Exercise 7.4.2 (The equality of the first convergents implies the equality of the first $a_i$)

Proof. Let $\xi$ and ${\xi }^{\prime }$ be irrational numbers such that

We write $h_i^{\prime }∕k_i^{\prime }$ the convergents of ${\xi }^{\prime }$ for all $i$.

By hypothesis, $h_i∕k_i=h_i^{\prime }∕k_i^{\prime }$ if $0\le i\le n$. Since these fractions are in lowest terms by Theorem 7.5, we have more precisely

First $a_0=h_0=h_0^{\prime }=a_0^{\prime }$ by (7.6), so $a_0=a_0^{\prime }$.

Reasoning by induction, assume that $(a_0,a_1,\dots ,a_i)=(a_0^{\prime },a_1^{\prime },\dots ,a_i^{\prime })$ for some $i<n$. Then $i+1\le n$, so $h_{i+1}∕k_{i+1}=h_{i+1}^{\prime }∕k_{i+1}^{\prime }$. By Theorem 7.4, we obtain

and the induction hypothesis shows that

By Theorem 7.3,

By Theorem 7.5, $h_i{k}_{i-1}-h_{i-1}{k}_i={(-1)}^{i-1}\ne 0$, so the homographic function $x\mapsto \frac{x{h}_i+h_{i-1}}{x{k}_i+k_{i-1}}$ is one-to-one. Therefore $a_{i+1}=a_{i+1}^{\prime }$, and the induction is done, up to $n$, so

The continued fraction expansions of $\xi$ and ${\xi }^{\prime }$ are identical up to $a_n$. □