Let $R_{1}$ and $R_{2}$ be two relations defined as follows :
$R_{1}=\left\{(a, b) \in \mathbf{R}^{2}: a^{2}+b^{2} \in Q\right\}$ and
$R_{2}=\left\{(a, b) \in \mathbf{R}^{2}: a^{2}+b^{2} \notin Q\right\}$, where $Q$ is the set of all
rational numbers. Then :
Correct Option: 1
For $R_{1}$ let $a=1+\sqrt{2}, b=1-\sqrt{2}, c=8^{1 / 4}$
$a R_{1} b \Rightarrow a^{2}+b^{2}=(1+\sqrt{2})^{2}+(1-\sqrt{2})^{2}=6 \in Q$
$b R_{1} c \Rightarrow b^{2}+c^{2}=(1-\sqrt{2})^{2}+\left(8^{1 / 4}\right)^{2}=3 \in Q$
$a R_{1} c \Rightarrow a^{2}+c^{2}=(1+\sqrt{2})^{2}+\left(8^{1 / 4}\right)^{2}=3+4 \sqrt{2} \notin Q$
$\therefore R_{1}$ is not transitive.
For $R_{2}$ let $a=1+\sqrt{2}, b=\sqrt{2}, c=1-\sqrt{2}$
$a R_{2} b \Rightarrow a^{2}+b^{2}=(1+\sqrt{2})^{2}+(\sqrt{2})^{2}=5+2 \sqrt{2} \notin Q$
$b R_{2} c \Rightarrow b^{2}+c^{2}=(\sqrt{2})^{2}+(1-\sqrt{2})^{2}=5-2 \sqrt{2} \notin Q$
$a R_{2} c \Rightarrow a^{2}+c^{2}=(1+\sqrt{2})^{2}+(1-\sqrt{2})^{2}=6 \in Q$
$\therefore R_{2}$ is not transitive.
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