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I would like to prove that given any $\alpha = [a_0; a_1 \ldots] \in \mathbb{R}$, with: \begin{equation} \alpha_n = [a_n; a_{n-1}, \ldots, a_0]\\ \frac{p_n}{q_n} = [a_0; a_1, \ldots, a_n] \end{equation} Then we have the following; \begin{equation} q_n \alpha - p_n = \frac{(-1)^n}{\alpha_{n+1}q_n +q_{n-1}} \end{equation}

I have started by deriving the relationship that $\alpha_n = \frac{q_n}{q_{n-1}}$, but I'm not sure if (or how) this might help. Also, I know that \begin{equation} p_n q_{n-1} - p_{n-1}q_n = (-1)^{n+1} \end{equation} Any suggestions are appreciated.

sc636
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    Are you sure the problem is written properly? As it is, the LHS could be irrational while the RHS is certainly rational. – preferred_anon Mar 28 '17 at 08:29
  • I've just checked the problem and it seems to be correct. That's a good point though, I didn't see that! Suppose $\alpha$ was a rational number then... – sc636 Mar 28 '17 at 08:39
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    Looking at it, the most likely thing for it to mean is $\alpha_{n}$ on the LHS. Perhaps try that? – preferred_anon Mar 28 '17 at 09:10
  • I have tried to prove that the equality will hold for $\alpha_n$ instead of $\alpha$ using induction, but I run into a problem. For a base induction step I try to prove it holds for $n=0$. Using the values $q_{-1}=0, q_{-2} =1, p_{-1}=1, p_{-2} =0$ and the formula I derived above, we end up having to divide by 0... (not sure if I'm doing it right though). – sc636 Mar 28 '17 at 10:53

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