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
$R_2=\left\{\right(a,b)\in R^2:a^2+b^2\notin Q\}$, where $Q$ is the set of all rational numbers. Then :

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

Answer 1

The correct option is C Neither $R_1$ nor $R_2$ is transitive.

For $R_1$
Let $a=1+\sqrt{2},b=1-\sqrt{2},c=8^{\frac{1}{4}}$
$a{R}_{1}b\Rightarrow a^2+b^2=6\in Q$
$b{R}_{1}c\Rightarrow b^2+c^2=3\in Q$
$a{R}_{1}c\Rightarrow a^2+c^2=3+4\sqrt{2}\notin Q$
$\therefore R_1$ is not transitive.

For $R_2$
Let $a=1+\sqrt{2},b=\sqrt{2},c=1-\sqrt{2}$
$a{R}_{2}b\Rightarrow a^2+b^2=5+2\sqrt{2}\notin Q$
$b{R}_{2}c\Rightarrow b^2+c^2=5-2\sqrt{2}\notin Q$
$a{R}_{2}c\Rightarrow a^2+c^2=6\in Q$
$\therefore R_2$ is not transitive.