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A relation R is reflexive if for every element $a$ in the set, $(a, a) \in R$. In this case, we need to check if $mRm$ for all $m \in N$. Since every natural number is a multiple of itself, $m$ is a multiple of $m$ for all $m \in N$. Therefore, R is reflexive.
A relation R is symmetric if whenever $(a, b) \in R$, then $(b, a) \in R$. In this case, we need to check if $mRn$ implies $nRm$. That is, if $m$ is a multiple of $n$, then $n$ is a multiple of $m$. Consider $m = 4$ and $n = 2$. $4$ is a multiple of $2$, so $mRn$. However, $2$ is not a multiple of $4$, so $nRm$ is false. Therefore, R is not symmetric.
A relation R is transitive if whenever $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$. In this case, we need to check if $mRn$ and $nRp$ implies $mRp$. That is, if $m$ is a multiple of $n$ and $n$ is a multiple of $p$, then $m$ is a multiple of $p$. If $m$ is a multiple of $n$, then $m = kn$ for some integer $k$. If $n$ is a multiple of $p$, then $n = lp$ for some integer $l$. Substituting $n = lp$ into $m = kn$, we get $m = k(lp) = (kl)p$. Since $kl$ is an integer, $m$ is a multiple of $p$. Therefore, R is transitive.
Final Answer: R is reflexive and transitive but not symmetric.
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