Let R_1 and R_2 be two relations defined as follows : R_1=a, b∈R^2:a^2+b^2∈Q and R_2=a, b∈R^2:a^2+b^2∉Q, where Q is the set of all rational numbers, then

Correct answer: (C) Neither R_1 nor R_2 is transitive.. For R_1 let a=1+√2, b=1-√2, c=8^14 aR_1b⇒a^2+b^2=1+2^2+1-2^2=6∈Q bR_1c⇒b^2+c^2=1-2^2+814^2=3∈Q aR_1c⇒a^2+c^2=1+2^2+814^2=3+4√2∉Q ∴R_1 is not transitive. For R_2 let a=1+√2, b=√2, c=1-√2 aR_2b⇒a^2+b^2=1+2^2+2^2=5+2√2∉Q bR_2c⇒b^2+c^2=2^2+1-2^2=5-2√2∉Q aR_2c⇒a^2+c^2=1+2^2+1-2^2=6∈Q ∴R_2 is not transitive.

Q42 / 75 · JEE Main 2020 (03 Sep Shift 2) · Mathematics

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

A
$R_{1}$ is transitive but $R_{2}$ is not transitive.
B
$R_{2}$ is transitive but $R_{1}$ is not transitive.
Correct answer
C
Neither $R_{1}$ nor $R_{2}$ is transitive.
D
$R_{1}$ and $R_{2}$ are both transitive.

Solution

For

R_{1}

let

a = 1 + \sqrt{2}, b = 1 - \sqrt{2}, c = 8^{\frac{1}{4}}

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

$b R_{1} c \Rightarrow b^{2} + c^{2} = {(1 - \sqrt{2})}^{2} + {(8^{\frac{1}{4}})}^{2} = 3 \in Q$

$a R_{1} c \Rightarrow a^{2} + c^{2} = {(1 + \sqrt{2})}^{2} + {(8^{\frac{1}{4}})}^{2} = 3 + 4 \sqrt{2} \notin Q$

$∴$ $R_{1}$ is not transitive.

For $R_{2}$ let $a = 1 + \sqrt{2}, b = \sqrt{2}, c = 1 - \sqrt{2}$

$a R_{2} b \Rightarrow a^{2} + b^{2} = {(1 + \sqrt{2})}^{2} + {(\sqrt{2})}^{2} = 5 + 2 \sqrt{2} \notin Q$

$b R_{2} c \Rightarrow b^{2} + c^{2} = {(\sqrt{2})}^{2} + {(1 - \sqrt{2})}^{2} = 5 - 2 \sqrt{2} \notin Q$

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

$∴$ $R_{2}$ is not transitive.

Hence, the correct answer is option: 'C'.

Let R1 and R2 be two relations defined as follows :R1=a, b∈R2:a2+b2∈Q and R2=a, b∈R2:a2+b2∉Q, where Q is the set of all rational numbers, then

Solution

Relevant Questions

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