First slide

Relations XII

Question

the relation R in the set R of real numbers, defined as $R = \{(a, b) : a \leq b^{2}\}$ is

Easy

A

Reflexive
B

Symmetric
C

Transitive
D

none

Solution

Given $R = \{(a, b) : a \leq b^{2}\}$
For $(\frac{1}{2}, \frac{1}{2}) \notin R a s \frac{1}{2} > {(\frac{1}{2})}^{2}$
Therefore, R is not reflexive
$Now, (1, 4) \in R as 1 < 4^{2}$
$But, 4 is not less than 1 .$
$Therefore, (4, 1) \notin R$
$Therefore, R is not symmetric,$
$Now, (3, 2), (2, 1.5) \in R (as 3 < 2^{2} = 4 and 2 < (1.5)^{2} = 2.25)$
$But, 3 > (1.5)^{2} = 2.25$
$Therefore, (3, 1.5) \notin R$
Therefore, R is not transitive. Hence, R is neither reflexive, nor symmetric, nor 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