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Consider the following relations :
$R=\{(x, y) \mid x, y$ are real numbers an $x=w y$ for some rational number $\mathbf{w}\}$; $S=\left\{\left(\frac{m}{n}, \frac{p}{q}\right) \mid m . n \cdot p\right.$ and $q$ are integers such that $n, q \neq 0$ and $q m=p n\}$. Then

1) neither $R$ nor $S$ is an equivalence relation
2) $S$ is an equivalence relation but $R$ is not an equivalence relation
3) $R$ and $\mathrm{S}$ both are equivalence relations
4) $R$ is an equivalence relation but $S$ is not an equivalence relation
in Mathematics by Diamond (75,948 points) | 9 views

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Answer

(2)

Explanation

$x R y$ need not implies $y R x$
$S: \frac{m}{n} s \frac{p}{q} \Leftrightarrow q m=p n$
$\frac{m}{n} s \frac{m}{n}$ reflexive
$\frac{m}{n} s \frac{p}{q} \Rightarrow \frac{p}{q} s \frac{m}{n}$ symmetric
$\frac{m}{n} s \frac{p}{q}, \frac{\mathrm{p}}{\mathrm{q}} \mathrm{s} \frac{\mathrm{r}}{\mathrm{s}} \Rightarrow q m=p n, p s=r q \Rightarrow m s=r n$ transitive.
$S$ is an equivalence relation.

by Diamond (75,948 points)

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