Question

Consider the following relations R= {(x, y): x, y$\in$ R} and x = $\omega$ y, for some rational no. $\omega$
S = {$(\frac{m}{n},\frac{p}{q})$ : m, n, pand q are integers such that n, q $\ne$ 0 and qm = pn}
Then

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

Answer 1

Answer: (C)

S is equivalence but R is not

For R xRy $\Rightarrow x=\omega y$, for reflexive xRx$\Rightarrow x=\omega y$ which is true when $\omega$ = 1
For symmetric consider x=0, y$\ne$ 0, xRy $\Rightarrow$ 0Ry $\Rightarrow$ 0= $\omega$ y, which is true when $\omega$ =0.
Now yRx $\Rightarrow yR0\Rightarrow y=\omega$.0. There is no rational value of $\omega$ for which $y=\omega \times 0\Rightarrow R$ is not symmetric and hence not equivalence.
For S $\frac{m}{n}S\frac{m}{n}\Rightarrow mn=nm\Rightarrow S$ is reflexive
For symmetric let $\frac{m}{n}S\frac{p}{q}\Rightarrow qm=np$
$\frac{p}{q}S\frac{m}{n}\Rightarrow$pn = mq, which is true.
$\Rightarrow$ S is symmetric
For transitive, let$\frac{m}{n}S\frac{p}{q}\Rightarrow$ qm = pn .......(i)
$\frac{p}{q}S\frac{r}{s}\Rightarrow ps=rq$ ......(ii)
From (i) and (ii), we conclude that ms = nr $\Rightarrow \frac{m}{n}S\frac{r}{s}$
$\Rightarrow$ S is transitive
Hence, S is an equivalence relation.