Exercise 7.4.3 (Convergents of $\beta$ when $\alpha

Question

Let $\alpha,\ \beta,\ \gamma$ be irrational numbers satisfying $\alpha < \beta < \gamma$. If $\alpha$ and $\gamma$ have identical convergents $h_0/k_0,\ h_1/k_1, \ldots$, up to $h_n/k_n$, prove that $\beta$ also has these same convergents up to $h_n/k_n$.

Richard Ganaye · Accepted Answer

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\begin{proof} 
Let $\alpha,\ \beta,\ \gamma$ be irrational numbers satisfying $\alpha < \beta < \gamma$, and
\begin{align*}
\alpha &= \langle a_0,a_1, a_2,\ldots angle,&(\xi_i = \langle a_i, a_{i+1},\ldots angle)\
\beta&= \langle a'_0,a'_1, a'_2,\ldots angle,&(\xi'_i = \langle a_i, a_{i+1},\ldots angle)\
\gamma &= \langle a''_0,a''_1, a''_2,\ldots angle&(\xi''_i = \langle a_i, a_{i+1},\ldots angle).
\end{align*}
We write $h'_i/k'_i$ the convergents of $\beta$, and $h''_i/k''_i$ the convergents of $\gamma$.

By hypothesis,  $\alpha$ and $\gamma$ have identical convergents up to rank $n$, so $h_i/k_i = h''_i/k''_i$. Since these fractions are in lowest terms,
\begin{align}
h_j= h''_j,\quad k_j= k''_i,\qquad (0 \leq j \leq n).
\end{align}
 Then Problem 2 shows that
\begin{align}
a_j = a''_j\qquad (0 \leq j\leq n).
\end{align}
Since $a_0 = a''_0$, we obtain
$$\lfloor \alpha floor = a_0 = a''_0 = \lfloor \gamma floor,$$
therefore $a_0 \leq \alpha < \beta < \gamma < a_0 + 1$, so $a'_0 = \lfloor \beta floor = a_0$.

Reasoning by induction, suppose that $(a_0, a_1,\ldots,a_i) = (a'_0, a'_1,\ldots,a'_i)$ for some $i<n$. Then, for all $j \in [\![1,i]\!]$,
\begin{align*}
\frac{h'_j}{k'_j} &= \langle a'_0,a'_1,\ldots, a'_j angle\
&=\langle a_0,a_1,\ldots, a_j angle\
&= \frac{h_j}{k_j}.
\end{align*}
Therefore $h_j/k_j = h'_j/k'_j$, where these fractions are in lowest terms, so
\begin{align}
h_j= h'_j,\quad k_j= k'_i,\qquad (0 \leq j \leq i).
\end{align}

By Theorem 7.10 and Theorem 7.3,
\begin{align*}
\alpha &= \langle a_0,a_1,\ldots,a_i,\xi_{i+1} angle\
&= \frac{\xi_{i+1} h_i + h_{i-1}}{\xi_{i+1} h_i + h_{i-1}},
\end{align*}
and similar equalities for $\beta$ and $\gamma$, so
\begin{align}
&\alpha = \frac{\xi_{i+1} h_i + h_{i-1}}{\xi_{i+1} k_i +k_{i-1}},\qquad \beta = \frac{\xi'_{i+1} h'_i + h'_{i-1}}{\xi_{i+1} k'_i + k'_{i-1}},\qquad \gamma = \frac{\xi''_{i+1} h''_i + h''_{i-1}}{\xi''_{i+1} k''_i + k''_{i-1}}.
\end{align}
Using (1) and (3), this gives
\begin{align}
\frac{\xi_{i+1} h_i + h_{i-1}}{\xi_{i+1} k_i + k_{i-1}} < \frac{\xi'_{i+1} h_i + h_{i-1}}{\xi_{i+1} k_i + k_{i-1}} < \frac{\xi''_{i+1} h_i + h_{i-1}}{\xi_{i+1} k_i + k_{i-1}}.
\end{align}
Consider the homographic function
$$
f
\left\{
\begin{array}{ccl}
\R^+ & 	o & \R^+\
x  &\mapsto &\frac{x h_i + h_{i-1}}{x k_i + k_{i-1}}
\end{array}
ight.,
$$
so that (5) becomes
\begin{align}
f(\xi_{i+1}) < f(\xi'_{i+1}) < f(\xi'_{i+1}).
\end{align}
Then for all $x \in \R^+$, $f'(x) = \frac{h_i k_{i-1} - k_i h_{i-1}}{(xk_i + k_{i-1})^2 }= \frac{(-1)^{i-1}}{(xk_i + k_{i-1})^2}$, so $f$ is strictly monotonic on $\R^+$. Then 
\begin{align}
\xi_{i+1} < \xi'_{i+1} < \xi''_{i+1}	ext{ or } \xi_{i+1} < \xi'_{i+1} < \xi''_{i+1}.
\end{align}
Since $i+1 \leq n$, $a''_{i+1} = a_{i+1}$ by (2), thus
$$\lfloor \xi''_{i+1}floor  = a''_{i+1} = a_{i+1} = \lfloor \xi_{i+1} floor.$$
Therefore, (7) implies
\begin{align*}
&a_{i+1} \leq \xi_{i+1} < \xi'_{i+1} < \xi''_{i+1} < a_{i+1}\
&\qquad 	ext{or}\
&a_{i+1} \leq \xi_{i+1} < \xi'_{i+1} < \xi''_{i+1} < a_{i+1}.
\end{align*}
In both cases, $a_{i+1} = \lfloor \xi'_{i+1} floor = a'_{I+1}$. The induction is done, which proves that  $a_i = a'_i$ for all $i \in [\![0,n]\!]$.

Then, for all $j \in  [\![0,n]\!]$,
$$\frac{h_j}{k_j} = \langle a_0,a_1,\ldots, a_j angle = \langle a'_0,a'_1,\ldots, a'_j angle = \frac{h'_j}{k'_j},$$
so  $\beta$ also has the same convergents as $\alpha$ and $\gamma$ up to $h_n/k_n$.
\end{proof}

Exercise 7.4.3 (Convergents of $\beta$ when $\alpha < \beta < \gamma$)

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Exercise 7.4.3 (Convergents of $\beta$ when $\alpha < \beta < \gamma$)

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