Question

Consider the following relations:
$R = ${$(x,y)|x,y$ are real numbers and $x = wy$ for some rational number $w$}
$S = ${$\left( {\dfrac{m}{n},\dfrac{p}{q}} \right):m,n,p,q$ are integers such that $n,q \ne 0,qm = pn$}, then:
A. neither $R$ nor $S$ is an equivalence relation.
B. $S$ is an equivalence relation but $R$ is not an equivalence relation.
C. $R$ and $S$ both are the equivalence relations.
D. $R$ is an equivalence relation but$S$is not an equivalence relation.

Answer 1

Hint: Consider each relation separately and check the type of relations satisfied by the selected relation. Do the same for the second given relation as well. A relation is said to be equivalence if it satisfies to be reflexive, symmetric and transitive.

Complete step-by-step answer:
Here both the sets of $R,S$ are the different relations.
Now if we solve for $R$, it is given that
$R = ${$(x,y)|x,y$ are real numbers and $x = wy$ for some rational number $w$}
Now we know if it is reflexive, then $aRa$exists. Here the relation is given $x = wy$
So for the relation $xRy$=
$x = wy$
Therefore for $xRx \Rightarrow x = wx$
We get $w = 1$ and one is the rational number hence $xRx$ is true so it is reflexive.
For the symmetric relation we know that if $xRy$ and $yRx$ must also be valid.
Now if $x = 0,y \ne 0$
Then $xRy$$ \Rightarrow x = wy$
$0 = wy$
Then $w = 0$ it is true.
Now for $yRx \Rightarrow y = wx$
$y = 0$
But we assume that $y \ne 0$
So it is not symmetric.
Hence it is not an equivalence relation.
Now we know that
$S = ${$\left( {\dfrac{m}{n},\dfrac{p}{q}} \right):m,n,p,q$are integers such that $n,q \ne 0,qm = pn$},
Here $S$ is the relation.
For reflexive $\dfrac{m}{n}S\dfrac{m}{n}$ must be valid
So as $mn = nm$
Hence it is a reflexive relation.
For symmetric
$\dfrac{m}{n}S\dfrac{p}{q} \Rightarrow qm = pn$
Then $\dfrac{p}{q}S\dfrac{m}{n} \Rightarrow pn = mq$
Hence it is symmetric also.
Let $\dfrac{m}{n}S\dfrac{p}{q} \Rightarrow qm = pn$$ - - - - - (1)$
$\dfrac{p}{q}S\dfrac{r}{s} \Rightarrow ps = rq$$ - - - - (2)$
Now if we multiply (1) and (2)
$(mq)(ps) = (pn)(rq)$
$ms = nr$
$\dfrac{m}{n}S\dfrac{r}{s}$
Hence it is transitive.
So it is an equivalence relation.

Hence option B is the correct answer.

Note: If $\{ (1,1),(2,2),(3,3)\} $, then it is a reflexive relation.
If $\{ (1,2),(2,1),(3,1),(1,3)\} $ then it is a symmetric relation.
If $\{ (1,2),(2,3),(1,3)\} $ then it is a transitive relation.
If all three satisfy, then it is an equivalence relation.