1. **Problem Statement:** (a) Prove that between any two real numbers $a$ and $b$ there is at least one rational number. (b) Evaluate: (i) $|3^2 - 5^2|$ (ii) Solve the inequality $\sqrt{2x + 3} > x$ (iii) Solve the inequality $|2x^2 - 13| < 5$ 2. **Part (a) Proof:** - Given two real numbers $a$ and $b$ with $a < b$. - We want to show there exists a rational number $r$ such that $a < r < b$. - Since $b - a > 0$, choose a natural number $n$ such that $\frac{1}{n} < b - a$. - By the Archimedean property, there exists an integer $m$ such that $m > a n$. - Then $\frac{m}{n} > a$. - Also, $\frac{m}{n} \leq a + \frac{1}{n} < a + (b - a) = b$. - Therefore, $a < \frac{m}{n} < b$, and $\frac{m}{n}$ is a rational number between $a$ and $b$. 3. **Part (b) Evaluations:** (i) Calculate $|3^2 - 5^2|$: $$3^2 = 9, \quad 5^2 = 25$$ $$|9 - 25| = |-16| = 16$$ (ii) Solve $\sqrt{2x + 3} > x$: - Domain: $2x + 3 \geq 0 \Rightarrow x \geq -\frac{3}{2}$. - Square both sides (valid since both sides are non-negative in domain): $$2x + 3 > x^2$$ - Rearrange: $$x^2 - 2x - 3 < 0$$ - Factor: $$(x - 3)(x + 1) < 0$$ - Inequality holds between roots: $$-1 < x < 3$$ - Combine with domain $x \geq -\frac{3}{2}$: $$-1 < x < 3$$ (iii) Solve $|2x^2 - 13| < 5$: - This means: $$-5 < 2x^2 - 13 < 5$$ - Add 13 to all parts: $$8 < 2x^2 < 18$$ - Divide by 2: $$4 < x^2 < 9$$ - Take square roots: $$-3 < x < -2 \quad \text{or} \quad 2 < x < 3$$ **Final answers:** (a) There is always a rational number between any two real numbers. (b)(i) $16$ (b)(ii) $-1 < x < 3$ (b)(iii) $-3 < x < -2$ or $2 < x < 3$