[proofplan]
The proof is a direct determinant computation. First we record the standard determinant identity for consecutive convergents from the defining recurrence. Then we compute $R_{n,m+1} - R_{n,m}$; the denominator is positive and the numerator is exactly the consecutive-convergent determinant, so its sign depends only on the parity of $n$. The endpoint identities follow by substituting $m=0$ and $m=a_{n+1}$ into the same recurrence.
[/proofplan]
[step:Establish positivity of the continued fraction denominators]
By the theorem statement, every partial quotient $a_k$ with $k \geq 1$ belongs to $\mathbb{N}=\{1,2,3,\dots\}$, so $a_k>0$. We first prove that $q_k > 0$ for every integer $k \geq 0$. The initial value is $q_0 = 1 > 0$. Also
\begin{align*}
q_1 = a_1 q_0 + q_{-1} = a_1 > 0.
\end{align*}
For $k \geq 2$, if $q_{k-1} > 0$ and $q_{k-2} > 0$, then, since $a_k \in \mathbb{N}$,
\begin{align*}
q_k = a_k q_{k-1} + q_{k-2} > 0.
\end{align*}
Induction gives $q_k > 0$ for all $k \geq 0$.
Therefore, if $n \geq 1$ and $0 \leq m \leq a_{n+1}$, then
\begin{align*}
m q_n + q_{n-1} > 0,
\end{align*}
because $q_{n-1} > 0$ and $m q_n \geq 0$. For $n=0$ and $1 \leq m \leq a_1$, the denominator is
\begin{align*}
m q_0 + q_{-1} = m > 0.
\end{align*}
[/step]
[step:Compute the determinant of consecutive convergents]
We use the defining recurrence relations for the continued-fraction convergents: $p_{-1}=1$, $p_0=a_0$, $p_k=a_kp_{k-1}+p_{k-2}$ for $k\geq 1$, and $q_{-1}=0$, $q_0=1$, $q_k=a_kq_{k-1}+q_{k-2}$ for $k\geq 1$.
[claim:Consecutive convergents have alternating determinant]
For every integer $k \geq 0$,
\begin{align*}
p_k q_{k-1} - p_{k-1} q_k = (-1)^{k-1},
\end{align*}
where the case $k=0$ means $p_0q_{-1}-p_{-1}q_0=-1$.
[/claim]
[proof]
For $k=0$,
\begin{align*}
p_0q_{-1}-p_{-1}q_0 = a_0 \cdot 0 - 1 \cdot 1 = -1 = (-1)^{-1}.
\end{align*}
Here $(-1)^{-1}=-1$, so the displayed formula holds.
Now let $k \geq 1$. Using the recurrence relations for $p_k$ and $q_k$, we compute
\begin{align*}
p_k q_{k-1} - p_{k-1} q_k = (a_k p_{k-1} + p_{k-2})q_{k-1} - p_{k-1}(a_k q_{k-1} + q_{k-2}).
\end{align*}
Expanding the products gives
\begin{align*}
p_k q_{k-1} - p_{k-1} q_k = a_k p_{k-1}q_{k-1} + p_{k-2}q_{k-1} - a_k p_{k-1}q_{k-1} - p_{k-1}q_{k-2}.
\end{align*}
Cancelling the two copies of $a_kp_{k-1}q_{k-1}$ yields
\begin{align*}
p_k q_{k-1} - p_{k-1} q_k = p_{k-2}q_{k-1} - p_{k-1}q_{k-2}.
\end{align*}
Rewriting the right-hand side gives
\begin{align*}
p_k q_{k-1} - p_{k-1} q_k = -\bigl(p_{k-1}q_{k-2} - p_{k-2}q_{k-1}\bigr).
\end{align*}
Thus each determinant is the negative of the preceding determinant. Since the determinant at $k=0$ is $-1$, induction gives
\begin{align*}
p_k q_{k-1} - p_{k-1}q_k = (-1)^{k-1}
\end{align*}
for every $k \geq 0$.
[/proof]
[guided]
The goal of this step is to isolate the only sign information we will need later. Define, for each integer $k \geq 0$, the determinant-like integer
\begin{align*}
D_k := p_k q_{k-1} - p_{k-1}q_k.
\end{align*}
This quantity measures the signed cross-difference between two consecutive convergents.
First compute the initial value. Since $p_{-1}=1$, $q_{-1}=0$, $p_0=a_0$, and $q_0=1$,
\begin{align*}
D_0 = p_0q_{-1}-p_{-1}q_0 = a_0 \cdot 0 - 1 \cdot 1 = -1.
\end{align*}
Now fix $k \geq 1$. The recurrence relations say
\begin{align*}
p_k = a_kp_{k-1}+p_{k-2}.
\end{align*}
Also,
\begin{align*}
q_k = a_kq_{k-1}+q_{k-2}.
\end{align*}
Substituting these into $D_k$ gives
\begin{align*}
D_k = p_k q_{k-1} - p_{k-1}q_k.
\end{align*}
Substituting the recurrence relations gives
\begin{align*}
D_k = (a_k p_{k-1}+p_{k-2})q_{k-1} - p_{k-1}(a_kq_{k-1}+q_{k-2}).
\end{align*}
Expanding gives
\begin{align*}
D_k = a_kp_{k-1}q_{k-1}+p_{k-2}q_{k-1} - a_kp_{k-1}q_{k-1}-p_{k-1}q_{k-2}.
\end{align*}
Cancelling the two copies of $a_kp_{k-1}q_{k-1}$ gives
\begin{align*}
D_k = p_{k-2}q_{k-1}-p_{k-1}q_{k-2}.
\end{align*}
By the definition of $D_{k-1}$, this is
\begin{align*}
D_k = -D_{k-1}.
\end{align*}
The terms containing $a_kp_{k-1}q_{k-1}$ cancel exactly. This cancellation is the reason the sign of the cross-difference is independent of the partial quotient $a_k$.
Since $D_0=-1$ and $D_k=-D_{k-1}$ for every $k \geq 1$, the signs alternate:
\begin{align*}
D_k = (-1)^{k-1}.
\end{align*}
Equivalently,
\begin{align*}
p_k q_{k-1} - p_{k-1}q_k = (-1)^{k-1}
\end{align*}
for every $k \geq 0$.
[/guided]
[/step]
[step:Compare successive intermediate fractions]
Fix $n \geq 1$ and let $m$ be an integer with $0 \leq m < a_{n+1}$. For each integer $j$ with $0 \leq j \leq a_{n+1}$, define the intermediate fraction $R_{n,j}\in\mathbb{R}$ by
\begin{align*}
R_{n,j} := \frac{jp_n+p_{n-1}}{jq_n+q_{n-1}}.
\end{align*}
The denominator positivity proved above shows that each $R_{n,j}$ is well-defined. Define the positive denominators
\begin{align*}
d_m := m q_n + q_{n-1}, \qquad d_{m+1} := (m+1)q_n + q_{n-1}.
\end{align*}
Then
\begin{align*}
R_{n,m+1}-R_{n,m} = \frac{(m+1)p_n+p_{n-1}}{d_{m+1}} - \frac{mp_n+p_{n-1}}{d_m}.
\end{align*}
Putting the two fractions over the common positive denominator $d_md_{m+1}$ gives
\begin{align*}
R_{n,m+1}-R_{n,m} = \frac{\bigl((m+1)p_n+p_{n-1}\bigr)d_m - \bigl(mp_n+p_{n-1}\bigr)d_{m+1}}{d_md_{m+1}}.
\end{align*}
Expanding the numerator and cancelling like terms gives
\begin{align*}
\bigl((m+1)p_n+p_{n-1}\bigr)d_m
- \bigl(mp_n+p_{n-1}\bigr)d_{m+1}
= p_nq_{n-1}-p_{n-1}q_n.
\end{align*}
By the determinant identity proved above,
\begin{align*}
R_{n,m+1}-R_{n,m}
= \frac{(-1)^{n-1}}{d_md_{m+1}}.
\end{align*}
Since $d_m d_{m+1}>0$, this difference is positive when $n$ is odd and negative when $n$ is even. Hence $(R_{n,m})_{m=0}^{a_{n+1}}$ is strictly increasing when $n$ is odd and strictly decreasing when $n$ is even.
[/step]
[step:Identify the endpoints for $n \geq 1$]
For $m=0$, the definition gives
\begin{align*}
R_{n,0}
= \frac{0 \cdot p_n+p_{n-1}}{0 \cdot q_n+q_{n-1}}
= \frac{p_{n-1}}{q_{n-1}}.
\end{align*}
For $m=a_{n+1}$, the recurrence relations give
\begin{align*}
p_{n+1} = a_{n+1}p_n+p_{n-1}.
\end{align*}
Also,
\begin{align*}
q_{n+1} = a_{n+1}q_n+q_{n-1}.
\end{align*}
Therefore
\begin{align*}
R_{n,a_{n+1}}
= \frac{a_{n+1}p_n+p_{n-1}}{a_{n+1}q_n+q_{n-1}}
= \frac{p_{n+1}}{q_{n+1}}.
\end{align*}
Because the finite sequence is strict monotone and these are its two endpoint values, every $R_{n,m}$ with $0<m<a_{n+1}$ lies strictly between $\frac{p_{n-1}}{q_{n-1}}$ and $\frac{p_{n+1}}{q_{n+1}}$.
[/step]
[step:Handle the initial case $n=0$]
Let $m$ be an integer with $1 \leq m < a_1$. For each integer $j$ with $1\leq j\leq a_1$, define the initial intermediate fraction $R_{0,j}\in\mathbb{R}$ by
\begin{align*}
R_{0,j} := \frac{jp_0+p_{-1}}{jq_0+q_{-1}}.
\end{align*}
Since $p_{-1}=1$, $q_{-1}=0$, $p_0=a_0$, and $q_0=1$, this gives
\begin{align*}
R_{0,m} = \frac{m a_0+1}{m}.
\end{align*}
The denominator $m$ is positive by the admissibility condition $m \geq 1$.
For successive admissible values,
\begin{align*}
R_{0,m+1}-R_{0,m} = \frac{(m+1)p_0+p_{-1}}{(m+1)q_0+q_{-1}} - \frac{mp_0+p_{-1}}{mq_0+q_{-1}}.
\end{align*}
Putting the two fractions over the common positive denominator $m(m+1)$ gives
\begin{align*}
R_{0,m+1}-R_{0,m} = \frac{p_0q_{-1}-p_{-1}q_0}{m(m+1)}.
\end{align*}
Using $p_0q_{-1}-p_{-1}q_0=-1$, we get
\begin{align*}
R_{0,m+1}-R_{0,m} = \frac{-1}{m(m+1)} < 0.
\end{align*}
Thus $(R_{0,m})_{m=1}^{a_1}$ is strictly decreasing. Finally, using the recurrence relations at $k=1$,
\begin{align*}
p_1 = a_1p_0+p_{-1}.
\end{align*}
Also,
\begin{align*}
q_1 = a_1q_0+q_{-1}.
\end{align*}
so
\begin{align*}
R_{0,a_1}
= \frac{a_1p_0+p_{-1}}{a_1q_0+q_{-1}}
= \frac{p_1}{q_1}.
\end{align*}
This proves the claimed monotonicity, denominator positivity, and endpoint identity in the initial case.
[/step]
