Let R be a relation defined in the set of real numbers by aRb⇔1+ab>0. Then R is
A
Equivalence relation
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B
Transitive
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C
Symmetric
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D
Anti-symmetric
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Solution
The correct option is C Symmetric i) Reflexive: Let R' be the set of real numbers Let a∈R′⇒1+a.a=1+a2>0 ⇒(a,a)∈R i.e R is reflexive
ii) Symmetric: a,b∈R′ Let (a,b)∈R⇒1+a.b>0⇒1+b.a>0⇒(b,a)∈R i.e R is symmetric
iii) Transitive: The relation R is not transitive because we find that (2,12)∈R and (12,−2)∈R but (2,−2)∉R since 1+2(−2)=−3 which is not positive Hence R is reflexive, symmetric but not transitive