Let A be set of all real numbers and R be a relation on A defined by \( R=\{(a,b):1+ab\gt 0\} \), then relation R is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Reflexive and symmetric
Reflexive: for every real a, \( 1+a\cdot a=1+a^2\gt 0 \), so (a, a) is in R.
Symmetric: \( 1+ab=1+ba \), so if (a, b) is in R then so is (b, a).
Transitive: take a = 1, b = −1/2, c = −3. Then \( 1+ab=\frac{1}{2}\gt 0 \) and \( 1+bc=\frac{5}{2}\gt 0 \), but \( 1+ac=-2\lt 0 \). So R is not transitive.
Hence R is reflexive and symmetric, but not transitive (so not an equivalence relation).