Let $R$ be the equivalence relation in the set $Z$ of integer given by $R=\left\{(a,b):2\text{ divides}a-b\right\}$. Show that relation $R$ is transitive. Write the equivalence class $[0]$ .

$R=\{(a,b):2\text{divides}a-b\}$

Transitive:
If $2$ divides $(a,-b)$, and $2$ divides $(b-c),$
So, $2$ divides $(a-b)+(b-c)$ also
So, $2$ divides $(a-c)$

$\therefore (a-b)\in R$ and $(b-c)\in R$, then $(a-c)\in R$

Therefore, $R$ is transitive.

Now, Equivalence class $[0]$:
 This means that one element is $0$, we need to find other elements which satisfy $R$

So, $(a,0)\in R$ where $a\in Z$
Because given that $R$ is in set $Z$, so both $a$ and $b$ belong to set $Z$

Now,
If $(a,0)\in R$
$2$ divides $a-$
i.e $2$ divides $a$
 
So, Possible values of $a$ are $0,\pm 2,\pm 4,\pm 6,\pm 8,....$
i.e all even numbers and $0$

We use $\pm$ because $2$ can divide $2$ and $-2$
So, Equivalence class $[0]=\{0,\pm 2,\pm 4,\pm 6,....\}$