$\mathrm{R}$ is reflexive, as $2$ divides $(\mathrm{a}-\mathrm{a})$ for all $\mathrm{a} \in \mathrm{Z}$. Further, if $(\mathrm{a}, \mathrm{b}) \in \mathrm{R} ,$ then $2$ divides $\mathrm{a}-\mathrm{b}$. Therefore, $2$ divides $\mathrm{b} – \mathrm{a}$. Hence, $(\mathrm{b}, \mathrm{a}) \in \mathrm{R}$, which shows that $\mathrm{R}$ is symmetric. Similarly, if $(\mathrm{a}, \mathrm{b}) \in \mathrm{R}$ and $(\mathrm{b}, \mathrm{c}) \in \mathrm{R} ,$ then $\mathrm{a}-\mathrm{b}$ and $\mathrm{b}-\mathrm{c}$ are divisible by $2$. Now, $\mathrm{a}-\mathrm{c}=(\mathrm{a}-\mathrm{b})+(\mathrm{b}-\mathrm{c})$ is even (Why?). So, $(\mathrm{a}-\mathrm{c})$ is divisible by $2 .$ This shows that $\mathrm{R}$ is transitive. Thus, $\mathrm{R}$ is an equivalence relation in $\mathrm{Z}$.