Let S be the relation on the set R of real numbers defined by \(S = \{(a, b) : 1 + ab \gt 0\}\). Then S is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → reflexive and symmetric but not transitive
Reflexive: for every real number a,
\(1 + a \cdot a = 1 + a^{2} \ge 1 \gt 0\)
Symmetric: \(ba = ab\), so \(1 + ab \gt 0\) gives \(1 + ba \gt 0\).
Transitive fails. Take \(a = -2\), \(b = 0\), \(c = 3\):
\(1 + (-2)(0) = 1 \gt 0\), so \((-2, 0) \in S\)
\(1 + (0)(3) = 1 \gt 0\), so \((0, 3) \in S\)
\(1 + (-2)(3) = -5 \lt 0\), so \((-2, 3) \notin S\)
So S is reflexive and symmetric but not transitive.