(D) Since $n|n$ for all $n \in N$,$R$ is reflexive.Since $2|6$ but $6$ does not divide $2$,$R$ is not symmetric.Let $nRm$ and $mRp$. This implies $n|m

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Reflexive,transitive,but not symmetric

(D) Since $n|n$ for all $n \in N$,$R$ is reflexive.Since $2|6$ but $6$ does not divide $2$,$R$ is not symmetric.Let $nRm$ and $mRp$. This implies $n|m$ and $m|p$. Since divisibility is transitive,$n|p$,which implies $nRp$. Therefore,$R$ is transitive.Thus,$R$ is reflexive and transitive,but not symmetric.

Class 12 Mathematics Relation and Function Types of Relations

Medium

On the set $N$ of natural numbers,the relation $R$ is defined by $nRm$ if $n$ 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

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Relation and Function

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Types of Relations

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Medium

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Let $R$ and $S$ be two non-void relations on a set $A$. Which of the following statements is false?

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On the set $N$ of natural numbers,the relation $R$ is defined by $nRm$ if $n$ is a factor of $m$ (i.e.,$n|m$). Then $R$ is:

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