4 Marks

Question

Let n be a fixed positive integer. Define a relation R on Z as follows: (a, b) a-b is divisible by n. Show that R is an equivalence relation on Z.

Vidyadip · Accepted Answer

We observe the following properties of R.

Reflexivity: Consider $\text{a}\in\text{N}$

Here, a - a = 0 = 0 × n

Implies that a - a is divisible by n

Implies that $\text{a, a}\in\text{R}$

Implies that $\text{a, a}\in\text{R}$ for all $\text{a}\in\text{Z}.$

So, R is reflexive on Z.

Symmetry: Consider $\text{a, b}\in\text{R}$

Here a - b is divisible by n

Implies that a - b = np for some $\text{p}\in\text{Z}$

Implies that b - a = n - p.

Implies that b - a is divisible by n $\big[\text{p}\in\text{Z}$ implies that $-\text{p}\in\text{Z}\big]$

implies that $\text{b, a}\in\text{R}$

So, R is symmetric on Z.

Transitivity: Consider a, b and b, c $\in\text{R}$

Here, a - b is divisible by n and b - c is divisible by n.

implies that a - b = np for some $\text{p}\in\text{Z}$ and b - c = nq for some $\text{q}\in\text{Z}$

Adding the above two

we get a - b + b - c = np + nq

Implies that a - c = n(p + q).

Here, $\text{p}+\text{q}\in\text{Z}$

Implies that $\text{a, c}\in\text{R}$ for all $\text{a, c}\in\text{Z.}$

So, R is transitive on Z.

Hence, R is an equivalence relation on Z.

Answer