Show that the relation R on the set Z of integers, given by

Solution:

We observe the following properties of relation R.

Reflexivity:

Let $a$ be an arbitrary element of the set $Z$. Then,

$a \in R$

$\Rightarrow a-a=0=0 \times 2$

$\Rightarrow 2$ divides $a-a$

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

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

Symmetry:

Let $(a, b) \in R$

$\Rightarrow 2$ divides $a-b$

$\Rightarrow \frac{a-b}{2}=p$ for some $p \in Z$

$\Rightarrow \frac{b-a}{2}=-p$

Here, $-p \in Z$

$\Rightarrow 2$ divides $b-a$

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

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

Transitivity:

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

$\Rightarrow 2$ divides $a-b$ and 2 divides $b-c$

$\Rightarrow \frac{a-b}{2}=p$ and $\frac{b-c}{2}=q$ for some $p, q \in Z$

Adding the above two, we get

$\frac{a-b}{2}+\frac{b-c}{2}=p+q$

$\Rightarrow \frac{a-c}{2}=p+q$

Here, $p+q \in Z$

$\Rightarrow 2$ divides $a-c$

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

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

Hence, R is an equivalence relation on Z.