Question

Let $R$ be a relation on the set $N$ of natural numbers defined by ${}_n{R}_m\iff n$ 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

Answer 1

The correct option is D

Reflexive, transitive but not symmetric

Explanation for the correct option:

Step-1: Checking reflexivity:

The relation is defined as

${}_n{R}_m\iff n|m$

Now consider $m=n$

${}_{\Rightarrow n}{R}_n\iff n|n$

Since $n=1\times n$

Therefore $n$ is a factor of itself.

Thus $R$ is reflexive.

Step-2: Checking Symmetricity:

Put $n=2\&m=4$

$\Rightarrow$${}_2{R}_4\iff 2|4$ is true

But ${}_4{R}_2\iff 4\cancel{|}2$

i.e $4$ is not factor of $2$

Thus $R$ is not symmetric.

Step-3: Checking transtivity:

Let ${}_n{R}_m\iff n|m$ then assume $m=nx,\forall x\in N$

And ${}_m{R}_p\iff m|p$ then assume $p=my,\forall y,m\in N$

Now,

$$\begin{gathered} {}_n{R}_p\iff n|p \\ \iff n|my \\ \iff n|nxy \end{gathered}$$

Therefore, $n$ is a factor of $p$

Thus R is transitive,

Hence, option (D) is correct.