Continued Fraction: Why do we get with $\gamma \in \mathbb{R}\setminus \mathbb{Q}$ the CF $\frac{1}{\gamma}=\langle0;a_0,a_1,\dotsc\rangle$

Question

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.

Answer 1

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 $\langle 0\;;a_0,a_1,\dots\rangle=\frac1\gamma$.