Consider a relation R defined as aRb if 2 + ab > 0 where a, b are...

Relation $R$ is reflexive, since $2 + a \times a > 0, \forall$ real number $a$.

It is symmetric, since $2 + ab > 0 ⇒ 2 + ba > 0.$
However, $R$ is not transitive.

Consider three real numbers $2, - \dfrac{1}{6}$ and $- 2$. We have
$\text{2} + 2 \times \left(- \dfrac{1}{6}\right) = \dfrac{5}{3}$> 0$<br />and \text{2} + \left(- \dfrac{1}{6}\right) \left(-2\right) = \dfrac{7}{3}$> 0$<br />So, 2R\left(- \dfrac{1}{6}\right)$ and \left(- \dfrac{1}{6}\right)$R \left(-2\right)$
But, $2 R \left(-2\right)$ since 2 + 2\left(-2\right) = - 2 ≯ 0