Q.2 Show that the relation R in the set R of real numbers, defined as
R = \{(a, b) : a \leq b^2 \} is neither reflexive nor symmetric nor transitive.

Answers (1)

R = \{(a, b) : a \leq b^2 \}

Taking  

\left ( \frac{1}{2},\frac{1}{2} \right )\notin R

and 

\left ( \frac{1}{2} \right )> \left ( \frac{1}{2} \right )^{2}

So,R is not reflexive.

Now,

\left ( 1,2 \right )\in R because    1< 4.

But, 4\nless 1  i.e. 4 is not less than 1 

So,\left ( 2,1 \right )\notin R

Hence, it is not symmetric.

\left ( 3,2 \right )\in R\, \, and \, \, \left ( 2,1.5 \right )\in R   as 3< 4\, \, and \, \, 2< 2.25

 Since \left ( 3,1.5 \right )\notin R  because 3\nless 2.25

Hence, it is not transitive.

Thus, we can conclude that it is neither reflexive, nor symmetric ,nor transitive. 

Preparation Products

Knockout NEET Sept 2020

An exhaustive E-learning program for the complete preparation of NEET..

₹ 15999/- ₹ 6999/-
Buy Now
Rank Booster NEET 2020

This course will help student to be better prepared and study in the right direction for NEET..

₹ 9999/- ₹ 4999/-
Buy Now
Knockout JEE Main Sept 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
Buy Now
Test Series NEET Sept 2020

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 11999/-
Buy Now
Exams
Articles
Questions