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(C) Reflexivity: For $x \in Z$, we check if $(x, x) \in R$. $x + 2x = 3x$, which is clearly divisible by $3$. Thus, $xRx$ holds for all $x \in Z$, so

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$R$ is an equivalence relation

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(C) Reflexivity: For $x \in Z$, we check if $(x, x) \in R$. $x + 2x = 3x$, which is clearly divisible by $3$. Thus, $xRx$ holds for all $x \in Z$, so $R$ is reflexive. Symmetry: Let $(x, y) \in R$, which means $x + 2y = 3\lambda$ for some integer $\lambda$. Then $x = 3\lambda - 2y$. We check $y + 2x$: $y + 2x = y + 2(3\lambda - 2y) = y + 6\lambda - 4y = 6\lambda - 3y = 3(2\lambda - y)$. Since $3(2\lambda - y)$ is divisible by $3$, $(y, x) \in R$. Thus, $R$ is symmetric. Transitivity: Let $(x, y) \in R$ and $(y, z) \in R$. Then $x + 2y = 3\lambda$ and $y + 2z = 3\mu$ for some integers $\lambda, \mu$. Adding these: $(x + 2y) + (y + 2z) = 3\lambda + 3\mu \Rightarrow x + 3y + 2z = 3(\lambda + \mu)$. $x + 2z = 3(\lambda + \mu) - 3y = 3(\lambda + \mu - y)$. Since $x + 2z$ is divisible by $3$, $(x, z) \in R$. Thus, $R$ is transitive. Conclusion: Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation.

Let $R$ be a relation defined on the set $Z$ of all integers such that $x R y$ if and only if $x+2y$ is divisible by $3$. Then:

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