Question

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

• A. Reflexive and symmetric but not transitive
• B. Reflexive and transitive but not symmetric
• C. Symmetric, transitive but not reflexive
• D. Reflexive, transitive and symmetric
• E. None of the above is true

Answer 1

The correct option is A

Reflexive and symmetric but not transitive

Explanation of correct answer :

Determining the relation $R$ on $S$.

Given, $R=\left\{\right(a,b):1+ab>0\}$ on $S$

If $(a,a)$ satisfies the relation, the relation $R$ will be reflexive.

$$\begin{gathered} 1+1\times 1>0 \\ 1+a\times a>0 \\ (a,a)\in S \end{gathered}$$

Hence, $R$ is reflexive on $S$.

If $(a,b)$ and $(b,a)$ satisfies the relation, the relation $R$ will be symmetric.

$$\begin{gathered} 1+ab>0 \\ 1+ba>0 \\ (a,b)\in S \\ (b,a)\in S \end{gathered}$$

Hence, $R$ is symmetric on $S$.

If $(a,b)$, $(b,a)$, $(c,a)$ satisfies the relation, the relation $R$ will be transitive.

$$\begin{gathered} \left(a,b\right)\in S \\ (b,a)\in S \\ (c,a)\in S \end{gathered}$$

It is not transitive.

Thus,$R$ is Reflexive and symmetric but not transitive on $S$.

Hence, the correct option is Option (A).