Let $R_{1}$ and $R_{2}$ be two relations defined as follows:
$R_{1} = {(a, b) \in R^{2} : a^{2} + b^{2} \in Q}$ and $R_{2} = {(a, b) \in R^{2} : a^{2} + b^{2} \notin Q}$ , where $Q$ is the set of all rational numbers. Then

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$R_{2}$ is transitive but $R_{1}$ is not transitive.

Neither $R_{1}$ nor $R_{2}$ is transitive.

$R_{1}$ is transitive but $R_{1}$ is not transitive.

$R_{1}$ and $R_{2}$ are both transitive.

answer is A.

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Detailed Solution

(a): For $R_{1}$ : Let $a = 2 + \sqrt{2}, b = 2 - \sqrt{2}, c = 2^{5 / 4}$
$∴ a R_{1} b \Rightarrow a^{2} + b^{2} = {(2 + \sqrt{2})}^{2} + {(2 - \sqrt{2})}^{2} = 12 \in Q$
$\begin{array}{r} b R_{1} c \Rightarrow b^{2} + c^{2} = {(2 - \sqrt{2})}^{2} + {(2^{5 / 4})}^{2} = 6 \in Q \\ a R_{1} c \Rightarrow a^{2} + c^{2} = {(2 + \sqrt{2})}^{2} + {(2^{5 / 4})}^{2} = 6 + 8 \sqrt{2} \notin Q \end{array}$
$∴ R_{1}$ is not transitive
For $R_{2}$ Let $a = 2^{3 / 2}, b = 3^{3 / 4}, c = 1$
Then, $a R_{2} b \Rightarrow a^{2} + b^{2} = {(2^{3 / 2})}^{2} + {(3^{3 / 4})}^{2} = 8 + 3 \sqrt{3} \notin Q$
$\begin{array}{r} b R_{2} c \Rightarrow b^{2} + c^{2} = {(3^{3 / 4})}^{2} + {(1)}^{2} = 3 \sqrt{3} + 1 \notin Q \\ a R_{2} c \Rightarrow a^{2} + c^{2} = {(2^{3 / 2})}^{2} + {(1)}^{2} = 8 + 1 = 9 \in Q \end{array}$
$∴ R_{2}$ is not transitive.