Consider the following relations: $R=\{(x, y) \mid x, y$ are real numbers and $x=$ wy for some rational…
Consider the following relations:
$R=\{(x, y) \mid x, y$ are real numbers and $x=$ wy for some rational number w $\}$;
$S=\left\{\left(\frac{m}{n}, \frac{p}{q}\right) \mid m, n, p\right.$ and $q$ are integers such that $n, q \neq 0$ and $\left.q m=p n\right\}$. Then
neither $\mathrm{R}$ nor $\mathrm{S}$ is an equivalence relation
$S$ is an equivalence relation but $R$ is not an equivalence relation
$R$ and $S$ both are equivalence relations
$\mathrm{R}$ is an equivalence relation but $\mathrm{S}$ is not an equivalence relation
Solution
xRy need not implies $y R x$
$\mathrm{S}: \frac{\mathrm{m}}{\mathrm{n}} \mathrm{s} \frac{\mathrm{p}}{\mathrm{q}} \Leftrightarrow \mathrm{qm}=\mathrm{pn}$ $\frac{m}{n} s \frac{m}{n}$ reflexive
$
\begin{array}{ll}
\frac{m}{n} s \frac{p}{q} & \Rightarrow \frac{p}{q} s \frac{m}{n} \text { symmetric } \\
\frac{m}{n} s \frac{p}{q}, \frac{p}{q} s \frac{r}{s} & \Rightarrow q m=p n, p s=r q \quad \Rightarrow m s=r n \text { transitive. }
\end{array}
$
$S$ is an equivalence relation.