The teacher hasn't uploaded a solution for this question yet.
A relation $R$ is reflexive if $a R a$ for all $a \in N$. Here, $m R n$ means $m$ is a multiple of $n$. Since every natural number $m$ is a multiple of itself (as $m = 1 \times m$), $m R m$ holds true for all $m \in N$. Thus, R is reflexive.
A relation $R$ is symmetric if $m R n \implies n R m$. Let $m=4$ and $n=2$. Here, $4 R 2$ is true because 4 is a multiple of 2. However, $2 R 4$ is false because 2 is not a multiple of 4. Since $m R n$ does not imply $n R m$, R is not symmetric.
A relation $R$ is transitive if $m R n$ and $n R p \implies m R p$. If $m$ is a multiple of $n$, then $m = kn$ for some $k \in N$. If $n$ is a multiple of $p$, then $n = jp$ for some $j \in N$. Substituting $n$, we get $m = k(jp) = (kj)p$. Since $kj \in N$, $m$ is a multiple of $p$. Thus, R is transitive.
Final Answer: The relation R is reflexive and transitive, but not symmetric.
AI generated content. Review strictly for academic accuracy.