A) Reflexive and symmetric
B) Symmetric and transitive
C) Only transitive
D) An equivalence relation
Correct Answer: A
Solution :
[a] we have, \[R=\{(a,b):1+ab>0,ab\in R\}\] Let \[a\in R\therefore {{a}^{2}}\ge 0\] or \[1+{{a}^{2}}>0\] or \[(a,a)\in R\] \[\therefore \] R is reflexive. Let \[(a,b)\in R,\Rightarrow 1+ab>0\Rightarrow 1+ba>0\] \[\Rightarrow (b,a)\in R\therefore R\] is symmetric. \[\left( 2,\frac{1}{3} \right)\in R\] because \[1+2\left( \frac{1}{3} \right)=\frac{5}{3}>0\] \[\left( \frac{1}{3},-1 \right)\in R\] because \[1+\frac{1}{3}(-1)=\frac{2}{3}>0\] Now, \[(2,-1)\in R\]if \[1+2(-1)=-1<0,\] which is not true. \[\therefore \] R is not transitive.
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