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

Reflexive and symmetric but not transitive

The correct answer is Option (1) → Reflexive and symmetric but not transitive

We observe the following properties:

Reflexivity: Let a be an arbitrary element of R. Then,

$a ∈ R$

$⇒ 1+ a. a=1+ a^2 > 0$    [$∵ a^2>0$ for all $a ∈ R$]

$⇒(a, a) ∈ R_1$   [By def. of $R_1$]

Thus, $(a, a) ∈ R_1$ for all $a ∈ R$. So, $R_1$ is reflexive on R.

Symmetry: Let $(a, b) ∈ R$. Then,

$(a, b) ∈ R_1$

$⇒ 1+ ab > 0$

$⇒ 1 + ba > 0$ [$∵ ab = ba$ for all $a, b ∈ R$]

$⇒ (b, a) ∈ R_1$   [By def. of $R_1$]

Thus, $(a, b) ∈ R_1⇒ (b, a) ∈ R_1$ for all $a, b ∈ R$.

So, $R_1$ is symmetric on R.

Transitivity: We observe that $(1, 1/2) ∈ R_1$ and $(1/2, -1) ∈ R_1$ but $(1, -1)∉ R_1$ because $1 + 1 × (−1) = 0≯0$.

So, $R_1$ is not transitive on R.