Given $S = {(x, y) : y = x + 1 ext{ and } 0 < x < 2}$. Since $x \ne x+1$ for any $x \in (0,2)$, we have $(x, x) \notin S$. Therefore, $S$ is not reflexive, and hence not an equivalence relation. Also $T = {(x, y) : x - y ext{ is an integer }}$. Since $x - x = 0$ is an integer for all $x \in R$, $T$ is reflexive. If $x - y$ is an integer, then $y - x$ is also an integer, so $T$ is symmetric. If $x - y$ is an integer and $y - z$ is an integer, then $(x - y) + (y - z) = x - z$ is also an integer. Therefore, $T$ is transitive. Hence $T$ is an equivalence relation.