Among the relations S = \ (a, b): a, b ∈ R - \ 0\ , 2 + (a)/(b) >...

Given,

$a, b \in I$ and relation $T$ is defined as ${a}^{2} - {b}^{2} \in I$

Now if ${a}^{2} - {b}^{2} \in I$ so ${b}^{2} - {a}^{2}$ will also be integer, for example if ${5}^{2} - {4}^{2} = 9$ is integer then ${4}^{2} - {5}^{2} = - 9$ is also a integer,

Now checking transitive,

If ${4}^{2} - {3}^{2} = 7$ is an integer, ${3}^{2} - {2}^{2} = 5$ is an integer then ${4}^{2} - {2}^{2} = 12$ is also an integer, hence we can say that $T$ is symmetric and transitive,

Now checking relation $S$ which is defined as $2 + \dfrac{a}{b} > 0$,

So, if we replace $\dfrac{a}{b}$ by $\dfrac{- 1}{9}$ then $2 + \dfrac{a}{b} > 0$ is true but we take $\dfrac{b}{a} = - 9$ for symmetric we get $2 + \dfrac{b}{a} = 2 - 9 = - 7 ≯ 0$ hence, the relation is not symmetric,

Now checking transitive, now if $2 + \dfrac{a}{b} > 0 ⇒ \dfrac{a}{b} > - 2$ and $\dfrac{b}{c} > - 2$ then we cannot say that $\dfrac{a}{c} > - 2$,

For example if we take $\dfrac{4}{1} > - 2, \dfrac{1}{- 1} > - 2$ then $\dfrac{4}{- 1} ≯ - 2$, hence it is not transitive,

Hence, we can say that $T$ is symmetric and $S$ is not.