Question

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

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

Answer 1

The correct option is C $S$ is an equivalence relation but $R$ is not an equivalence relation

R: $x=wy$ for a function to be a reflexive function

$(a,a)\in R\Rightarrow a=\omega a$

$w=1$ (only) $\therefore R$ is not a reflexive function.

$\begin{array}{cc}S:\{(\frac{m}{n},\frac{p}{q})& mq=np\\ \Rightarrow \frac{q}{n}& =\frac{p}{m}\Rightarrow \frac{m}{n}=\frac{p}{q}\end{array}$

If $(\frac{a}{b},\frac{a}{b})\in S$, then $S$ will be reflexive function If $S\in (\frac{a}{b},\frac{c}{d})$ and $\in (\frac{c}{d},\frac{a}{b})\Rightarrow S$ is symmetric for a function to be transitive,

$\begin{array}{cc}S\in (\frac{1}{2},\frac{2}{4}),& S\in (\frac{2}{4},\frac{4}{8})\\ s\in (a,b),S\in (b,c)& \Rightarrow S\in (a,c)\\ \therefore \frac{1}{2}=\frac{4}{8}=\frac{1}{2}& \Rightarrow S\in (\frac{1}{2},\frac{4}{8})\\ & \therefore \text{S is transitive}\end{array}$

$\begin{array}{c}\text{Hence,}S\text{is an equivalence relation.}\\ \Rightarrow \text{(c) is the correct option}\end{array}$