Mathematics - Sets and Relations Question with Solution | TestHub

MathematicsSets and RelationsQuestions on Symmetric Transitive and Reflexive PropertiesEasy2 minPYQ_2013

MathematicsEasysingle choice

Let $R$ be the relation on the set $R$ of all real numbers, defined by $a R b$ If $∣ a - b ∣ \leq 1$ . Then, $R$ is

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Solution:

Since, $∣ a - a ∣ = 0 \leq 1,$ so $a R a, \forall a \in R$ $∴ R$ is reflexive. Now, $a R b \Rightarrow ∣ a - b ∣ \leq 1 \Rightarrow ∣ b - a ∣ \leq 1 \Rightarrow b R a$ $∴ R$ is symmetric. But $R$ is not transitive as $1 R 2, 2 R 3$ but $1 \neq R^{'} 3$ $[∵ ∣1 - 3∣ = 2 > 1]$

Stream:BITSATSubject:MathematicsTopic:Sets and RelationsSubtopic:Questions on Symmetric Transitive and Reflexive Properties

⏱ 2mℹ️ Source: PYQ_2013

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