A relation R on the set R of real numbers defined as R = {(a,b) : 1 + ab > 0} is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Reflexive and symmetric but not transitive
Reflexive. For every real \( a \), \( 1+a\cdot a=1+a^{2}\gt 0 \), so \( (a,a)\in \mathrm{R} \).
Symmetric. If \( 1+ab\gt 0 \), then \( 1+ba\gt 0 \) because \( ab=ba \).
Transitive. This fails. Take \( a=1,\ b=0,\ c=-2 \).
Both \( (1,0) \) and \( (0,-2) \) belong to R, since \( 1+0=1\gt 0 \) in each case.
But \( 1+(1)(-2)=-1 \), which is not greater than 0, so \( (1,-2)\notin \mathrm{R} \).
Hence R is reflexive and symmetric but not transitive.