Let $R_1$ and $R_2$ be two relations defined as follows: $R_1=\left\{(a,b)\in R^2:a^2+b^2\in Q\right\}$ and $R_2=\{(a,b)\in R^2:a^2+b^2\notin Q\}$, where Q is the set of all rational numbers. Then 

If $a=1+\sqrt{2},b=1-\sqrt{2}$ and $c=8^{\frac{1}{4}}$ then

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

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

$c^2+a^2={\left(8^{\frac{{\displaystyle 1}}{{\displaystyle 4}}}\right)}^2={(1+\sqrt{2})}^2=2\sqrt{2}+1+2+2\sqrt{2}\notin Q\Rightarrow not a{R}_{1}c$

$\therefore R_1$ is not transitive

If $a=1+\sqrt{2}, b=\sqrt{2}, c=1-\sqrt{2}$ then

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

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

$c^2+a^2={(1-\sqrt{2})}^2+{(1+\sqrt{2})}^2=2(1+2)=6\in Q\Rightarrow not a{R}_{2}C$

$\therefore R_2$ is not transitive