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

Let R_{1} and R_{2} be two relations defined as follows... | Relation (Mathematics)

Exam ( $3^{rd\ }$ Sept $2^{nd\ }$ Shift 2020) 3rd September Shift 2

Mathematics

Let

R_{1}

and

R_{2}

be two relations defined as follows:

R_{1} = \left\{ (a,b) \in R^{2}:a^{2} + b^{2} \in Q \right\}

and

R_{2} = \left\{ (a,b) \in R^{2}:a^{2} + b^{2} \notin Q \right\}

, where

Q

is the set of all rational numbers. Then

Neither

R_{1}

nor

R_{2}

is transitive.

R_{2}

is transitive but

R_{1}

is not transitive.

R_{1}

is transitive but

R_{2}

is not transitive.

R_{1}

and

R_{2}

are both transitive.

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Correct Answer:

Explanation

(a) : For $R_{1}:$ Let $a = 2 + ,b = 2 - ,c = 2^{5/4}$ $\\ \therefore aR_{1}b \Rightarrow a^{2} + b^{2} = (2 + )^{2} + (2 - )^{2} \\= 12 \in QbR_{1}c \Rightarrow b^{2} + c^{2} = (2 - )^{2} + \left( 2^{5/4} \right)^{2} \\= 6 \in QaR_{1}c \Rightarrow a^{2} + c^{2} = (2 + )^{2} + \left( 2^{3/4} \right)^{2} \\= 6 + 8Q$ $\\ \therefore R_{1}$ is not transitive $\\$ For $R_{2}\\$ : Let $a = 2^{3/2},b = 3^{3/4},c = 1\\$ Then, $aR_{2}b \Rightarrow a^{2} + b^{2} = \left( 2^{3/2} \right)^{2} + \left( 3^{3/4} \right)^{2} \\= 8 + 3 \notin Q$ $\\bR_{2}c \Rightarrow b^{2} + c^{2} = \left( 3^{3/4} \right)^{2} + (1)^{2} \\= 3 + 1 \notin QaR_{2}c \Rightarrow a^{2} + c^{2} \\= \left( 2^{3/2} \right)^{2} + (1)^{2} = 8 + 1 = 9$ in Q $\\ \therefore R_{2}$ is not transitive.

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Explanation

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