JEE PYQ: Sets And Relations Question 7
Question 7 - 2020 (03 Sep Shift 2)
Let $R_1$ and $R_2$ be two relations defined as follows:
$R_1 = {(a, b) \in \mathbf{R}^2 : a^2 + b^2 \in Q}$ and
$R_2 = {(a, b) \in \mathbf{R}^2 : a^2 + b^2 \notin Q}$, where $Q$ is the set of all rational numbers. Then:
(1) Neither $R_1$ nor $R_2$ is transitive.
(2) $R_2$ is transitive but $R_1$ is not transitive.
(3) $R_1$ is transitive but $R_2$ is not transitive.
(4) $R_1$ and $R_2$ are both transitive.
Show Answer
Answer: (1)
Solution
For $R_1$, let $a = 1 + \sqrt{2}, b = 1 - \sqrt{2}, c = 8^{1/4}$
$aR_1b \Rightarrow a^2 + b^2 = (1+\sqrt{2})^2 + (1-\sqrt{2})^2 = 6 \in Q$
$bR_1c \Rightarrow b^2 + c^2 = (1-\sqrt{2})^2 + (8^{1/4})^2 = 3 \in Q$
$aR_1c \Rightarrow a^2 + c^2 = (1+\sqrt{2})^2 + (8^{1/4})^2 = 3 + 4\sqrt{2} \notin Q$
$\therefore R_1$ is not transitive.
For $R_2$, let $a = 1 + \sqrt{2}, b = \sqrt{2}, c = 1 - \sqrt{2}$
$aR_2b \Rightarrow a^2 + b^2 = (1+\sqrt{2})^2 + (\sqrt{2})^2 = 5 + 2\sqrt{2} \notin Q$
$bR_2c \Rightarrow b^2 + c^2 = (\sqrt{2})^2 + (1-\sqrt{2})^2 = 5 - 2\sqrt{2} \notin Q$
$aR_2c \Rightarrow a^2 + c^2 = (1+\sqrt{2})^2 + (1-\sqrt{2})^2 = 6 \in Q$
$\therefore R_2$ is not transitive.
BETA
