NCERT Solutions of Class 12 Maths | CBSE Textbook Solutions | Chapter 1 | Relations and Functions | Exercise 1.1 | Question 2 |

Question Details
BookNCERT Textbook
Chapter1 [Relations and Functions]
Question No.2
Question TypeExercise

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

Expert Answer

(i) \(\mathrm{R}\) is not reflexive, \(\because\) a is not less than or equal to \(\mathrm{a}^2\) for all \(\mathrm{a} \in \mathrm{R}\), e.g., \(\frac{1}{2}\) is not less than \(\frac{1}{4}\).
(ii) \(\mathrm{R}\) is not symmetric since if \(\mathrm{a} \leq \mathrm{b}^2\) then \(\mathrm{b}\) is not less than or equal to \(\mathrm{a}^2\), e.g. \(2<5^2\) but 5 is not less than \(2^2\).
(iii) \(\mathrm{R}\) is not transitive : If \(\mathrm{a} \leq \mathrm{b}^2, \mathrm{~b} \leq \mathrm{c}^2\), then a is not less than \(\mathrm{c}^2\), e.g. \(2<(-2)^2,-2<(-1)^2\), but 2 is not less than \((-1)^2\).

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