Reciprocal of a continued fraction

I have to prove the following:

Let $\alpha=[a_0;a_1,a_2,...,a_n]$ and $\alpha>0$, then $\dfrac1{\alpha}=[0;a_0,a_1,...,a_n]$

I started with

$$\alpha=[a_0;a_1,a_2,...,a_n]=a_0+\cfrac1{a_1+\cfrac1{a_2+\cfrac1{a_3+\cfrac1{a_4+\cdots}}}}$$

and

$$\frac1{\alpha}=\frac1{[a_0;a_1,a_2,...,a_n]}=\cfrac1{a_0+\cfrac1{a_1+\cfrac1{a_2+\cfrac1{a_3+\cfrac1{a_4+\cdots}}}}}$$

But now I don't know how to go on. In someway I have to show, that $a_0$ is replaced by $0$, $a_1$ by $a_0$ and so on.

Any help is appreciated.

• Just add '$0+$' in front of your expression for $\frac{1}{\alpha}$. – Joe Johnson 126 Nov 27 '11 at 13:38
• $$\frac1{\alpha}=0+\cfrac1{a_0+\cfrac1{a_1+\cfrac1{a_2+\cfrac1{a_3+\cfrac1{a_4+ \cdots }}}}}$$ – J. M. is a poor mathematician Nov 27 '11 at 13:39
• What do you get when you expand $[0;a_0,a_1,a_2,\ldots,a_n]$? – Michael Slone Nov 27 '11 at 13:40
• That's all? And what is the explanation I'm allowed to do it? – ulead86 Nov 27 '11 at 13:41
• Was it ever said that $a_0$ cannot be $0$? – J. M. is a poor mathematician Nov 27 '11 at 13:43

You can just add '$+0$' to the expression for $\frac{1}{\alpha}$.