JEE Main 2010 — Sets And Relations Question with Solution

Let's evaluate each relation for the properties of an equivalence relation: reflexivity, symmetry, and transitivity.

Relation R : $R=(x,y)\mid x,y$ are real numbers and $x=wy$ for some rational number $w$.

• Reflexivity : For all $x$ in $R$, $x=1x$. Since 1 is a rational number, every element is related to itself.


• Symmetry : For all $x,y$ in $R$, if $x=wy$ for some rational $w$, then $y=\frac{1}{w}x$. However, if $w=0$, then $\frac{1}{w}$ is undefined, and therefore, $R$ doesn't satisfy symmetry.


• Transitivity : If $x=wy$ and $y=vz$ for some rational numbers $w$ and $v$, then $x=(wv)z$. Since the product of rational numbers is rational, if $x$ is related to $y$ and $y$ is related to $z$, then $x$ is related to $z$.

Therefore, $R$ is not an equivalence relation on $R$ since it does not satisfy the symmetry property.

Relation S : $S=\{\left(\frac{m}{n},\frac{p}{q}\right)\mid m,n,p$ and $q$ are integers such that $n,q\ne 0$ and $qm=pm\}$

• Reflexivity : For all $\frac{m}{n}$, $\frac{m}{n}=\frac{m}{n}$. Since $n\ne 0$ and $m=m$, every element is related to itself.


• Symmetry : For all $\frac{m}{n}$, $\frac{p}{q}$, if $qm=pn$, then $np=mn$. So if $\frac{m}{n}$ is related to $\frac{p}{q}$, then $\frac{p}{q}$ is related to $\frac{m}{n}$.


• Transitivity :
  
  $$\begin{array}{r}\frac{m}{n}R\frac{p}{q}\text{ and }\frac{p}{q}R\frac{r}{s}\\ \\ \Rightarrow mq=np\text{ and }ps=rq\\ \\ \Rightarrow mq\cdot ps=np\cdot rq\\ \\ \Rightarrow ms=nr\\ \\ \Rightarrow \frac{m}{n}=\frac{r}{s}\Rightarrow \frac{m}{n}R\frac{r}{s}\end{array}$$
  
  So if $\frac{m}{n}$ is related to $\frac{p}{q}$ and $\frac{p}{q}$ is related to $\frac{r}{s}$, then $\frac{m}{n}$ is related to $\frac{r}{s}$. The relation $S$ is transitive.

Therefore, $S$ is an equivalence relation on the set of all fractions where denominator is not zero.

In conclusion, the correct answer is

Option C : $S$ is an equivalence relation but $R$ is not an equivalence relation.