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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
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$R=\{(x, y) \mid x, y$ 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
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Verified Answer
The correct answer is:
$\mathrm{R}$ is an equivalence relation but $\mathrm{S}$ is not an equivalence relation
$\mathrm{R}$ is an equivalence relation but $\mathrm{S}$ is not an equivalence relation
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.
$\mathrm{S}: \frac{\mathrm{m}}{\mathrm{n}} \mathrm{s} \frac{\mathrm{p}}{\mathrm{q}} \Leftrightarrow \mathrm{qm}=\mathrm{pn}$ $\frac{m}{n} s \frac{m}{n}$ 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.
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