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}$> 0$
So, 2R \left(- \dfrac{1}{6}\right)and \left(- \dfrac{1}{6}\right)R \left(-2\right)
But, $2 R \left(-2\right)$ since $2 + 2\left(-2\right) = - 2 ≯ 0$