Consider the relations R = {(x ,y) | x ,y are real numbers and x = wy for some rational number w};
$\mathrm{S}=\left\{\left(\frac{\mathrm{m}}{\mathrm{n}},\frac{\mathrm{p}}{\mathrm{q}}\right)\mid \mathrm{m},\mathrm{n},\mathrm{p}\text{ and }\mathrm{q}\text{ are integers such that }\mathrm{n},\mathrm{q}\ne 0\text{ and }\mathrm{qm}=\mathrm{pn}\right\}.$ Then 

(1) Since x R y need not implies y R x $\Rightarrow$R is not symmetric. Hence  R is not an equivalence relation.

$$\begin{gathered} let a be a non zero real number \\ (0,a) \in R for w=0 ,where as (a,0) \notin R for any w. \end{gathered}$$

(2) $\mathrm{S}:\frac{\mathrm{m}}{\mathrm{n}}\mathrm{s}\frac{\mathrm{p}}{\mathrm{q}}\iff \mathrm{qm}=\mathrm{pn}$

(i) $\frac{\mathrm{m}}{\mathrm{n}}\mathrm{s}\frac{\mathrm{m}}{\mathrm{n}}$reflexive since mn=nm 

(ii)$\frac{\mathrm{m}}{\mathrm{n}}\mathrm{s}\frac{\mathrm{p}}{\mathrm{q}}\Rightarrow \frac{\mathrm{p}}{\mathrm{q}}\mathrm{s}\frac{\mathrm{m}}{\mathrm{n}}$symmetric 

(iii) $\frac{\mathrm{m}}{\mathrm{n}}\mathrm{s}\frac{\mathrm{p}}{\mathrm{q}},\frac{\mathrm{p}}{\mathrm{q}}\mathrm{s}\frac{\mathrm{r}}{\mathrm{s}}\Rightarrow \mathrm{qm}=\mathrm{pn},\mathrm{ps}=\mathrm{rq}\Rightarrow \mathrm{ms}=\mathrm{rn}$ transitive.
$\therefore$S is an equivalence relation.