R= {(a, b): a ≤ b²} is neither reflexive nor symmetric nor transitive

Question 1:-  Show that the relation R on the set R of real numbers, defined as R= {(a, b): a ≤ b²} is neither reflexive nor symmetric nor transitive.

Solution: We have, R=\{(a,b): a\leq b^2\} where a, b ∈ R

Reflexivity:- Obiviously, \frac{1}{2} is a real number and \frac{1}{2}\leq (\frac{1}{2})^2 is not true.

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

Therefore, R is not reflexive.

Symmetric:- Let the real numbers 1 and 2

such that (1,2)\in R \Rightarrow 1\leq 2^2

\Rightarrow 2 \leq 1^2 is not true

Thus (2,1)\notin R

Hence, R is not symmetric.

Transitive:- Let a = 2, b = -2, c = 1

(2,-2)\in R and (-2, 1)\in R \quad \Rightarrow 2 \leq (-2)^2 and -2\leq (1)^2

\quad \Rightarrow 2 \leq (1)^2 is not true

\Rightarrow (2,1)\notin R

Hence, R is not transitive.

 

Class 12 ncert solution math exercise 1.1

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