Let S be set of all real numbers. Then relation R = {(a, b) : 1 + ab > 0} on S is

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1.Relation and Function

hard

Let $S$ be the set of all real numbers. Then the relation $R = \{(a, b) : 1 + ab > 0\}$ on $S$ is

A

Reflexive and symmetric but not transitive

B

Reflexive and transitive but not symmetric

C

Symmetric, transitive but not reflexive

D

Reflexive, transitive and symmetric

Solution

(a) Since $1 + a.a = 1 + {a^2} > 0\,,\forall a \in S$, $(a,\,a) \in R$

$R$ is reflexive.

Also$(a,b) \in R$ ==> $1 + ab > 0$ ==> $1 + ba > 0$ ==> $(b,\,a) \in R$,

$R$ is symmetric.

and $(b,\,c) \in R$ need not imply $(a,\,c) \in R$. Hence, $R$ is not transitive.

Standard 12

Mathematics

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