Consider the following relations R={(x,y) | x,y are real numbers and x=wy for some relational number w};S=mn,pq/m,n,p and q are integers such that n,q≠0 and qm=pn. Then

xRy need not imply yRx.mnSpq⇔qm=pnmnSmn⇒S is reflexivemnSpq⇒pqSmn⇒S is symmetricmnSpq,pqSrs⇒qm=pn,ps=rq⇒ms=rn⇒mnSrs⇒R is transitiveHence, S is an equivalence relation

Consider the following relations R={(x,y) | x,y are real numbers and x=wy for some relational number w};S=mn,pq/m,n,p and q are integers such that n,q≠0 and qm=pn. Then

xRy need not imply yRx.mnSpq⇔qm=pnmnSmn⇒S is reflexivemnSpq⇒pqSmn⇒S is symmetricmnSpq,pqSrs⇒qm=pn,ps=rq⇒ms=rn⇒mnSrs⇒R is transitiveHence, S is an equivalence relation

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Consider the following relations $R = {(x, y) | x, y a r e r e a l n u m b e r s a n d x = w y f o r s o m e r e l a t i o n a l n u m b e r w};$
$S = \{(\frac{m}{n}, \frac{p}{q}) / m, n, p a n d q a r e int e g e r s s u c h t h a t n, q \neq 0 a n d q m = p n\} . T h e n$

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R and S both are equivalence relations.

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

Neither R nor S is an equivalence relation

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

answer is D.

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Detailed Solution

$\begin{array}{l} x R y n e e d n o t i m p l y y R x . \\ \frac{m}{n} S \frac{p}{q} \Leftrightarrow q m = p n \\ \frac{m}{n} S \frac{m}{n} \Rightarrow S i s r e f l e x i v e \\ \frac{m}{n} S \frac{p}{q} \\ \Rightarrow \frac{p}{q} S \frac{m}{n} \Rightarrow S i s s y m m e t r i c \\ \frac{m}{n} S \frac{p}{q}, \frac{p}{q} S \frac{r}{s} \\ \Rightarrow q m = p n, p s = r q \Rightarrow m s = r n \\ \Rightarrow \frac{m}{n} S \frac{r}{s} \Rightarrow R i s t r a n s i t i v e \\ H e n c e, S i s a n e q u i v a l e n c e r e l a t i o n \end{array}$