| Board | CBSE |
| Book | NCERT Textbook |
| Class | 12 |
| Subject | Mathematics |
| Chapter | 1 [Relations and Functions] |
| Exercise | 1.1 |
| Question No. | 2 |
| Question Type | Exercise |
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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