Question

Let $S$ be the set of all real numbers. Then, the relation $R = \{(a$, $b)$ : $1 + ab > 0\}$ on $S$ is

• A. Reflexive and symmetric but not transitive
• B. Reflexive and transitive but not symmetric
• C. Symmetric and transitive but not reflexive
• D. Reflexive, transitive and symmetric

Answer 1

Answer: (A)

Reflexive : As $1+a \cdot a=1+a^{2} > 0$, $a \in S$.
$\therefore (a$, $b) \in R\,$
$ \therefore R$ is reflexive.
Symmetric : $(a$, $b)\in R$
$ \Rightarrow 1 + ab > 0$
$\Rightarrow 1+ba > 0$
$ \Rightarrow (b$, $a) \in R$.
$\therefore R$ is symmetric.
Transitive : $(a$, $b)\in R$ and $(b$, $c) \in R$ need not imply $(a$, $c) \in R$.
Hence, $R$ is not transitive.