Between any two different real numbers there are infinitely many rationals, even though the rationals have "gaps." The proof is a short, concrete recipe built on the Archimedean property, and it is a model for many analysis proofs: choose a scale, then count steps.
Key idea
Density of $\mathbb{Q}$. If $a < b$, there is a rational $\frac{m}{n}$ with $a < \frac{m}{n} < b$. Recipe: choose a positive integer $n$ with $\frac{1}{n} < b - a$; then let $m$ be the smallest integer greater than $na$.
Why it works
Steps of size $\frac{1}{n}$ are shorter than the gap $b - a$, so they cannot jump over the whole interval $(a, b)$. Precisely: $m$ is the smallest integer with $m > na$, so $m - 1 \le na$. Dividing by $n$: $a < \frac{m}{n}$ and $\frac{m}{n} \le a + \frac{1}{n} < a + (b - a) = b$. So $\frac{m}{n}$ is strictly between $a$ and $b$. The integer $m$ exists because the integers have no gaps: $m = \lfloor na \rfloor + 1$.
Applying the result again to $\left(a, \frac{m}{n}\right)$ gives another rational, and so on: there are infinitely many. The same idea shows the irrationals are dense too: find a rational $q$ between $\frac{a}{\sqrt{2}}$ and $\frac{b}{\sqrt{2}}$ with $q \ne 0$; then $q\sqrt{2}$ is irrational and lies between $a$ and $b$.
Worked examples
Worked example: Following the recipe
Find a fraction between $a = 0.31$ and $b = 0.34$ using the recipe.
- $b - a = 0.03$, and $\frac{1}{n} < 0.03$ means $n > 33.\overline{3}$, so $n = 34$.
- $na = 34 \cdot 0.31 = 10.54$, so the smallest integer above it is $m = 11$.
- Check: $\frac{11}{34} \approx 0.3235$, which is between $0.31$ and $0.34$.
Answer: $\frac{11}{34}$
Worked example: Next to an irrational
Use the recipe for $a = \sqrt{2}$ and $b = 1.42$.
- $b - a \approx 0.00579$, and $\frac{1}{0.00579} \approx 172.8$, so $n = 173$.
- $na = 173\sqrt{2} = \sqrt{59858}$. Since $244^{2} = 59536$ and $245^{2} = 60025$, $\lfloor 173\sqrt{2} \rfloor = 244$, so $m = 245$.
- $\frac{245}{173} \approx 1.41618$, which is indeed between $\sqrt{2} \approx 1.41421$ and $1.42$.
Answer: $\frac{245}{173}$
Common mistake
Choosing $n$ with $\frac{1}{n} \le b - a$, or $m$ with $m \ge na$. With equality allowed, the fraction can land exactly on $b$ (or on $a$). For $a = 0$ and $b = \frac{1}{2}$, the choice $n = 2$ gives $\frac{1}{n} = b - a$ and $m = 1$, so $\frac{m}{n} = \frac{1}{2} = b$, not strictly inside. The strict inequalities in the recipe are what make the final check $a < \frac{m}{n} < b$ work.
The recipe is not the only route. The fraction with the smallest denominator between $0.31$ and $0.34$ is $\frac{1}{3}$; the proof only needs some fraction, and it chooses one that is easy to justify.