1.

Let N be the set of natural numbers and relation R on N be defined as R = {(x,y):x, y ∈ N and x divides y}. Examine whether R is reflexive, symmetric & transitive.

Answer»

Relation R is defined as

R = {(x,y) : x, y \(\in\) N and x divides y}

Reflexive:

Let x \(\in\) N and every natural number divides itself.

\(\therefore\) x divides x.

⇒ x R x \(\forall\) x \(\in\) N

\(\therefore\) Relation R is reflexive relation.

Symmetric: \(\because\) 2 divides 6

i.e., (2, 6) \(\in\) R

But 6 does not divide 2

\(\because\) (6, 2) \(\notin\) R

Hence, relation R is not symmetric.

Transitive:

Let x R y, x R z

i.e., x divides y & y divides z.

Let y = ax & z = by where a & b are arbitary constant such that x divides y & y divides z.

⇒ z = by = b(ax)

= ab x

\(\therefore\) x divides z.

⇒ x R z.

So, R is transitive relation.

For reflexive 

x divides x i.e., x Rx, ∀ x ∈ N

R is reflexive

For symmetric

2 divides 6 2R6 i.e., (2,6) ∈ R but (6,2) ∈! R as 6 does not divide 2

Also, R is not symmetric

For transitive 

Let x divides y and y divides z 

i.e., xRy and yRz

kj1x = y & k2y = z (k1,k2 are positive integer)

k1k2x = k1y = z

x divides z 

xRz

So, R is transitive.


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