Question

Let $\mathrm{S}$ be the set of all real numbers. A relation $\mathrm{R}$ has been defined on $\mathrm{S}$ by $a\mathrm{Rb}=|a-b|\le 1$, then $R$ is.

• A. Symmetric and transitive but not reflexive
• B. Reflexive and transitive but not symmetric
• C. Reflexive and symmetric but not transitive
• D. an equivalence relation

Answer 1

The correct option is C

Reflexive and symmetric but not transitive

Explanation for the correct option.

Step 1: For reflexive

Given, $a\mathrm{Rb}=|a-b|\le 1$

For $aRa$ then,

$\begin{array}{rcl}\left|a-a\right|& \le & 1\\ 0& \le & 1\end{array}$

So, it is reflexive.

Step 2: For symmetric

Let $a\mathrm{Rb}=|a-b|\le 1$ then,

$\begin{array}{rcl}bRa& \Rightarrow & \left|b-a\right|\le 1\\ \left|-\left(a-b\right)\right|& \le & 1\\ \left|a-b\right|& \le & 1\end{array}$

So, it is symmetric.

Step 3: For transitive

Let, $a\mathrm{Rb}=|a-b|\le 1$ and $bRc=|b-c|\le 1$ then,

$\left|a-c\right|\le 1$ is not always true.

So, it is not transitive.

Hence option C is correct.