Show that the relation R defined by

Solution:

We observe the following relations of relation R.

Reflexivity:

Let $a$ be an arbitrary element of $R$. Then,

$a-a=0=0 \times 3$

$\Rightarrow a-a$ is divisible by 3

$\Rightarrow(a, a) \in R$ for all $a \in Z$

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

Symmetry:

Let $(a, b) \in R$

$\Rightarrow a-b$ is divisible by 3

$\Rightarrow a-b 3 p$ for some $p \in Z$

$\Rightarrow b-a=3(-p)$

Here, $-p \in Z$

$\Rightarrow b-a$ is divisible by 3

$\Rightarrow(b, a) \in R$ for all $a, b \in Z$

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

Transitivity:

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

$\Rightarrow a-b$ and $b-c$ are divisible by 3

$\Rightarrow a-b=3 p$ for some $p \in Z$

and $b-c=3 q$ for some $q \in Z$

Adding the above two, we get

$a-b+b-c=3 p+3 q$

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

Here, $p+q \in Z$

$\Rightarrow a-c$ is divisible by 3

$\Rightarrow(a, c) \in R$ for all $a, c \in Z$

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

Hence, R is an equivalence relation on Z.