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

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

$R_{1}$ is transitive But $R_{2}$ is not transitive

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

answer is A.

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Detailed Solution

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

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

$\begin{array}{l} b^{2} + c^{2} = (1 - \sqrt{2})^{2} + {(8^{1 / 4})}^{2} = 1 + 2 - 2 \sqrt{2} + 2 \sqrt{2} = 3 \in Q \Rightarrow b R_{1} c \\ c^{2} + a^{2} = {(8^{1 / 4})}^{2} + (1 + \sqrt{2})^{2} = 2 \sqrt{2} + 1 + 2 + 2 \sqrt{2} \notin Q \Rightarrow not a R_{1} c . ∴ R_{1} is not transitive. \end{array}$

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

$\begin{array}{l} a_{\cdot}^{2} + b^{2} = (1 + \sqrt{2})^{2} + (\sqrt{2})^{2} = 1 + 2 + 2 \sqrt{2} + 2 \notin Q \Rightarrow a R_{2} b \\ b^{2} + c^{2} = (\sqrt{2})^{2} + (1 - \sqrt{2})^{2} = 2 + 1 + 2 - 2 \sqrt{2} \notin Q \Rightarrow b R_{2} c \\ c^{2} + a^{2} = (1 - \sqrt{2})^{2} + (1 + \sqrt{2})^{2} = 2 (1 + 2) = 6 \in Q \Rightarrow not a R_{2} C . ∴ R_{2} is not transitive. \end{array}$