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Let R be a relation on \mathbb{R}, given by
$$ R=\{(a, b): 3 a-3 b+\sqrt{7} \text { is an irrational number }\} .
Then R is

Option: 1

an equivalence relation


Option: 2

reflexive and symmetric but not transitive


Option: 3

reflexive but neither symmetric nor transitive


Option: 4

reflexive and transitive but not symmetric


Answers (1)

best_answer

$$ \begin{aligned} (\mathrm{a}, \mathrm{a}) \in & \mathrm{R} \Rightarrow 3 \mathrm{a}-3 \mathrm{a}+\sqrt{7} \\ & =\sqrt{7} \text { (irrational) } \end{aligned}
\Rightarrow \mathrm{R} is reflexive
Let a=\frac{2 \sqrt{7}}{3}and  b=\frac{\sqrt{7}}{3}
$$ (a, b) \in R \Rightarrow 2 \sqrt{7}-\sqrt{7}+\sqrt{7}

$$ \begin{aligned} & =2 \sqrt{7} \text { (irration) } \\ (\mathrm{b}, \mathrm{a}) \in \mathrm{R} & \Rightarrow \sqrt{7}-2 \sqrt{7}+\sqrt{7} \\ & =0(\text { rational) } \end{aligned}
\Rightarrow \mathrm{R} is no symmetric
Let \mathrm{a}=\frac{2 \sqrt{7}}{3}, \mathrm{~b}=\frac{\sqrt{7}}{3}, \mathrm{C}=\frac{3 \sqrt{7}}{3}
(\mathrm{a} ; \mathrm{b}) \in \mathrm{R} \Rightarrow 2 \sqrt{7} (irrational)
(b ; c) \in \mathrm{R} \Rightarrow \sqrt{7} (irrational)
$$ (\mathrm{a}, \mathrm{c}) \in \mathrm{R} \Rightarrow 2 \sqrt{7}-3 \sqrt{7}+\sqrt{7}
=0 (rational)
\mathrm{R} is not transitive
\Rightarrow \mathrm{R} is reflexive but neither symmetric nor transitive

Posted by

jitender.kumar

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