Question

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

(A) Reflexive and symmetric but not transitive

(C) Symmetric, transitive but not reflexive

(B) Reflexive, transitive but not symmetric

(D) Reflexive, transitive and symmetric

Answer 1

SLUTION :  $1+a \cdot a=1+a^2>0, \forall a \in S, \therefore(a, a) \in R$

$\therefore \quad \mathrm{R}$ is reflexive

$(a, b) \in R \Rightarrow 1+a b>0 \Rightarrow \quad 1+b a>0 \Rightarrow \quad(b, a) \in R$

$\therefore \quad \mathrm{R}$ is symmetric.

$\because \quad(\mathrm{a}, \mathrm{b}) \in \mathrm{R}$ and $(\mathrm{b}, \mathrm{c}) \in \mathrm{R}$ need not imply $(\mathrm{a}, \mathrm{c}) \in \mathrm{R}$

Hence, $R$ is not transitive.

Above Correct Answer is Option A.