Mark the correct alternative in the following question:

We have,

$R=\{(m, n): n$ divides $m ; m, n \in \mathbf{N}\}$

As, $m$ divides $m$

$\Rightarrow(m, m) \in R \forall m \in \mathbf{N}$

So, $R$ is reflexive

Since, $(2,1) \in R$ i. e. 1 divides 2

but 2 cannot divide 1 i. e. $(2,1) \notin R$

So, $R$ is not symmetric

Let $(m, n) \in R$ and $(n, p) \in R$. Then,

$n$ divides $m$ and $p$ divides $n$

$\Rightarrow p$ divides $m$

$\Rightarrow(m, p) \in R$

So, $R$ is transitive

Hence, the correct alternative is option (d).