Question

Let the relation $R$ defined on the set of natural numbers $N$ be : $R=\left\{\left(a,b):bisdivisibleby\forall a,b\in N\right)\right\}$ Then $R$ is

• A. Reflexive and symmetric only
• B. Symmetric and transitive only
• C. Reflexive and transitive only
• D. An equivalence relation

Answer 1

The correct option is C

Reflexive and transitive only

Explanation for correct answer:

The correct option is $\left(C\right)$ :Reflexive and transitive only.

The relation $R$ defined on the set of natural numbers $N$ is : $R=\left\{\left(a,b):bisdivisibleby\forall a,b\in N\right)\right\}$

$R$ is reflexive as every natural number is divisible by itself. So $(a,a)\in R$

$R$ is transitive as

$$\begin{gathered} (a,b)\in R\Rightarrow b=ak \\ (b,c)\in R\Rightarrow c=bq \\ \Rightarrow c=a\left(kq\right) \\ \Rightarrow \mathrm{c}\mathrm{is}\mathrm{divisible}\mathrm{by}\mathrm{a} \\ \Rightarrow (\mathrm{a},\mathrm{c})\in \mathrm{R} \end{gathered}$$

$(a,b)\in Rand(b,c)\in R\Rightarrow (a,c)\in R$

Therefore the correct option is $\left(C\right)$ :Reflexive and transitive only.

Explanation for the incorrect answers:

Option A: Reflexive and symmetric only

$R$ is reflexive but $R$ is not symmetric as $(a,b)\in R$does not imply $(b,a)\in R$, i.e., if $b$ is divisible by $a$, then $a$ is not divisible by $b$.

Option B :Symmetric and transitive only

$R$ is transitive but not symmetric as $(a,b)\in R$does not imply $(b,a)\in R$, i.e., if $b$ is divisible by $a$, then $a$ is not divisible by $b$.

Option D: An equivalence relation

$R$ is Reflexive and transitive but not symmetric, so it is not an equivalence relation.

Hence, the option (C) is the correct answer