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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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    $\begingroup$ Presumably ${\mathbb R} \backslash {\mathbb Q}$ was meant. $\endgroup$ Dec 8, 2011 at 20:24
  • $\begingroup$ Absolutly right. I edited it. Thanks. $\endgroup$
    – ulead86
    Dec 8, 2011 at 20:26

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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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  • $\begingroup$ Ok thanks, but why is here no difference to the "normal" CF of the golden ratio in $\mathbb{R}$. I don't see the reason why we use $\gamma \in \mathbb{R} \setminus \mathbb{Q}$ $\endgroup$
    – ulead86
    Dec 8, 2011 at 20:51
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    $\begingroup$ @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. $\endgroup$ Dec 8, 2011 at 20:57

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