

**Solution:**

Given ρ = {(a, b)

∈ N × N : a is divisor of b}.

Let m, n, p ∈ N and (m, n) ∈ ρ and (n, p ) ∈ ρ. Then

(m, n)

∈ ρ and (n, p ) ∈ ρ

⇒ m is divisor of n and n

is divisor of p

⇒ m is divisor of p

⇒ (m, p) ∈ ρ

Therefore, (m, n) ∈ ρ and (n, p) ∈ ρ ⇒ (m, p) ∈ ρ.

So,

R is transitive relation.

● **Set Theory**

● **Sets**

● **Subset**

● **Practice Test on Sets and Subsets**

● **Problems on Operation on Sets**

● **Practice Test on Operations on Sets**

● **Venn Diagrams in Different Situations**

● **Relationship in Sets using Venn Diagram**

● **Practice Test on Venn Diagrams**

**From Transitive Relation on Set to HOME PAGE**

about The Math.