Q61 / 180 · KCET 2014 · Mathematics

Let \( S \) be the set of all real numbers. A relation \( R \) has been defined on \( S \) by \( a R b \Leftrightarrow |a-b| \leq 1 \),
then \( R \) is

  • A
    reflexive and transitive but not symmetric
  • B
    an equivalence relation
  • C
    symmetric and transitive but not reflexive
  • Correct answer
    D
    reflexive and symmetric but not transitive

Solution

Given that $a R b \Leftrightarrow|a-b| \leq 1$ Now, $a R a \Leftrightarrow|a-a|=0 \leq 1$ Therefore, $\mathrm{R}$ is reflexive Again, $a R b \Leftrightarrow|a-b| \leq 1$ Then, $b R a \Leftrightarrow|b-a| \leq 1$ $\Rightarrow|a-b| \leq 1$, which is true. Therefore, $\mathrm{R}$ is symmetric Take $\mathrm{a} \mathrm{b}==12$, Then, $|a-b|=|1-2|=1=1$ Take $\mathrm{b} \mathrm{c}==23$ and . Then, $|b-c|=|2-3|=1$ But aRc $|a-c|=|1-3|=2>1$ Therefore, $\mathrm{R}$ not transitive Hence, $\mathrm{R}$ is reflexive and symmetric but not transitive.

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

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