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In order to show $R$ is an equivalence relation, we need to show $R$ is Reflexive, Symmetric and Transitive.

Given that, $\forall a, b \in Z, R=\{(a, b):(a+b)$ is even $\}$

Now,

$\underline{R}$ is Reflexive if $(a, a) \in \underline{R} \underline{\forall} \underline{a} \in \underline{Z}$

For any $a \in A$, we have

$a+a=2 a$, which is even.

$\Rightarrow(a, a) \in R$

Thus, $R$ is reflexive.

$\underline{R}$ is Symmetric if $(a, b) \in \underline{R} \Rightarrow \underline{(b, a)} \in \underline{R} \underline{\forall} \underline{a, b} \in \underline{Z}$

$(a, b) \in R$

$\Rightarrow a+b$ is even.

$\Rightarrow \mathrm{b}+\mathrm{a}$ is even.

$\Rightarrow(b, a) \in R$

Thus, $R$ is symmetric.

$\underline{R}$ is Transitive if $(a, b) \in \underline{R}$ and $(b, c) \in \underline{R} \Rightarrow \underline{(a, c)} \in \underline{R} \underline{\forall} \underline{a}, b, c \in \underline{Z}$

Let $(a, b) \in R$ and $(b, c) \in R \forall a, b, c \in Z$

$\Rightarrow a+b=2 P$ and $b+c=2 Q$

Adding both, we get

$a+c+2 b=2(P+Q)$

$\Rightarrow \mathrm{a}+\mathrm{c}=2(\mathrm{P}+\mathrm{Q})-2 \mathrm{~b}$

$\Rightarrow \mathrm{a}+\mathrm{c}$ is an even number

$\Rightarrow(\mathrm{a}, \mathrm{c}) \in \mathrm{R}$

Thus, $R$ is transitive on $Z$.

Since $R$ is reflexive, symmetric and transitive it is an equivalence relation on $Z$.