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(D) $1$. Reflexivity: For every $a \in Z$,$a - a = 0$. Since $n | 0$ is true for any positive integer $n$,$aRa$ holds. Thus,$R$ is reflexive.$2$. Symm

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All of the above $(a), (b),$ and $(c)$

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(D) $1$. Reflexivity: For every $a \in Z$,$a - a = 0$. Since $n | 0$ is true for any positive integer $n$,$aRa$ holds. Thus,$R$ is reflexive.$2$. Symmetry: If $aRb$,then $n | (a - b)$,which means $a - b = nk$ for some integer $k$. Then $b - a = -(a - b) = n(-k)$,which implies $n | (b - a)$. Thus,$bRa$ holds,and $R$ is symmetric.$3$. Transitivity: If $aRb$ and $bRc$,then $n | (a - b)$ and $n | (b - c)$. This implies $a - b = nk_1$ and $b - c = nk_2$ for some integers $k_1, k_2$. Adding these equations,$(a - b) + (b - c) = a - c = n(k_1 + k_2)$. Since $n | (a - c)$,$aRc$ holds,and $R$ is transitive.Conclusion: Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation. Therefore,the correct option is $(D)$.

Let $n$ be a fixed positive integer. The relation $R$ on the set of integers $Z$ is defined by $aRb \Leftrightarrow n | (a - b)$. Then $R$ is:

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