Among the relations S=(a,b):a,b∈ R-0,2+ a b>0 and T=(a,b):a,b∈ R,a^2-b^2∈ Z,

Given,

$a,b\in I$a,b∈I and relation $T$T is defined as ${a}^{2}-{b}^{2}\in I$a2−b2∈I

Now if ${a}^{2}-{b}^{2}\in I$a2−b2∈I so ${b}^{2}-{a}^{2}$b2−a2 will also be integer, for example if ${5}^{2}-{4}^{2}=9$52−42=9 is integer then ${4}^{2}-{5}^{2}=-9$42−52=−9 is also a integer,

Now checking transitive,

If ${4}^{2}-{3}^{2}=7$42−32=7 is an integer, ${3}^{2}-{2}^{2}=5$32−22=5 is an integer then ${4}^{2}-{2}^{2}=12$42−22=12 is also an integer, hence we can say that $T$T is symmetric and transitive,

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

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

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

For example if we take $\frac{4}{1}>-2,\frac{1}{-1}>-2$14​>−2,−11​>−2 then $\frac{4}{-1}≯-2$−14​≯−2, hence it is not transitive,

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