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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, $\forall a, b \in Z, a R b$ if and only if $a-b$ is divisible by $5 .$

Now,

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

$\mathrm{aRa} \Rightarrow(\mathrm{a}-\mathrm{a})$ is divisible by $5 .$

$a-a=0=0 \times 5[$ since 0 is multiple of 5 it is divisible by 5$]$

$\Rightarrow \mathrm{a}-\mathrm{a}$ is divisible by 5

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

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

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

$(a, b) \in R \Rightarrow(a-b)$ is divisible by 5

$\Rightarrow(\mathrm{a}-\mathrm{b})=5 \mathrm{z}$ for some $\mathrm{z} \in \mathrm{Z}$

$\Rightarrow-(\mathrm{b}-\mathrm{a})=5 \mathrm{z}$

$\Rightarrow \mathrm{b}-\mathrm{a}=5(-\mathrm{z})[\because \mathrm{z} \in \mathrm{Z} \Rightarrow-\mathrm{z} \in \mathrm{Z}]$

$\Rightarrow(\mathrm{b}-\mathrm{a})$ is divisible by 5

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

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

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

$(a, b) \in R \Rightarrow(a-b)$ is divisible by 5

$\Rightarrow a-b=5 z_{1}$ for some $z_{1} \in z$

$(b, c) \in R \Rightarrow(b-c)$ is divisible by 5

$\Rightarrow \mathrm{b}-\mathrm{c}=5 \mathrm{z}_{2}$ for some $\mathrm{z}_{2} \in \mathrm{Z}$

Now,

$a-b=5 z_{1}$ and $b-c=5 z_{2}$

$\Rightarrow(a-b)+(b-c)=5 z_{1}+5 z_{2}$

$\Rightarrow a-c=5\left(z_{1}+z_{2}\right)=5 z_{3}$ where $z_{1}+z_{2}=z_{3}$

$\Rightarrow a-c=5 z_{3}\left[\because z_{1}, z_{2} \in z \Rightarrow z_{3} \in z\right]$

$\Rightarrow(a-c)$ is divisible by $5 .$

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

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

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