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

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.