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

reflexive and symmetric but not transitive

$\rho$ is reflexive since $|a-a|=0<\frac{1}{2}$ for all $a \in R$

$\rho$ is symmetric since $|a-b|<\frac{1}{2}$

$\Rightarrow|b-a|<\frac{1}{2}$

$\rho$ is not transitive. For if we take three numbers $\frac{3}{4}, \frac{1}{3}, \frac{1}{8}$, then

$\left|\frac{3}{4}-\frac{1}{3}\right|=\frac{5}{12}<\frac{1}{2}$ and $\left|\frac{1}{3}-\frac{1}{8}\right|=\frac{5}{24}<\frac{1}{2}$

but $\left|\frac{3}{4}-\frac{1}{8}\right|=\frac{5}{8}>\frac{1}{2}$

Thus $\frac{3}{4} \rho \frac{1}{3}$ and $\frac{1}{3} \rho \frac{1}{8}$ but $\frac{3}{4}(\sim \rho) \frac{1}{8}$

Hence (1) is the correct answer.