Mathematics - Sets and Relations Question with Solution | TestHub

For$R_1$let$a=1+\sqrt{2}, b=1-\sqrt{2}, c=8^{\frac{1}{4}}$

$a{R}_{1}b\Rightarrow a^2+b^2={\left(1+\sqrt{2}\right)}^2+{\left(1-\sqrt{2}\right)}^2=6\in Q$

$b{R}_{1}c\Rightarrow b^2+c^2={\left(1-\sqrt{2}\right)}^2+{\left(8^{\frac{1}{4}}\right)}^2=3\in Q$

$a{R}_{1}c\Rightarrow a^2+c^2={\left(1+\sqrt{2}\right)}^2+{\left(8^{\frac{1}{4}}\right)}^2=3+4\sqrt{2}\notin Q$

$\therefore$$R_1$ is not transitive.

For $R_2$ let $a=1+\sqrt{2}, b=\sqrt{2}, c=1-\sqrt{2}$

$a{R}_{2}b\Rightarrow a^2+b^2={\left(1+\sqrt{2}\right)}^2+{\left(\sqrt{2}\right)}^2=5+2\sqrt{2}\notin Q$

$b{R}_{2}c\Rightarrow b^2+c^2={\left(\sqrt{2}\right)}^2+{\left(1-\sqrt{2}\right)}^2=5-2\sqrt{2}\notin Q$

$a{R}_{2}c\Rightarrow a^2+c^2={\left(1+\sqrt{2}\right)}^2+{\left(1-\sqrt{2}\right)}^2=6\in Q$

$\therefore$$R_2$ is not transitive.