Exercise 7.3.4 (Value of $\langle a_n,a_{n-1}, \ldots,a_0 \rangle$)

Proof. (of Lemma.)

(i)

By 7.6, $\left(\begin{array}{cc}\hfill a_0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill h_0\hfill & \hfill h_{-1}\hfill \\ \hfill k_0\hfill & \hfill k_{-1}\hfill \end{array}\right).$

Assume that the equality (1) is true for some integer $n$. Then

$$\begin{array}{llllll}\hfill \left(\begin{array}{cc}\hfill a_0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{cc}\hfill a_1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)\cdots \left(\begin{array}{cc}\hfill a_n\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{cc}\hfill a_{n+1}\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)& =\left(\begin{array}{cc}\hfill h_n\hfill & \hfill h_{n-1}\hfill \\ \hfill k_n\hfill & \hfill k_{n-1}\hfill \end{array}\right)\left(\begin{array}{cc}\hfill a_{n+1}\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)& \hfill & & \hfill & \\ \hfill & =\left(\begin{array}{cc}\hfill a_{n+1}{h}_n+h_{n-1}\hfill & \hfill h_n\hfill \\ \hfill a_{n+1}{k}_n+k_{n-1}\hfill & \hfill k_n\hfill \end{array}\right)& \hfill & & \hfill & \\ \hfill & =\left(\begin{array}{cc}\hfill h_{n+1}\hfill & \hfill h_n\hfill \\ \hfill k_{n+1}\hfill & \hfill k_n\hfill \end{array}\right)& \hfill \text{by (7.6)}.& & \hfill & & \hfill \end{array}$$

The induction is done, which proves (i).

(ii)

Put $\left(\begin{array}{cc}\hfill u_n\hfill & \hfill r_n\hfill \\ \hfill v_n\hfill & \hfill s_n\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill a_0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{cc}\hfill a_1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)\cdots \left(\begin{array}{cc}\hfill a_n\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)$.

Then part (i) shows that $\left(\begin{array}{cc}\hfill h_n\hfill & \hfill h_{n-1}\hfill \\ \hfill k_n\hfill & \hfill k_{n-1}\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill u_n\hfill & \hfill r_n\hfill \\ \hfill v_n\hfill & \hfill s_n\hfill \end{array}\right)$, thus $u_n=h_n,v_n=k_n$. By Theorem 7.4.,

$$\begin{array}{lll}\hfill ⟨a_0,a_1,\dots ,a_n⟩=\frac{h_n}{k_n}=\frac{u_n}{v_n}.& & \hfill \text{(3)}\end{array}$$

Proof. (of Problem 7.3.4)

(a)

Since we don’t know if $a_0$ is positive, we take away $a_0$: by (1), $$\begin{array}{llll}\hfill \left(\begin{array}{cc}\hfill a_1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)\cdots \left(\begin{array}{cc}\hfill a_n\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)& ={\left(\begin{array}{cc}\hfill a_0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)}^{-1}\left(\begin{array}{cc}\hfill h_n\hfill & \hfill h_{n-1}\hfill \\ \hfill k_n\hfill & \hfill k_{n-1}\hfill \end{array}\right)& \hfill & \\ \hfill & =\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -a_0\hfill \end{array}\right)\left(\begin{array}{cc}\hfill h_n\hfill & \hfill h_{n-1}\hfill \\ \hfill k_n\hfill & \hfill k_{n-1}\hfill \end{array}\right)& \hfill & \\ \hfill & =\left(\begin{array}{cc}\hfill k_n\hfill & \hfill k_{n-1}\hfill \\ \hfill h_n-a_0{k}_n\hfill & \hfill h_{n-1}-a_0{k}_{n-1}\hfill \end{array}\right)& \hfill & \\ \hfill & =\left(\begin{array}{cc}\hfill k_n\hfill & \hfill k_{n-1}\hfill \\ \hfill \ast \hfill & \hfill \ast \hfill \end{array}\right)& \hfill & \end{array}$$

The transpose of both members gives

$$\begin{array}{llll}\hfill \left(\begin{array}{cc}\hfill a_n\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)\cdots \left(\begin{array}{cc}\hfill a_1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)& =\left(\begin{array}{cc}\hfill k_n\hfill & \hfill \text{∗}\hfill \\ \hfill k_{n-1}\hfill & \hfill \text{∗}\hfill \end{array}\right).& \hfill & \end{array}$$

Since all $a_i(1\le i\le n)$ are positive, the part (ii) of the Lemma, applied tho the sequence $(a_n,a_{n-1},\cdots ,a_1)$, shows that, for all $n\in \mathbb{N}$,

$$⟨a_n,a_{n-1},\dots a_1⟩=\frac{k_n}{k_{n-1}}.$$

(b)

Similarly, the transpose of both members of (1) gives $$\left(\begin{array}{cc}\hfill a_n\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{cc}\hfill a_{n-1}\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)\cdots \left(\begin{array}{cc}\hfill a_0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill h_n\hfill & \hfill k_n\hfill \\ \hfill h_{n-1}\hfill & \hfill k_{n-1}\hfill \end{array}\right).$$

If $a_0>0$, then the Lemma shows that

$$⟨a_n,a_{n-1},\dots ,a_0⟩=\frac{h_n}{h_{n-1}}.$$

(If $a_0=0$, then $\left(\begin{array}{cc}\hfill a_0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)=I_2$, so $⟨a_n,a_{n-1},\dots ,a_1⟩=\frac{h_n}{h_{n-1}}$.)

In conclusion, if $a_0>0$, then