Question
Prove that the relation R on Z defined by $(\text{a, b})\in\text{R}\Leftrightarrow\ \text{a}-\text{b}$ is divisible by 5 is an equivalence relation on Z.

Answer

We observe the following properties of relation R.
Reflexivity: Let a be an arbitrary element of R. Then,
⇒ a - a = 0 = 0 × 5
⇒ a - a is divisible by 5
$\Rightarrow\ (\text{a, a})\in\text{R}$ for all $\text{a}\in\text{Z}$
So, R is reflexive on Z.
Symmetry: Let $(\text{a, b})\in\text{R}$
⇒ a - b is divisible by 5
⇒ a - b = 5p for some $\text{p}\in\text{Z}$
⇒ b - a = 5(-p)
Here, $-\text{p}\in\text{Z}$ [Since $\text{p}\in\text{Z}$]
⇒ b - a is divisible by 5
$\Rightarrow\ (\text{b, a})\in\text{R}$ for all $\text{a, b}\in\text{Z}$
So, R is symmetric on Z.
Transitivity: Let a, b and $\text{b},\text{c}\in\text{R}$
⇒ a - b is divisible by 5
⇒ a - b = 5p for some Z
Also, b - c is divisible by 5
⇒ b - c = 5q for some Z
Adding the above two, we get
a - b + b - c = 5p + 5q
⇒ a - c = 5(p + q)
⇒ a - c is divisible by 5
Here, $\text{p}+\text{q}\in\text{Z}$
$\Rightarrow\ \text{a, c}\in\text{R}$ for all $\text{a, c}\in\text{Z}$
So, R is transitive on Z.
Hence, R is an equivalence relation on Z.

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