**1 Answer**

written 7.5 years ago by |

Consider any $a,b,c \in Z.$

Since $a-a=0=3.0\Rightarrow(a-a)$ is divisible by 3.

$\Rightarrow(a,a)\in R \Rightarrow$ is reflexive.

Let $(a,b)\in R \Rightarrow (a-b)$ is divisible by 3.

$\Rightarrow a-b=3q$ for some $q \in Z\Rightarrow b-a=3(-q)$

$\Rightarrow (b-a)$ is divisible by 3 $\hspace{8cm}(\because q\in Z \Rightarrow -q \in Z \Rightarrow -q \in Z)$

Thus, $(a,b)\in R \Rightarrow (b,a)\in R \Rightarrow R$ is symmetric.

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

$\Rightarrow(a-b)$ is divisible by 3 and $(b-c)$ is divisible by 3

$\Rightarrow a-b=3q$ and $b-c=3q'$ for some $q,q' \in Z$

$\Rightarrow (a-b)+(b-c)=3(q+q') \Rightarrow a-c=3(q+q')$

$\Rightarrow(a-c)$ is divisible by 3 $\hspace{8cm} (\because q.q' \in Z \Rightarrow q+q' \in Z)$

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

Thus, $(a,b)\in R$ and $(b,c) \in R \Rightarrow (a,c) \in R \Rightarrow R$ is transitive.

Therefore, the relation R is reflexive, symmetric and transitive, and hence it is an equivalence relation.