Prove that the relation R on Z defined by

Solution:

We observe the following properties of relation R.

Reflexivity:

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

$\Rightarrow a-a=0=0 \times 5$

$\Rightarrow a-a$ is divisible by 5

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

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

Symmetry:

Let $(a, b) \in R$

$\Rightarrow a-b$ is divisible by 5

$\Rightarrow a-b=5 p$ for some $p \in Z$

$\Rightarrow b-a=5(-p)$

Here, $-p \in Z$                [Since $p \in Z]$

$\Rightarrow b-a$ is divisible by 5

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

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

Transitivity:

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

$\Rightarrow a-b$ is divisible by 5

$\Rightarrow a-b=5 p$ for some $Z$

Also, $b-c$ is divisible by 5

$\Rightarrow b-c=5 q$ for some $Z$

Adding the above two, we get

$a-b+b-c=5 p+5 q$

$\Rightarrow a-c=5(p+q)$

$\Rightarrow a-c$ is divisible by 5

Here, $p+q \in Z$

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

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

Hence, R is an equivalence relation on Z.