Question

The relation “congruence modulo $m$” on the set $Z$ of all integers is a relation of type

• A. reflexive only
• B. symmetric only
• C. transitive only
• D. equivalence

Answer 1

Answer: (D)

(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 $Z$.