Consider the following relations: $R=\{(x, y) \mid x, y$ are real numbers and $x=w y$ for some ratio

Question

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

Professor PC Thomas Classes · Answer

(c) : We have $(x, x) \in R$ for $w=1$ implying that $R$ is reflexive.\ For $a 
eq 0,(a, 0) 
otin R$ for any $w$ but $(0, a) \in R$. Thus $R$ is not symmetric. Hence $R$ is not an equivalence relation.\ As $\left(\frac{m}{n}, \frac{m}{n}ight) \in S$\ $\Rightarrow \quad m n=m n, S$ is reflexive.\ $\left(\frac{m}{n}, \frac{p}{q}ight) \in S \Rightarrow q m=p n$\ But this can be written as $n p=m q$, giving $\left(\frac{p}{q}, \frac{m}{n}ight) \in S$. Thus $S$ is symmetric.\ Again $\left(\frac{m}{n}, \frac{p}{q}ight) \in S$ and $\left(\frac{p}{q}, \frac{a}{b}ight) \in S$\ means $q m=p n$ and $b p=a q$.\ i.e., $\frac{m}{n}=\frac{p}{q}$ and $\frac{p}{q}=\frac{a}{b}$\ i.e., $\frac{m}{n}=\frac{a}{b}$ Thus $\left(\frac{m}{n}, \frac{a}{b}ight) \in S$ This means $S$ is transitive.\