Question
Let $$\rho $$ be the relation on the set $$R$$ of all real numbers defined by setting $$a\rho b$$ iff $$\left| {a - b} \right| \leqslant \frac{1}{2}.$$ Then $$\rho $$ is :
A.
reflexive and symmetric but not transitive
B.
symmetric and transitive but not reflexive
C.
transitive but neither reflexive nor symmetric
D.
none of these
Answer :
reflexive and symmetric but not transitive
Solution :
$$\eqalign{
& \rho \,{\text{is reflexive, since }}\left| {a - a} \right| = 0 < \frac{1}{2}{\text{ for all }}a\, \in \,R \cr
& \rho \,{\text{is symmetric, since }} \Rightarrow \left| {b - a} \right| < \frac{1}{2} \cr
& \rho \,{\text{is not transitive}}{\text{.}} \cr
& {\text{For, if we take three numbers }}\frac{3}{4},\frac{1}{3},\frac{1}{8} \cr
& {\text{Then,}}\,\,\left| {\frac{3}{4} - \frac{1}{3}} \right| = \frac{5}{{12}} < \frac{1}{2}{\text{ and }}\left| {\frac{1}{3} - \frac{1}{8}} \right| = \frac{5}{{24}} < \frac{1}{2} \cr
& {\text{But, }}\left| {\frac{3}{4} - \frac{1}{8}} \right| = \frac{5}{8} > \frac{1}{2} \cr
& {\text{Thus, }}\frac{3}{4}\rho \frac{1}{3}{\text{ and }}\frac{1}{3}\rho \frac{1}{8}{\text{ but }}\frac{3}{4}\left( { \sim \rho } \right)\frac{1}{8} \cr} $$