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।

Step-by-step solution

1. 1

   Given Relation

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

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

   Definitions of Properties

   Reflexive: $(a,a) \in R$ for all $a$. 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

   Strategy: Counterexamples

   To prove a property does not hold, find a single counterexample.
4. 4

   Check Reflexivity

   For $a = 1/2$, $a \le a^2$ means $1/2 \le (1/2)^2 = 1/4$, which is false.

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

   Check Symmetry

   For $a=1, b=2$, $1 \le 2^2$ is true. But $2 \le 1^2$ is false.

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

   Check Transitivity

   Consider $(3,2)$ and $(2,1.5)$. $3 \le 2^2 = 4$ (true). $2 \le (1.5)^2 = 2.25$ (true).

   $$3 \le 2^2 \text{ and } 2 \le (1.5)^2$$
7. 7

   Transitivity Counterexample

   However, $3 \le (1.5)^2 = 2.25$ is false. So, R is not transitive.

   $$3 \not\le (1.5)^2$$
8. 8

   Conclusion

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