Let $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ be two relations defined as follows:
$\mathrm{R}_{1}=\left\{(\mathrm{a}, \mathrm{b}) \in \mathrm{R}^{2}: \mathrm{a}^{2}+\mathrm{b}^{2} \in \mathrm{Q}\right\}$ and
$\mathrm{R}_{2}=\left\{(\mathrm{a}, \mathrm{b}) \in \mathrm{R}^{2}: \mathrm{a}^{2}+\mathrm{b}^{2} \notin \mathrm{Q}\right\}$
where $Q$ is the set of all rational numbers. Then:
Correct Option: , 4
Let $a^{2}+b^{2} \in Q \& b^{2}+c^{2} \in Q$
eg. $\quad a=2+\sqrt{3} \& b=2-\sqrt{3}$
$\mathrm{a}^{2}+\mathrm{b}^{2}=14 \in \mathrm{Q}$
Let $\quad \mathrm{c}=(1+2 \sqrt{3})$
$b^{2}+c^{2}=20 \in Q$
But $\quad a^{2}+c^{2}=(2+\sqrt{3})^{2}+(1+2 \sqrt{3})^{2} \notin Q$
for $R_{2}$ Let $a^{2}=1, b^{2}=\sqrt{3} \& c^{2}=2$
$\mathrm{a}^{2}+\mathrm{b}^{2} \notin \mathrm{Q} \& \mathrm{~b}^{2}+\mathrm{c}^{2} \notin \mathrm{Q}$
But $a^{2}+c^{2} \in Q$
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