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I am really confused about the proof of this theorem: For any continued fraction, $$q_n\alpha - p_n = \frac{(-1)^n}{\alpha_{n+1}q_n + q_{n-1}}$$

I got the base induction case for $n=0$ but I can't figure out how to finish the induction step. Thanks for your help!

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  • $\begingroup$ Do you mean, $\alpha_n=\frac{p_n}{q_n}$? What does the lonely $\alpha$ mean? $\endgroup$ – chubakueno May 22 '14 at 3:42
  • $\begingroup$ $\alpha$ is the irrational number approximated by the continued fraction and each $\alpha_i$ is the error for the $i$th convergent. $\endgroup$ – Elliot Gorokhovsky May 22 '14 at 4:03
  • $\begingroup$ So $\alpha_0$ is $\alpha - a_0$ $\endgroup$ – Elliot Gorokhovsky May 22 '14 at 4:03
  • $\begingroup$ Equivalently, for a continued fraction $\alpha = [a_0,a_1 \ldots a_i \ldots]$, $\alpha_i = [a_{i+1}, a_{i+2}, \ldots]$ $\endgroup$ – Elliot Gorokhovsky May 22 '14 at 4:05
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    $\begingroup$ @RenéG Firstly, the left hand side should be $q_n \alpha-p_n$, not $q_n \alpha +p_n$. Next, it really much depends on what standard properties of continued fractions you are allowed to assume, or what restrictions are placed on how you prove this result. For instance, you may find it easier to first prove by induction that $p_{n-1}q_n-p_nq_{n-1}=(-1)^n$ and that $[a_0, a_1, \ldots, a_{m-1}, x]=\frac{xp_{m-1}+p_{m-2}}{xq_{m-1}+q_{m-2}}$, then combining these with $\alpha=[a_0, a_1, \ldots, a_n, \alpha_{n+1}]$. $\endgroup$ – Ivan Loh May 22 '14 at 20:56
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Well then. I'm not sure how much you know about continued fractions, so if need be, refer to here where I set up some standard properties of continued fractions, albeit modulo different notation.

In particular, we have the equalities

$$p_{n-1}q_n-p_nq_{n-1}=(-1)^n$$ $$[a_0; a_1, \ldots, a_{m-1}, x]=\frac{xp_{m-1}+p_{m-2}}{xq_{m-1}+q_{m-2}}$$ $$\alpha=[a_0; a_1, \ldots, a_n, \alpha_{n+1}]$$

Thus

\begin{align} q_n\alpha-p_n=q_n[a_0; a_1, \ldots, a_n, \alpha_{n+1}]-p_n& =q_n\frac{\alpha_{n+1}p_n+p_{n-1}}{\alpha_{n+1}q_n+q_{n-1}}-p_n\\ &=\frac{p_{n-1}q_n-p_nq_{n-1}}{\alpha_{n+1}q_n+q_{n-1}}\\ &=\frac{(-1)^n}{\alpha_{n+1}q_n+q_{n-1}} \end{align}

For a proof of the equalities I use (which are standard well known properties in continued fractions), refer to my linked answer above, albeit with different notation.

If that doesn't prove useful, you might want to consider picking up a book which covers continued fractions and learn from there.

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