Among the relations
$S = {(a, b) : a, b \in R - {0}, 2 + \frac{a}{b} > 0}$ and $T = {(a, b) : a, b \in R, a^2 - b^2 \in Z}$,
(A)
$S$ is transitive but $T$ is not
(B)
both $S$ and $T$ are symmetric
(C)
neither $S$ nor $T$ is transitive
(D)
$T$ is symmetric but $S$ is not
MEDIUM
Correct Answer:D
Explanation
For relation $T$: if $(a, b) \in T$, then $a^2 - b^2 = I$ where $I \in Z$.
Then, $(b, a)$ on relation $T$ means $b^2 - a^2 = -I$.
Since $-I \in Z$, $T$ is symmetric.
For relation $S = {(a, b) : a, b \in R - {0}, 2 + \frac{a}{b} > 0}$, $2 + \frac{a}{b} > 0 \implies \frac{a}{b} > -2$.
If $(b, a) \in S$ then $2 + \frac{b}{a}$ is not necessarily positive. Therefore, $S$ is not symmetric.