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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.

  • 5
    Just add '$0+$' in front of your expression for $\frac{1}{\alpha}$.2011-11-27
  • 1
    $$\frac1{\alpha}=0+\cfrac1{a_0+\cfrac1{a_1+\cfrac1{a_2+\cfrac1{a_3+\cfrac1{a_4+ \cdots }}}}}$$2011-11-27
  • 0
    What do you get when you expand $[0;a_0,a_1,a_2,\ldots,a_n]$?2011-11-27
  • 0
    That's all? And what is the explanation I'm allowed to do it?2011-11-27
  • 0
    Was it ever said that $a_0$ cannot be $0$?2011-11-27
  • 0
    Don't think so ;)2011-11-27

1 Answers 1

5

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