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Relations — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSRelations5 Marks
Question
Let n be a fixed positive integer. Define a relation R on Z as follows:
$(\text{a, b})\in\text{R}\Leftrightarrow\ \text{a}-\text{b}$ is divisible by n. Show that R is an equivalence relation on Z.
✓

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.

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