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I'm trying to prove that $\sqrt 3 = [1;2,1,2,1,\ldots]$

I have found the recursive relation $$\sqrt 3 +1= 2+\frac{2}{\sqrt 3 +1}$$

and it's quite close to a continued fraction, but the problem is that the numerators here are $2$'s and not $1$'s.

blueplusgreen
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1 Answers1

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Just apply Gauss' iteration until you find a loop. The integer part of $\sqrt{3}$ is $1$ and $$ \frac{1}{\sqrt{3}-1}=\frac{\sqrt{3}+1}{2}=1+\frac{\sqrt{3}-1}{2}=1+\frac{1}{\frac{2}{\sqrt{3}-1}}=1+\frac{1}{\sqrt{3}+1}=1+\frac{1}{2+(\sqrt{3}-1)}$$ hence $\sqrt{3}=[1;\overline{1,2}]$.

Jack D'Aurizio
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