# Let the function

Question:

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:

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

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

3. $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ are both transitive

4. Neither $\mathrm{R}_{1}$ nor $\mathrm{R}_{2}$ is transitive

Correct Option: , 4

Solution:

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 $a^{2}+c^{2} \in Q$