Question

The following relations are defined on the set of real numbers check them for Reflexivity,Symmetry,Transitivity.

$1+ab>0$

Answer 1

R, S both $\because 1+a^2>0$ and $1+ba>0$ when

$1+ab>0\ne T.$ Consider $a=2,b=-\frac{1}{6},c=-2.$

aRb as $1+2(-\frac{1}{6})=1-\frac{1}{3}=\frac{2}{3}>0$

bRc as $1+(-\frac{1}{6})(-2)=1+\frac{1}{3}=\frac{4}{3}>0$

but aRc $=1+2(-2)=-3\ngtr 0.$

Therefore a is not related to c and hence no transitive.