Question

Let R be the relation on the set of all real numbers defined by a R b iff $|a-b|\le \frac{1}{2}.$ Then R is

• A. reflexive and symmetric but not transitive
• B. symmetric and transitive but not reflexive
• C. transitive but neither reflexive nor symmetric
• D. Equivalence

Answer 1

The correct option is A reflexive and symmetric but not transitive

$\text{Now,}|a-a|=0\le \frac{1}{2}$

$\therefore aRa,\forall a\in R$

Hence, $R$ is reflecive

$\text{again,}aRb\Rightarrow |a-b|⩽\frac{1}{2}$

$\Rightarrow |b-a|\le \frac{1}{2}\cdots (\begin{array}{c}\text{using property}\\ \text{of moolulus)}\end{array}$

$\Rightarrow bRa$

Hence, $R$ is symmetric
$\frac{1}{2}R1$ ie $|\frac{1}{2}-1|=\frac{1}{2}$

$1R\frac{3}{2}$ ie $|1-\frac{3}{2}|=\frac{1}{2}$

but $\frac{1}{2}R\frac{3}{2}ie,|\frac{1}{2}-\frac{3}{2}|=1$

50, $R$ is not Transitive