1. **Problem statement:** (i) Prove that between any two real numbers $a$ and $b$ there is at least one rational number. (ii) Evaluate the following expressions and inequalities: (i) $|3^2 - 5|$ (ii) $\sqrt{2x + 3} > x$ (iii) $|2x^2 - 13| < 5$ 2. **Proof for (i):** We want to show that for any real numbers $a < b$, there exists a rational number $r$ such that $a < r < b$. 3. **Key idea:** Since rationals are dense in the real numbers, between any two real numbers there is always a rational number. 4. **Proof steps:** - Because $b - a > 0$, choose a natural number $n$ such that $\frac{1}{n} < b - a$. - Consider the set of rational numbers of the form $\frac{m}{n}$ where $m$ is an integer. - There exists an integer $m$ such that $a < \frac{m}{n} < b$ because the multiples of $\frac{1}{n}$ are spaced by $\frac{1}{n}$ and cover the real line. 5. **Conclusion:** Thus, there is at least one rational number between $a$ and $b$. 6. **Evaluate (ii)(i):** Calculate $|3^2 - 5| = |9 - 5| = |4| = 4$. 7. **Evaluate (ii)(ii):** Solve inequality $\sqrt{2x + 3} > x$. - 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 $x \geq -\frac{3}{2}$, final solution: $$-1 < x < 3$$ 8. **Evaluate (ii)(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:** (i) There is always a rational number between any two real numbers $a$ and $b$. (ii)(i) $4$ (ii)(ii) $-1 < x < 3$ (ii)(iii) $-3 < x < -2$ or $2 < x < 3$