Among the relations $\mathrm{S}=\left\{(\mathrm{a},\mathrm{b}):\mathrm{a},\mathrm{b}\in \mathrm{ℝ}-\left\{0\right\},2+\frac{\mathrm{a}}{\mathrm{b}}>0\right\}$ and $\mathrm{T}=\left\{(\mathrm{a},\mathrm{b}):\mathrm{a},\mathrm{b}\in \mathrm{ℝ},{\mathrm{a}}^2-{\mathrm{b}}^2\in \mathrm{Z}\right\}$

$\mathrm{T},{\mathrm{a}}^2-{\mathrm{b}}^2\in \mathrm{Z},(\mathrm{a},\mathrm{b})\in \mathrm{T}\Rightarrow (\mathrm{b},\mathrm{a})\in \mathrm{T}$
T is symmetric
T is transitive also as $(\mathrm{a},\mathrm{b})\in \mathrm{T},(\mathrm{b},\mathrm{c})\in \mathrm{T}$
${\mathrm{a}}^2-{\mathrm{b}}^2={\mathrm{Z}}_1,{\mathrm{b}}^2-{\mathrm{c}}^2={\mathrm{Z}}_2$
Adding ${\mathrm{a}}^2-{\mathrm{c}}^2={\mathrm{Z}}_1+{\mathrm{Z}}_2\in \mathrm{T}$
For S,$2+\frac{\mathrm{a}}{\mathrm{b}}>0\Rightarrow \frac{\mathrm{a}}{\mathrm{b}}>-2$
But $\frac{\mathrm{b}}{\mathrm{a}}\text{ cannot be }>-2$so S is not symmetric
& also If $2+\frac{\mathrm{a}}{\mathrm{b}}>0\mathrm{\&}2+\frac{\mathrm{b}}{\mathrm{c}}>0\Rightarrow$
Then  $2+\frac{\mathrm{a}}{\mathrm{c}}$ can be positive (or) negative S is not Transitive