A) reflexive and symmetric but not transitive
B) symmetric and transitive but not reflexive
C) transitive but neither reflexive nor symmetric
D) reflexive, symmetric and transitive
Correct Answer: A
Solution :
\[R\]is reflexive since \[|a-a|=0\le \frac{1}{2}\forall \in R\] Also, R is symmetric as \[|a-b|\,\,\le \frac{1}{2}\Rightarrow |b-a|\,\,\le \frac{1}{2}\] e.g.: If we take three numbers \[\frac{3}{4},\,\,\frac{1}{3},\,\,\frac{1}{8}\] then \[\left| \frac{3}{4}-\frac{1}{3} \right|=\frac{5}{12}\le \frac{1}{2}\] and \[\left| \frac{1}{3}-\frac{1}{8} \right|=\frac{5}{24}\le \frac{1}{2}\] But \[\left| \frac{3}{4}-\frac{1}{8} \right|=\frac{5}{8}>\frac{1}{2}\] Thus, \[\frac{3}{4}R\frac{1}{3}\]and\[\frac{1}{3}R\frac{1}{8}\]but \[\frac{3}{7}R\frac{1}{8}\].
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