First slide

Relations XII

Question

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

Easy

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

Since $1 + a . a = 1 + a^{2} > 0, \forall a \in S, ∴ (a, a) \in R$
$∴ R is reflexive . Also (a, b) \in R \Rightarrow 1 + ab > 0 \Rightarrow 1 + ba > 0 \Rightarrow (b, a) \in R,$
$∴$ R is symmetric.
$∵ (a, b) \in R and (b, c) \in R$ need not imply $(a, c) \in R .$
Hence, R is not transitive.

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App