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I have a question concerning continued fractions:

If we have $\gamma \in \mathbb{R} \setminus \mathbb{Q}$ and $\gamma=\langle a_0;a_1,a_2,\dotsc\rangle$. Why do we get $\frac1\gamma = \langle 0;a_0,a_1,\dotsc\rangle$

Furthermore, how can I get the continued fraction of $\langle 0;\overline1 \rangle$?

Any help is appreciated.

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    Absolutly right. I edited it. Thanks.2011-12-08

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For the second question, let $x=\langle 0\;;\overline{1}\rangle$. Then $x=\frac1{1+\frac1{1+\frac1{\quad\ddots}}}=\frac1{1+x}\,,$ so $x(1+x)=1$. Solving this quadratic equation and noting that clearly $x>0$, we have $\langle 0\;;\overline{1}\rangle=\frac{-1+\sqrt5}2\,;$ this is $\varphi-1$, where $\varphi$ is the so-called golden ratio.

Now see if you can apply the same technique to show that if $\gamma=\langle a_0;a_1,a_2,\dots\rangle$. Then $\langle 0\;;a_0,a_1,\dots\rangle=\frac1\gamma$.

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    @Daniel: I made a silly mistake. It’s not $\varphi$: it’s $\varphi-1$. Taking $\gamma\in\mathbb{R}\setminus\mathbb{Q}$ merely ensures that the CF expansion is infinite.2011-12-08