Verified Solution • 2025–26 Session moderate

QJEEM10Ma3665: JEE Main 2010 Mathematics - Class 11 JEEM10Ma3665 Solution: Consider the following relations are real numbers and for some…

Consider the following relations

R={(x,y)∣x,yR=\{(x, y) \mid x, y are real numbers and x=wyx=w y for some rational number w}w\} ;

S={(mn,pq)∣m,n,pS=\left\{\left(\frac{m}{n}, \frac{p}{q}\right) \mid m, n, p\right. and qq are integers such that n,q≠0n, q \neq 0 and qm=pm}q m=p m\} . Then

Multiple Choice Options

  • A RR is an equivalence relation but SS is not an equivalence relation
  • B Neither RR nor SS is an equivalence relation
  • C SS is an equivalence relation but RR is not an equivalence relation
  • D RR and SS both are equivalence relations

Step-by-Step Verified Solution

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

Relation R : R=(x,y)∣x,yR=(x, y) \mid x, y are real numbers and x=wyx=w y for some rational number ww .

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

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

  • Transitivity : If x=wyx = w y and y=vzy = v z for some rational numbers ww and vv , then x=(wv)zx = (w v)z . Since the product of rational numbers is rational, if xx is related to yy and yy is related to zz , then xx is related to zz .

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

Relation S : S={(mn,pq)∣m,n,pS=\left\{\left(\frac{m}{n}, \frac{p}{q}\right) \mid m, n, p\right. and qq are integers such that n,q≠0n, q \neq 0 and qm=pm}q m=p m\}

  • Reflexivity : For all mn\frac{m}{n} , mn=mn\frac{m}{n} = \frac{m}{n} . Since n≠0n \neq 0 and m=mm = m , every element is related to itself.

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

  • Transitivity :

    mnRpq and pqRrs⇒mq=np and ps=rq⇒mq⋅ps=np⋅rq⇒ms=nr⇒mn=rs⇒mnRrs\begin{array}{rlr}\frac{m}{n} R \frac{p}{q} \text { and } \frac{p}{q} R \frac{r}{s} \\\\\Rightarrow m q=n p \text { and } p s=r q \\\\\Rightarrow m q \cdot p s=n p \cdot r q \\\\\Rightarrow \quad m s=n r \\\\\Rightarrow \frac{m}{n}=\frac{r}{s} \Rightarrow \frac{m}{n} R \frac{r}{s}\end{array}

    So if mn\frac{m}{n} is related to pq\frac{p}{q} and pq\frac{p}{q} is related to rs\frac{r}{s} , then mn\frac{m}{n} is related to rs\frac{r}{s} . The relation SS is transitive.

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

In conclusion, the correct answer is

Option C : SS is an equivalence relation but RR is not an equivalence relation.

Related Learning Resources & Exam Preparation