Exercise 7.3.5 (Inequalities for a rational number)

Proof. By hypothesis,

We define $h_i,k_i$ as in (7.6) for $0\le i\le n$, and

so that $r_n=u_0∕u_1$.

We may use the results of Theorem 7.4 and 7.5 up to $n$ with the same proofs. So for every index $i$ such that $0\le i\le n$,

(i)

$r_i=\frac{h_i}{k_i}$.

(ii)

$h_i{k}_{i-1}-h_{i-1}{k}_i={(-1)}^{i-1}$ and $r_i-r_{i-1}=\frac{{(-1)}^i}{k_i{k}_{i+1}}.$

(iii)

$h_i{k}_{i-2}-h_{i-2}{k}_i={(-1)}^i{a}_i$ and $r_i-r_{i-2}=\frac{{(-1)}^i{a}_i}{k_i{k}_{i-2}}.$

Therefore the finite sequence ${(r_{2j})}_{0\le 2j\le n}$ is strictly increasing and ${(r_{2j+1})}_{0\le 2j\le n}$ is strictly decreasing. Hence, for $0\le i<n$, $r_i<r_n<r_{i+1}$ or $r_{i+1}<r_n<r_i$ (following that $i$ is even or odd), so in both cases

• if $i=n-1$, then

  $$\begin{array}{llllll}\hfill \left|r_i-\frac{u_0}{u_1}\right|& =\left|r_{n-1}-r_n\right|& \hfill & & \hfill & \\ \hfill & =\frac{1}{k_n{k}_{n-1}}& \hfill (\text{by (ii)})& & \hfill & & \hfill \\ \hfill & =\frac{1}{k_{i+1}{k}_i}.& \hfill & & \hfill & \end{array}$$

• If $0\le i<n-1$, then

  $$\begin{array}{llllll}\hfill \left|r_i-\frac{u_0}{u_1}\right|& =|r_i-r_n|& \hfill & & \hfill & \\ \hfill & <|r_i-r_{i+1}|& \hfill (\text{by (1)})& & \hfill & & \hfill \\ \hfill & =\frac{1}{k_i{k}_{i+1}}& \hfill (\text{by (ii)}).& & \hfill & & \hfill \end{array}$$

In conclusion, if $0\le i<n$, then $|r_i-u_0∕u_1|\le 1∕(k_i{k}_{i+1})$, with equality if and only if $i=n-1$. □