1.Relation and Function
medium

Let $R$ be a relation on the set $N$ of natural numbers defined by $nRm $$\Leftrightarrow$ $n$ is a factor of $m$ (i.e.,$ n|m$). Then $R$ is

A

Reflexive and symmetric

B

Transitive and symmetric

C

Equivalence

D

Reflexive, transitive but not symmetric

Solution

(d) Since $n | n$ for all $n \in N$, therefore $R$ is reflexive.

Since $2 | 6$ but $6 |2$, therefore $R$ is not symmetric.

Let $n\ R\ m$ and $m\ R\ p$ ==> $n|m$ and $m|p ==> n|p ==> nRp$.

So, $R$ is transitive.

Standard 12
Mathematics

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