Question

Consider the following relations:
$R=\left\{\right(x,y)\mid x,y$ are real numbers and $x=wy$ for some rational number $w\}$;
$S=\left\{\left(\frac{m}{n},\frac{p}{q}\right)\right\}:m,n,p$ 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 relation

Answer 1

The correct option is C

$S$ is an equivalence relation but $R$ is not an equivalence relation

Step 1. Check the relation $R$ :

$R=\left\{\right(x,y)\mid x,y$ are real numbers and $x=wy$ for some rational number $w\}$;

We have$(x,x)\in R$for $w=1$ implying that $R$ is reflexive. For $a\ne 0,(a,0)\notin R$for any $w$ but $(0,a)\in R.$ Thus $R$ is not symmetric i.e. $xRy$need not implies $yRx$. Thus, $R$ is not an equivalence relation

Step2. Check the relation S :
$S=\left\{\left(\frac{m}{n},\frac{p}{q}\right)\right\}:m,n,p$ and $q$ are integers such that $n,q\ne 0$and $qm=pn\}.$

As $\left(\frac{m}{n},\frac{m}{n}\right)\in S$ since $mn=mn$,$S$ is reflexive.

$\left(\frac{m}{n},\frac{p}{q}\right)\in S\Rightarrow qm=pn.$

But this can be written as $np=mq$,
giving $\left(\frac{p}{q},\frac{m}{n}\right)\in S$. Thus $S$is symmetric.

Again $\left(\frac{m}{n},\frac{p}{q}\right)\in S,\left(\frac{p}{q},\frac{a}{b}\right)\in S$

means $qm=pnandbq=aq.$

Thus $\left(\frac{m}{n},\frac{a}{b}\right)\in S$

This means $S$is transitive.

Hence the correct option is C.