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Q.
The relation “congruence modulo $m$” on the set $Z$ of all integers is a relation of type
Relations and Functions - Part 2
Solution:
(i) Let $a \in Z $ then $a - a = 0 = 0 \times m$
$\Rightarrow a - a $ is divisible by $m$
$\Rightarrow a\equiv a $ (mod $m$)
$\Rightarrow R $ is reflexive.
(ii) $a, b \in Z$ such that $a \equiv b$ ( mod $m$)
as $a \equiv b $ (mod $m$)
$\Rightarrow a - b $ is divisible by $m$
$\Rightarrow a - b = \lambda m \forall \lambda \in Z$
$\Rightarrow (b - a) = (- \lambda) m $
$\Rightarrow (b - a) $ is divisible by $m$.
$\Rightarrow b \equiv a $(mod $m$)
$\Rightarrow R$ is symmetric on $Z$.
(iii) Let $a, b, c \in Z$ such that
$a \equiv b$ (mod $m), b \equiv c$ (mod $m$)
$\therefore a \equiv b$ (mod $m$)
$\Rightarrow a - b$ is divisible by $m$.
$\therefore a - b = \lambda_1 m$ for some $\lambda_1, \in Z$
Similarly $b - c = \lambda_2 m$ for some $\lambda_2 \in Z$
By (i) and (ii), we have
$a - c = (\lambda_1 + \lambda_2)m = km$ for some $k \in Z$
$\therefore a - c$ is divisible by $m$
$\therefore a \equiv c $(mod $m$)
$\therefore $ Congruence modulo $m$ is transitive on $Z$.
As the congruence modulo is reflexive, symmetric and transitive so it is an equivalence relation on $Z$.