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For any $x \in Z$, we observe that $(x - x) = 0$. Since $0$ is divisible by any positive integer $n$ (as $0 = n \times 0$), the pair $(x, x) \in R$. Thus, $R$ is reflexive.
Let $(x, y) \in R$. This implies $(x - y) = kn$ for some integer $k$. Then, $(y - x) = -(x - y) = -(kn) = (-k)n$. Since $-k$ is an integer, $(y - x)$ is also divisible by $n$. Thus, $(y, x) \in R$, and $R$ is symmetric.
Let $(x, y) \in R$ and $(y, z) \in R$. Then $(x - y) = kn$ and $(y - z) = mn$ for some integers $k, m$. Adding these equations: $$(x - y) + (y - z) = kn + mn$$ $$(x - z) = (k + m)n$$ Since $(k + m)$ is an integer, $(x - z)$ is divisible by $n$. Thus, $(x, z) \in R$, and $R$ is transitive.
Final Answer: Yes, R is an equivalence relation as it satisfies reflexivity, symmetry, and transitivity.
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