Mathematics - Sets and Relations Question with Solution | TestHub

Given, $\left(a,b\right)R\left(c,d\right)\Rightarrow ad\left(b-c\right)=bc\left(a-d\right)$

$\Rightarrow \left(a,b\right)R\left(c,d\right)\iff \frac{b-c}{bc}=\frac{a-d}{ad}$

$\Rightarrow \frac{1}{c}-\frac{1}{b}=\frac{1}{d}-\frac{1}{a}$

$\Rightarrow \frac{1}{a}-\frac{1}{b}=\frac{1}{d}-\frac{1}{c}$

Reflexive:

$\left(a,b\right)R\left(a,b\right)\Rightarrow \frac{1}{a}-\frac{1}{b}=\frac{1}{b}-\frac{1}{a}$

It is false. So, given relation is not reflexive.

Symmetric:

$\left(a,b\right)R\left(c,d\right)\Rightarrow \frac{1}{a}-\frac{1}{b}=\frac{1}{d}-\frac{1}{c}$

$\Rightarrow \frac{1}{c}-\frac{1}{d}=\frac{1}{b}-\frac{1}{a}$

$\therefore \left(c,d\right)R\left(a,b\right)$

It is symmetric.

Transitive:

$\left(a,b\right)R\left(c,d\right)\Rightarrow \frac{1}{a}-\frac{1}{b}=\frac{1}{d}-\frac{1}{c}$

$\left(c,d\right)R\left(e,f\right)=\frac{1}{c}-\frac{1}{d}=\frac{1}{e}-\frac{1}{f}$

$\frac{1}{a}-\frac{1}{b}=\frac{1}{e}-\frac{1}{f}$

$\Rightarrow \left(a,b\right)\cancel{R}\left(e,f\right)$

So, not transitive.