Question

Consider the following relations $R=\left\{\right.\left(\right.x,y\left.\right)|x,y$ are real numbers and $x=wy$ for some rational number $w\left.\right\}$
$S=\left\{\right.\left(\frac{m}{n} , \frac{p}{q}\right)$ where $m,n,p$ and $q$ are integers such that $n,q\neq 0$ and $qm=pn\left.\right\}$ . Then

• A. both $R,S$ are equivalence relations
• B. $R$ is an equivalence relation, but not $S$ .
• C. $S$ is an equivalence relation, but not $R$ .
• D. neither $R$ nor $S$ is an equivalence relation.

Answer 1

Answer: (C)

We have,
$R=\left\{\right.\left(\right.x,y\left.\right)|x,y$ are real numbers and $x=wy$ for some rational number $w\left.\right\}$
$R:x=wy$
$\left(0 , a\right)\in R$
$a=wa$
$\Rightarrow w=1$ , but $w\in Q$ that means $a=wa$ is not true for all $w\in Q$ . Hence, $R$ is not reflexive.
And,
$S=\left\{\right.\left(\frac{m}{n} , \frac{p}{q}\right)$ where $m,n,p$ and $q$ are integers such that $n,q\neq 0$ and $qm=pn\left.\right\}$ . $S=\left(\frac{m}{n} , \frac{p}{q}\right)$
$mq=np$
so, $\frac{q}{n}=\frac{p}{m}$
$\frac{m}{n}=\frac{p}{q}$
$\left(\frac{a}{b} , \frac{a}{b}\right)\in S$
so, $S$ is reflexive $S\in \left(\frac{a}{b} , \frac{c}{d}\right)$
$S\in \left(\frac{c}{d} , \frac{a}{b}\right)$
$\frac{c}{d}=\frac{a}{b}$
$bc=ad$
$\therefore $ $S$ is symmetrical
Checking for transitive: $S\in \left(\frac{1}{2} , \frac{2}{4}\right),S\in \left(\frac{2}{4} , \frac{4}{8}\right)$
$S\in \left(a , b\right),S\in \left(b , c\right)$
$S\in \left(a , c\right)$
$\frac{1}{2}=\frac{4}{8}=\frac{1}{2}$
$S\in \left(\frac{1}{2} , \frac{4}{8}\right)$
$S$ is transitive
$S$ is equivalence equation, option $2$ correct