Transitive Relation on Set

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 < b and b < c imply a < c, that is, aRb and bRc ⇒ aRc.

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.