Question

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

(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 $S$ both are equivalence relations

(4) $R$ is an equivalence relation but $S$ is not an equivalence relation

Answer 1

SOLUTION : $(x, x) \in R$ for $w=1$

$\therefore \quad \mathrm{R}$ is reflexive

If $x \neq 0$, then $(0, x) \in R$ for $w=0$ but $(x, 0) \notin R$ for any $w$

$\therefore \quad \mathrm{R}$ is not symmetric

$\Rightarrow \quad \mathrm{R}$ is not equivalence relation

$\left(\frac{m}{n}, \frac{p}{q}\right) \in S \quad \Rightarrow q m=p n \Rightarrow \frac{m}{n}=\frac{p}{q}$

(i) $\frac{m}{n}=\frac{m}{n} \quad \Rightarrow\left(\frac{m}{n}, \frac{m}{n}\right) \in S \Rightarrow$ Reflexive

(ii) $\frac{m}{n}=\frac{p}{q} \quad \Rightarrow \frac{p}{q}=\frac{m}{n} \quad \Rightarrow$ symmetric

(iii) $\frac{m}{n}=\frac{p}{q} \quad$ and $\frac{p}{q}=\frac{x}{y}$

$\Rightarrow \frac{m}{n}=\frac{x}{y} \quad \Rightarrow$ transitive

$\Rightarrow \mathrm{S}$ is equivalence relation

Above correct Answer is Option is (2).