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Q. Consider the following relations:-
$\mathrm{R}=\{(\mathrm{x}, \mathrm{y}) | \mathrm{x}, \text { y are real numbers and } \mathrm{x}=\mathrm{~ w y ~ f o r ~ s o m e ~ r a t i o n a l ~ n u m b e r ~ w \} ~}$
$\mathrm{S}=\left\{\left(\frac{\mathrm{m}}{\mathrm{n}}, \frac{\mathrm{p}}{\mathrm{q}}\right) | \mathrm{m}, \mathrm{n}, \mathrm{p} \text { and } \mathrm{q} \text { are integers such that } \mathrm{n}, \mathrm{q} \neq 0 \text { and } \mathrm{qm}=\mathrm{pn}\right\}$
Then :
(1) R is an equivalence relation but S is not an equivalence relation
(2) Neither R nor S is an equivalence relation
(3) S is an equivalence relation but R is not an equivalence relation
(4) R and S both are equivalence relations.
[AIEEE - 2010]
Ans. (3)
For $\mathrm{R}, \mathrm{xRy} \Rightarrow \mathrm{x}=\mathrm{wy}$
For reflexive
$\mathrm{xRx} \Rightarrow \mathrm{x}=\mathrm{wx}$
Which is true then $\mathrm{w}=1$
For symmetric
consider $\mathrm{x}=0, \mathrm{y} \neq 0$
$\mathrm{xRy} \Rightarrow$ oRy $\Rightarrow 0=\mathrm{wy}$
which is true when w=0
Now
y Rx $\Rightarrow$ yR $0 \Rightarrow y=w \times 0$
There is no rational value of $w$
for which $y=w \times 0$
Hence relation is not symmetric and hence not an equivalence relation
Now for $S$
For reflexive
$\frac{m}{n} S \frac{m}{n} \Rightarrow m n=n m$
which is true
For symmetric
Let $\frac{m}{n} S \frac{m}{n} \Rightarrow q m=n p$
$\frac{p}{q} S \frac{m}{n} \Rightarrow p n=m q$
which is true
Relation is symmetric
For transitive
Let $\frac{\mathrm{m}}{\mathrm{n}} \mathrm{S} \frac{\mathrm{p}}{\mathrm{q}} \Rightarrow \mathrm{qm}=\mathrm{pn} \quad \ldots(1)$
$\frac{\mathrm{p}}{\mathrm{q}} \mathrm{S} \frac{\mathrm{r}}{\mathrm{s}} \Rightarrow \mathrm{ps}=\mathrm{rq} \quad \quad \ldots$ (2)
From Equation ( 1) and equation ( 2)
$\Rightarrow \mathrm{ms}=\mathrm{nr}$
$\therefore \frac{\mathrm{m}}{\mathrm{n}} \mathrm{S} \frac{\mathrm{r}}{\mathrm{s}}$
S is transitive
$\therefore$ S is equivalence.
Q. Let R be the set of real numbers.
Statement-1: $\mathrm{A}=\{(\mathrm{x}, \mathrm{y}) \in \mathrm{R} \times \mathrm{R}: \mathrm{y}-\mathrm{x} \text { is an integer' is an equivalence relation on } \mathrm{R} .$
Statement- $2: \mathrm{B}=\{(\mathrm{x}, \mathrm{y}) \in \mathrm{R} \times \mathrm{R}: \mathrm{x}=\alpha \mathrm{y} \text { for some rational number } \alpha\}$ is an equivalence relation on $\mathrm{R} .$
(1) Statement-1 is true, Statement-2 is false.
(2) Statement-1 is false, Statement-2 is true
(3) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
(4) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
[AIEEE - 2011]
Ans. (1)
$y-x=$ integer $\Rightarrow x-y=$ integer reflex
$x-x=0=$ integer symmetric
$y-x=l,$ and $z-y=12$
$z-x=I_{1}+I_{2} \Rightarrow x R z$ transitive
A is equivalence
$x=\alpha y$ is equivalence only for $\alpha=1$ not for other values.
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