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

Since 1+a.a=1+a2>0, ∀a∈S, ∴ (a, a)∈R∴ R is reflexive. Also a, b∈R⇒1+ab>0⇒1+ba>0⇒(b, a)∈R,∴R is symmetric.∵(a, b)∈R and (b, c)∈R need not imply (a, c)∈R.Hence, R is not transitive.