Question

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

• A. Neither $R_1$ nor $R_2$ is a transitive relation
• B. $R_1$ and $R_2$ both are transitive relations
• C. $R_1$ is transitive but $R_2$ is not a transitive relation
• D. $R_2$ is transitive but $R_1$ is not a transitive relation

Answer 1

The correct option is A Neither $R_1$ nor $R_2$ is a transitive relation

$R_1=\left\{\right(a,b)\in R^2:a^2+b^2\in Q\}$

Let $a=2+\sqrt{3}$ and $b=2-\sqrt{3}$
$\Rightarrow a^2+b^2=14\in Q$
Let $c=(1+2\sqrt{3})$
$\Rightarrow b^2+c^2=20\in Q$
But $a^2+c^2=(2+\sqrt{3}{)}^2+(1+2\sqrt{3}{)}^2\notin Q$

Since $(2+\sqrt{3},2-\sqrt{3})$ and $(2-\sqrt{3},1+2\sqrt{3})\in R_1$ but $(2+\sqrt{3},1+2\sqrt{3})\notin R_1.$
So, $R_1$ is not a transitive relation $\cdots \left(1\right)$

$R_2=\left\{\right(a,b)\in R^2:a^2+b^2\notin Q\}.$
Let $a^2=1,b^2=\sqrt{3}$ and $c^2=2$
$\Rightarrow a^2+b^2\notin Q$ and $b^2+c^2\notin Q$
But $a^2+c^2\in Q$
Since $(1,\sqrt{\sqrt{3}})$ and $(\sqrt{\sqrt{3}},\sqrt{2})\notin R_2$ but $(1,\sqrt{2})\in R_2.$
So, $R_2$ is not a transitive relation $\cdots \left(2\right)$

$\therefore$ From $\left(1\right)$ and $\left(2\right)$ we can say that Neither $R_1$ nor $R_2$ is a transitive relation.