I'm trying to understand the proof of Lemma 2.1 in the paper Analyse de Fourier des fractions continues à quotients restreints by Queffelec and Ramare. (Link: https://ramare-olivier.github.io/Maths/kaufmanbis.pdf) The original is in French. I will translate below and then ask my questions.
Notation.
Let $x$ be a real number with simple continued fraction expansion $x=[0;a_1(x),a_2(x),\ldots]$. Define $x_{k+1} = [a_{k+1}(x);a_{k+2}(x),a_{k+3}(x),\ldots]$. Define $P_k(x)$ and $Q_k(x)$ to be the unique relatively prime positive integers such that $$[0;a_1(x),a_2(x),\ldots,a_k(x)] = \dfrac{P_k(x)}{Q_{k}(x)}$$
Lemma 2.1
Let $N \geq 2$ be an integer. Let $x$ and $y$ be two real numbers with continued fractions expansions $x=[0;a_1(x),a_2(x),\ldots]$ and $y=[0;a_1(y),a_2(y),\ldots]$.
If $a_i(x), a_i(y) \in \{1,\ldots,N\}$ for every $i$ and $a_i(x) = a_i(y)$ for $1 \leq i \leq k$, then
\begin{align}
\label{1}\tag{1}
|x-y| &\leq \dfrac{N^2}{Q_{k+1}(x)^2}
\end{align}
If, in addition, $a_{k+1}(x) \neq a_{k+1}(y)$, then
\begin{align}
\label{2}\tag{2}
|x-y| &\geq \dfrac{1}{N(N+2)Q_{k+1}(x)^2}
\end{align}
Proof
We have $Q_i(x) = Q_i(y)$ for $1 \leq i \leq k$, so we write $Q_i := Q_i(x) = Q_i(y)$ for $1 \leq i \leq k$. We have \begin{align}\label{3}\tag{3} x-y = \dfrac{(-1)^k (y_{k+1} - x_{k+1})}{(x_{k+1} Q_k + Q_{k-1})(y_{k+1} Q_k + Q_{k-1})}, \end{align} \begin{align}\label{4}\tag{4} y_{k+1} Q_k + Q_{k-1} \leq \left( \dfrac{y_{k+1}}{a_{k+1}(x)} + 1 \right) ( a_{k+1}(x) Q_k + Q_{k-1} ) \leq (N+2) Q_{k+1}(x), \end{align} \begin{align}\label{5}\tag{5} y_{k+1} Q_k + Q_{k-1} \geq \dfrac{1}{a_{k+1}(x)} ( a_{k+1}(x) Q_k + Q_{k-1} ) \geq \dfrac{1}{N} Q_{k+1}(x). \end{align} It remains to lower bound $|y_{k+1} - x_{k+1}|$ under the hypothesis $a_{k+1}(x) \neq a_{k+1}(y)$. The worst that can happen is that $x_{k+1}$ is as large as possible compared to $a_{k+1}(x)$, that $a_{k+1}(y) = a_{k+1}(x) + 1$, and that $y_{k+1}$ is as small as possible. Then \begin{align}\label{6}\tag{6} |y_{k+1} - x_{k+1}| \geq 1 - \dfrac{\sqrt{N^2 + 1}}{N+1} + \dfrac{\sqrt{2}}{N+1} \geq \dfrac{1}{N} \end{align}
Questions
- How to get the first inequality in \eqref{6}? I don't understand the argument.
- If I combine \eqref{3}, \eqref{4}, \eqref{6}, I get: \begin{align} |x-y| \geq \dfrac{1}{N(N+2)Q_{k+1}(x)} \cdot \dfrac{1}{x_{k+1} Q_k + Q_{k-1}} \end{align} To obtain \eqref{2}, it seems like I am supposed to have $x_{k+1} Q_k + Q_{k-1} \leq Q_{k+1}(x)$. But that isn't true because $x_{k+1} Q_k + Q_{k-1} > a_{k+1}(x) Q_k + Q_{k-1} = Q_{k+1}(x)$. How am I supposed to deduce \eqref{2} from what's stated in the proof?