Exercise 7.5.6 ($-\xi$ continued fraction expansion)

Proof.

• Suppose that $a_1>1$.

  Put ${\xi }_n=⟨a_n,a_{n+1},a_{n+2}\dots ⟩$ for any integer $n\in \mathbb{N}$. By (7.7), we have

  $$\xi =a_0+\frac{1}{{\xi }_1},\text{ and }{\xi }_1=a_1+\frac{1}{{\xi }_2}.$$

  Therefore

  $$\begin{array}{llll}\hfill -\xi & =-⟨a_0,a_1,a_2,\dots ⟩& \hfill & \\ \hfill & =-\left(a_0+\frac{1}{{\xi }_1}\right)& \hfill & \\ \hfill & =-a_0-1+\left(1-\frac{1}{{\xi }_1}\right)& \hfill & \\ \hfill & =-a_0-1+\frac{1}{1+\frac{1}{{\xi }_1-1}}& \hfill & \\ \hfill & =-a_0-1+\frac{1}{1+\frac{1}{a_1-1+\frac{1}{{\xi }_2}}}.& \hfill & \end{array}$$

  Since $a_1-1\in {\mathbb{N}}^{\text{∗}}$ and $-a_0-1\in \mathbb{Z}$, we obtain

  $$\begin{array}{llll}\hfill & a_1-1+\frac{1}{{\xi }_2}=⟨a_1-1,{\xi }_2⟩,& \hfill & \\ \hfill & 1+\frac{1}{a_1-1+\frac{1}{{\xi }_2}}=1+\frac{1}{⟨a_1-1,{\xi }_2⟩}=⟨1,a_1-1,{\xi }_2⟩,& \hfill & \\ \hfill & -a_0-1+\frac{1}{1+\frac{1}{a_1-1+\frac{1}{{\xi }_2}}}=-a_0-1+\frac{1}{⟨1,a_1-1,{\xi }_2⟩}=⟨-a_0-1,1,a_1-1,{\xi }_2⟩.& \hfill & \end{array}$$

  Hence

  $$\begin{array}{llll}\hfill -\xi & =⟨-a_0-1,1,a_1-1,{\xi }_2⟩& \hfill & \\ \hfill & =⟨-a_0-1,1,a_1-1,⟨a_2,a_3,a_4,\dots ⟩⟩& \hfill & \\ \hfill & =⟨-a_0-1,1,a_1-1,a_2,a_3,a_4,\dots ⟩& \hfill & \end{array}$$

• Suppose that $a_1=1$. Then ${\xi }_1=1+\frac{1}{{\xi }_2}$, thus

  $$\begin{array}{llll}\hfill -\xi & =-a_0-1+\frac{1}{1+\frac{1}{{\xi }_1-1}}& \hfill & \\ \hfill & =-a_0-1+\frac{1}{1+{\xi }_2}& \hfill & \\ \hfill & =-a_0-1+\frac{1}{1+a_2+\frac{1}{{\xi }_3}}& \hfill & \\ \hfill & & \hfill \end{array}$$

  As previously,

  $$\begin{array}{llll}\hfill & 1+a_2+\frac{1}{{\xi }_3}=⟨1+a_2,{\xi }_3⟩,& \hfill & \\ \hfill & -a_0-1+\frac{1}{1+a_2+\frac{1}{{\xi }_3}}=-a_0-1+\frac{1}{⟨1+a_2,{\xi }_3⟩}=⟨-a_0-1,1+a_2,{\xi }_3⟩.& \hfill & \\ \hfill & & \hfill \end{array}$$

  Therefore

  $$\begin{array}{llll}\hfill -\xi & =⟨-a_0-1,1+a_2,{\xi }_3⟩& \hfill & \\ \hfill & =⟨-a_0-1,1+a_2,⟨a_3,a_4,a_5,\dots ⟩⟩& \hfill & \\ \hfill & =⟨-a_0-1,1+a_2,a_3,a_4,a_5,\dots ⟩.& \hfill & \end{array}$$

In conclusion,

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