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BITSAT 2013
Exam
BITSAT
Year
2013
Subject
Mathematics
Chapter
Sets and Relations

BITSAT 2013 · Mathematics · Sets and Relations

Let R be the relation on the set R of all real numbers, defined by...

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

Options

  1. Areflexive and symmetric only
  2. Breflexive and transitive only
  3. Cequivalence
  4. DNone of the above

Correct answer

A. reflexive and symmetric only

Step-by-step solution

Since, ∣a−a∣=0≤1,|a-a|=0 \leq 1, so aRa,∀a∈R∴Ra R a, \forall a \in \mathbb{R}\therefore R is reflexive. Now, aRb⇒∣a−b∣≤1⇒∣b−a∣≤1⇒bRa∴Ra R b \Rightarrow|a-b| \leq 1 \Rightarrow|b-a| \leq 1 \Rightarrow b R a\therefore R is symmetric. But RR is not transitive as 1R2,2R31 R 2,2 R 3 but 1̸R′31 \not R^{\prime} 3 [\because|1-3|=2>1]
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