Exercise 7.4.7 ($k_n|k_{n-1}\xi - h_{n-1}| + k_{n-1} |k_n \xi - h_n| = 1$)

Proof. By (7.6), $k_{-1}|k_{-2}\xi -h_{-2}|+k_{-2}|k_{-1}\xi -h_{-1}|=1.$ and $k_0|k_{-1}\xi -h_{-1}|+k_{-1}|k_0\xi -h_0|=1,$ so the property is true for $n=-1$ and $n=0$.

Now we assume that $n\ge 1$. Then $k_n>0$ and $k_{n-1}>0$. Therefore $k_n|k_{n-1}\xi -h_{n-1}|+k_{n-1}|k_n\xi -h_n|=1$ is equivalent to

It remains to prove (1).

• If $n$ is odd, by Theorem 7.10,

  $$\frac{h_{n-1}}{k_{n-1}}<\xi <\frac{h_n}{k_n}.$$

  Hence

  $$\begin{array}{llllll}\hfill \left|\xi -\frac{h_{n-1}}{k_{n-1}}\right|+\left|\xi -\frac{h_n}{k_n}\right|& =\left(\xi -\frac{h_{n-1}}{k_{n-1}}\right)-\left(\xi -\frac{h_n}{k_n}\right)& \hfill & & \hfill & \\ \hfill & =\frac{h_n}{k_n}-\frac{h_{n-1}}{k_{n-1}}& \hfill & & \hfill & \\ \hfill & =\frac{h_n{k}_{n-1}-k_n{h}_{n-1}}{k_n{k}_{n-1}}& \hfill & & \hfill & \\ \hfill & =\frac{{(-1)}^{n-1}}{k_n{k}_{n-1}}& \hfill (\text{by Theorem 7.5})& & \hfill & & \hfill \\ \hfill & =\frac{1}{k_n{k}_{n-1}}& \hfill (\text{since }n\text{ is odd}.& & \hfill & & \hfill \end{array}$$

• If $n$ is even,

  $$\frac{h_n}{k_n}<\xi <\frac{h_{n-1}}{k_{n-1}}.$$

  Hence

  $$\begin{array}{llllll}\hfill \left|\xi -\frac{h_{n-1}}{k_{n-1}}\right|+\left|\xi -\frac{h_n}{k_n}\right|& =-\left(\xi -\frac{h_{n-1}}{k_{n-1}}\right)+\left(\xi -\frac{h_n}{k_n}\right)& \hfill & & \hfill & \\ \hfill & =\frac{h_{n-1}}{k_{n-1}}-\frac{h_n}{k_n}& \hfill & & \hfill & \\ \hfill & =\frac{-(h_n{k}_{n-1}-k_n{h}_{n-1})}{k_n{k}_{n-1}}& \hfill & & \hfill & \\ \hfill & =\frac{{(-1)}^n}{k_n{k}_{n-1}}& \hfill (\text{by Theorem 7.5})& & \hfill & & \hfill \\ \hfill & =\frac{1}{k_n{k}_{n-1}}& \hfill (\text{since }n\text{ is even}.& & \hfill & & \hfill \end{array}$$

In either case, (1) is true, thus for all $n\ge -1$,

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