# Transitive Relation

Let A be any set. A relation R on A is said to be a transitive relation if and only if,

(a,b) $\in$ R and (b,c) $\in$ R
$\Rightarrow$ (a,c) $\in$ R for all a,b,c $\in$ A.
that means aRb and bRc
$\Rightarrow$ aRc for all a,b,c $\in$ A.

## Examples on Transitive Relation

Example :1 Prove that the relation R on the set N of all natural numbers defined by (x,y) $\in$ R $\Leftrightarrow$ x divides y, for all x,y $\in$ N is transitive.
Solution : Let x, y, z $\in$ N such that (x,y) $\in$ R and (y,z) $\in$ R. Then
(x,y) $\in$ R and (y,z)$\in$ R
$\Rightarrow$ x divides y and y divides z
$\Rightarrow$ there exists p,q $\in$ N such that y = xp and z = yq
$\Rightarrow$ z = (xp)q
$\Rightarrow$ z = x(pq)
$\Rightarrow$ x divides z [ since pq $\in$ N]
$\Rightarrow$ (x,z) $\in$ R
Thus, (x,y) $\in$ R and (y,z)$\in$ R $\Rightarrow$ (x,z) $\in$ R for all x,y,z $\in$ N.
Hence, R is a transitive-relation on N.

Example 2: Let L be the set of all straight lines in a plane. Then the relation 'is parallel to' on L is a transitive-relation .
Solution : According to the relation l1 || l2 and l2 || l3 $\Rightarrow$ l1 || l3
And Let l1,l2,l3 $\in$ L.
$\Rightarrow$ L is a transitive-relation.

Example 3 : If R,S are both transitive,then prove that R∪S is not transitive.
Solution : Let A ={1, 2, 3}, let R = {(1, 2)}, let S = {(2, 1)}.
So R∪S = {(1, 2), (2, 1)}, which is not transitive, because, for instance,
1 is related to 2 and
2 is related to 1
but 1 is not related to 1.
∴ R∪S is not transitive.

12th grade math

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