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

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

(A) neither $R$ nor $S$ is an equivalence relation

(C) R and S both are equivalence relations

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

JEE Main · 2010 · 11 Apr 2010 · Offline · Sets Relations and Functions

Consider the following relations:

R = {(x, y) ∣ x, y

are real numbers and

x = w y

for some rational number

w}

;

S = {(\frac{m}{n}, \frac{p}{q}) m, n, p

and q are integers such that

n, q \neq = 0

and

qm = pn}

. Then

(A)neither $R$ nor $S$ is an equivalence relation
(B) $S$ is an equivalence relation but $R$ is not an equivalence relation✓ Correct
(C)R and S both are equivalence relations
(D) $R$ is an equivalence relation but $S$ is not an equivalence relation

Answer: (B)

For

R

, where

x R y

means

x = w y

for some rational

w

, symmetry fails: take

x = 0, y \neq = 0

, so

0 = 0 \cdot y

holds but

y = w \cdot 0

cannot, so

y R x

is false, and

R

is not an equivalence relation. For

S

, where

\frac{m}{n} S \frac{p}{q}

means

q m = p n

, which is just the statement that the two fractions are equal, reflexivity, symmetry and transitivity all hold. So

S

is an equivalence relation while

R

is not.

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option