Let R be a relation on the set N of natural n | vedclass

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(D) $1$. Reflexivity: For any $n \in N$,$n$ is a factor of $n$ (since $n = n 	imes 1$). Thus,$nRn$ holds for all $n \in N$. Therefore,$R$ is reflexiv

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Reflexive,transitive but not symmetric

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(D) $1$. Reflexivity: For any $n \in N$,$n$ is a factor of $n$ (since $n = n 	imes 1$). Thus,$nRn$ holds for all $n \in N$. Therefore,$R$ is reflexive.$2$. Symmetry: Consider $n=2$ and $m=6$. Since $2|6$,$2R6$ is true. However,$6$ is not a factor of $2$ $(6 
mid 2)$,so $6R2$ is false. Therefore,$R$ is not symmetric.$3$. Transitivity: Let $nRm$ and $mRp$. This means $n|m$ and $m|p$. By the definition of divisibility,there exist integers $k_1, k_2$ such that $m = nk_1$ and $p = mk_2$. Substituting $m$,we get $p = (nk_1)k_2 = n(k_1k_2)$. Since $k_1k_2$ is an integer,$n|p$,which implies $nRp$. Therefore,$R$ is transitive.Conclusion: $R$ is reflexive and transitive,but not symmetric.

Let $R$ be a relation on the set $N$ of natural numbers defined by $nRm \iff n$ is a factor of $m$ (i.e.,$n|m$). Then $R$ is

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