[proofplan] We use the Archimedean property of $\mathbb{R}$ twice: first to find an integer $n$ with $n(b - a) > 1$ (ensuring the interval $(a, b)$ is wide enough to contain a fraction with denominator $n$), then to locate the correct numerator $m$ so that $a < m/n < b$. The floor function (or equivalently the well-ordering of $\mathbb{Z}$ above any real number) identifies $m$. [/proofplan] [step:Apply the Archimedean property to find $n \in \mathbb{N}$ with $n(b - a) > 1$] Since $a < b$, the difference $b - a > 0$. By the [Archimedean property](/theorems/1232) of $\mathbb{R}$ (form (3): for $x = b - a > 0$ and $y = 1$, there exists $n \in \mathbb{N}$ with $n(b-a) > 1$), choose such $n$. This ensures \begin{align*} \frac{1}{n} < b - a, \end{align*} so the interval $(a, b)$ has length greater than $1/n$. [guided] The Archimedean property guarantees the existence of $n$. The condition $1/n < b - a$ is geometrically transparent: if we tile the real line with intervals of length $1/n$, at least one lattice point $m/n$ must fall inside $(a, b)$, because $(a, b)$ is wider than the tile spacing. The next step makes this precise. [/guided] [/step] [step:Find the integer $m$ with $a < m/n < b$ using the floor function] By the [Existence of the Floor Function](/theorems/1235), there exists a unique $m_0 \in \mathbb{Z}$ with $m_0 \le na < m_0 + 1$. Set $m := m_0 + 1 \in \mathbb{Z}$. Then \begin{align*} m = m_0 + 1 > na, \quad \text{so} \quad \frac{m}{n} > a. \end{align*} For the upper bound, $m_0 \le na$ gives $m = m_0 + 1 \le na + 1 < nb$ (using $1 < n(b - a)$, i.e., $na + 1 < nb$). Therefore \begin{align*} \frac{m}{n} < b. \end{align*} Combining: $a < m/n < b$. [guided] The integer $m_0 = \lfloor na \rfloor$ is the greatest integer not exceeding $na$. Then $m = m_0 + 1$ is the smallest integer strictly greater than $na$, giving $m/n > a$. The upper bound $m/n < b$ follows from: \begin{align*} m = m_0 + 1 \le na + 1 < na + n(b - a) = nb, \end{align*} where the strict inequality uses $1 < n(b - a)$. Dividing by $n > 0$ preserves the inequality, yielding $m/n < b$. [/guided] [/step] [step:Conclude that $r = m/n \in \mathbb{Q}$ satisfies $a < r < b$] The number $r := m/n$ is rational (since $m \in \mathbb{Z}$ and $n \in \mathbb{N}$) and satisfies $a < r < b$. This completes the proof. [/step]