1. **Problem statement:** (a) Prove that between any two real numbers $a$ and $b$ there is at least one rational number. 2. **Formula and important rules:** We use the density property of rational numbers which states: Between any two real numbers, there exists a rational number. 3. **Proof:** Assume without loss of generality that $a < b$. Since $b - a > 0$, choose a natural number $n$ such that $$\frac{1}{n} < b - a.$$ This is possible by the Archimedean property. 4. Consider the set of numbers $$\left\{\frac{k}{n} : k \in \mathbb{Z}\right\}.$$ There exists an integer $k$ such that $$\frac{k}{n} > a$$ because the multiples of $\frac{1}{n}$ are dense. 5. Since $$\frac{k-1}{n} \leq a < \frac{k}{n},$$ and $$\frac{1}{n} < b - a,$$ it follows that $$\frac{k}{n} < a + \frac{1}{n} < b.$$ Thus, $$a < \frac{k}{n} < b,$$ and $$\frac{k}{n}$$ is a rational number between $a$ and $b$. 6. **Final conclusion:** Between any two real numbers $a$ and $b$, there exists at least one rational number. --- Since the user asked multiple questions but per instructions we only solve the first, we stop here.