First slide

Relations XII

Question

Consider the following relations. R = {(x, y) | x, y are real numbers and x = wy for some rational number w}
$S = \{(\frac{m}{n}, \frac{p}{q})\} m, n, p, q$ are integer such that $n . q \neq 0$ and $q m = p n}$
Statement-1: S is an equivalence relation but R is not an equivalence relation.
Statement-2: R and S both are symmetric.

Moderate

A

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for ST
B

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1
C

STATEMENT-1 is True, STATEMENT-2 is False
D

STATEMENT-1 is False, STATEMENT-2 is True

Solution

Since $(0, 1) \in R$ but $(1, 0) \in R, R$ is not symmetric and hence is not an equivalence relation so statement-2 is false.
Next, For the relation $S, q m = p n \Rightarrow \frac{m}{n} = \frac{p}{q}$
Thus $(\frac{m}{n}, \frac{p}{q}) \in S \Rightarrow \frac{m}{n} = \frac{p}{q}$ which shows that S is reflexive and symmetric
Again , $(\frac{m}{n}, \frac{p}{q}) \in S and (\frac{p}{q}, \frac{r}{s}) \in S$
$\Rightarrow \frac{m}{n} = \frac{p}{q} = \frac{r}{s} \Rightarrow (\frac{m}{n}, \frac{r}{s} \in S)$
Thus S is transitive and hence S is an equivalence relation. So, statement 1 is true

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