Sets And Relations Question 6
Question 6 - 31 January - Shift 2
Among the relations
$S={(a, b): a, b \in \mathbb{R}-{0}, 2+\frac{a}{b}>0}$ and $T={(a, b): a, b \in \mathbb{R}, a^{2}-b^{2} \in Z}$,
(1) $S$ is transitive but $T$ is not
(2) $T$ is symmetric but $S$ is not
(3) Neither $S$ nor $T$ is transitive
(4) Both $S$ and $T$ are symmetric
Show Answer
Answer: (2)
Solution:
Formula: Symmetric relation (v), Transitive relation (vi)
For relation $T=a^{2}-b^{2}=-I$
Then, $(b, a)$ on relation $R$
$\Rightarrow b^{2}-a^{2}=-I$
$\therefore T$ is symmetric
$S={(a, b): a, b \in R-{0}, 2+\frac{a}{b}>0}$
$2+\frac{a}{b}>0 \Rightarrow \frac{a}{b}>-2, \Rightarrow \frac{b}{a}<\frac{-1}{2}$
If $(b, a) \in S$ then
$2+\frac{b}{a}$ not necessarily positive
a
$\therefore S$ is not symmetric
