Let n be a fixed positive integer. Define a r | vedclass

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(D) relation $R$ is defined on the set of integers $Z$ as $aRb \Leftrightarrow n | (a - b)$.$1$. Reflexive: For any $a \in Z$,$(a - a) = 0$. Since $n

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(D) relation $R$ is defined on the set of integers $Z$ as $aRb \Leftrightarrow n | (a - b)$.$1$. Reflexive: For any $a \in Z$,$(a - a) = 0$. Since $n | 0$ for any positive integer $n$,$aRa$ holds. Thus,$R$ is reflexive.$2$. Symmetric: If $aRb$,then $n | (a - b)$,which means $(a - b) = nk$ for some integer $k$. Then $(b - a) = -(a - b) = -nk = n(-k)$. Since $-k$ is an integer,$n | (b - a)$,so $bRa$. Thus,$R$ is symmetric.$3$. Transitive: 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,$(a - c) = (a - b) + (b - c) = n(k_1 + k_2)$. Since $(k_1 + k_2)$ is an integer,$n | (a - c)$,so $aRc$. Thus,$R$ is transitive.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. Define a relation $R$ on the set $Z$ of integers by $aRb \Leftrightarrow n | (a - b)$. Then $R$ is

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