सिद्ध कीजिए कि वास्तविक संख्याओं के समुच्चय R में R = (a, b): a ≤ b², द्वारा परिभाषित सम्बन्ध R, न तो स्वतुल्य है, न सममित है और न ही संक्रामक है। (2024)

Step-by-step solution

1. 1

   Given Relation

   Relation $R$ on $\mathbb{R}$ is $R = \{(a, b): a \le b^2\}$

   $$R = \{(a, b): a \le b^2\}$$
2. 2

   Definitions

   Reflexive: $(a, a) \in R$ for all $a \in \mathbb{R}$. Symmetric: $(a, b) \in R \implies (b, a) \in R$. Transitive: $(a, b) \in R, (b, c) \in R \implies (a, c) \in R$.
3. 3

   Proof Strategy

   To prove 'not', find a single counterexample for each property.
4. 4

   Not Reflexive

   For $a = 1/2$, is $1/2 \le (1/2)^2$? No, $1/2 \not\le 1/4$. So $(1/2, 1/2) \notin R$.

   $$\frac{1}{2} \not\le \left(\frac{1}{2}\right)^2$$
5. 5

   Not Symmetric

   Take $(1, 2)$. $1 \le 2^2$ (i.e., $1 \le 4$) is true. So $(1, 2) \in R$. Now check $(2, 1)$. Is $2 \le 1^2$? No, $2 \not\le 1$. So $(2, 1) \notin R$.

   $$1 \le 2^2 \text{ but } 2 \not\le 1^2$$
6. 6

   Not Transitive

   Take $(3, 2)$. $3 \le 2^2$ (i.e., $3 \le 4$) is true. So $(3, 2) \in R$. Take $(2, 1.5)$. $2 \le (1.5)^2$ (i.e., $2 \le 2.25$) is true. So $(2, 1.5) \in R$. Now check $(3, 1.5)$. Is $3 \le (1.5)^2$? No, $3 \not\le 2.25$. So $(3, 1.5) \notin R$.

   $$3 \le 2^2 \text{ and } 2 \le (1.5)^2 \text{ but } 3 \not\le (1.5)^2$$
7. 7

   Conclusion

   The relation $R$ is neither reflexive, nor symmetric, nor transitive.