Show that the relation $R$ in the set $Z$ of integers given by $R = \{(a, b) : 2 \text{ divides } a - b\}$ is an equivalence relation.

(N/A) relation $R$ is an equivalence relation if it is reflexive,symmetric,and transitive.
$1$. Reflexive: For any $a \in Z$,$a - a = 0$. Since $2$ divides $0$,$(a, a) \in R$. Thus,$R$ is reflexive.
$2$. Symmetric: Let $(a, b) \in R$. This means $2$ divides $a - b$,so $a - b = 2k$ for some integer $k$. Then $b - a = -(a - b) = -2k = 2(-k)$. Since $-k$ is an integer,$2$ divides $b - a$. Thus,$(b, a) \in R$,so $R$ is symmetric.
$3$. Transitive: Let $(a, b) \in R$ and $(b, c) \in R$. This means $a - b = 2k_1$ and $b - c = 2k_2$ for some integers $k_1, k_2$. Adding these,$(a - b) + (b - c) = 2k_1 + 2k_2$,which simplifies to $a - c = 2(k_1 + k_2)$. Since $k_1 + k_2$ is an integer,$2$ divides $a - c$. Thus,$(a, c) \in R$,so $R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.