Among the relations S = a , b : a , b ∈ R - 0 , 2 + a b > 0 and T = a , b : a , b ∈ R , a 2…

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 $2+\frac{a}{b}>0$,

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

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

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

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