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,

Question

Among the relations

$S={(a,b):a,b\in R-{0},2+\frac{a}{b}>0}$

and $T={(a,b):a,b\in R,{a}^{2}-{b}^{2}\in Z}$,

Accepted Answer

For relation $\mathrm{T}=\mathrm{a}^{2}-\mathrm{b}^{2}=-\mathrm{I}$

Then, $(\mathrm{b}, \mathrm{a})$ on relation $\mathrm{R}$

$\Rightarrow \mathrm{b}^{2}-\mathrm{a}^{2}=-\mathrm{I}$

$	herefore \mathrm{T}$ is symmetric

$\mathrm{S}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathrm{R}-\{0\}, 2+\frac{\mathrm{a}}{\mathrm{b}}>0ight\}$

$2+\frac{\mathrm{a}}{\mathrm{b}}>0 \Rightarrow \frac{\mathrm{a}}{\mathrm{b}}>-2, \Rightarrow \frac{\mathrm{b}}{\mathrm{a}}<\frac{-1}{2}$

If $(b, a) \in \mathbf{S}$ then

$2+\frac{\mathrm{b}}{\mathrm{a}}$ not necessarily positive

$	herefore \mathrm{S}$ is not symmetric

Solutions · 1