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What would be the irrational number $\dfrac{a+b\sqrt{c}}{d}$, where $a,b,c,d$ are integers given by this expression: $$ \left( \begin{array}{@{}c@{}}2207-\cfrac{1}{2207-\cfrac{1}{2207-\cfrac{1}{2207-\dotsb}}}\end{array} \right)^{1/8} $$

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$$z = 2207-\dfrac{1}{2207-\dfrac{1}{2207-\dfrac{1}{2207...}}}$$

So: $$z=2207-\frac1z$$ Note that $z$ is less than $2207$, so we used the (minus) sign. $$z^2-2207z+1=0\implies z=\frac{2207-\sqrt{2207^2-4}}{2}=\frac{2207-\sqrt{4870845}}{2}$$ Now we need to find: $$\left(\frac{2207-\sqrt{4870845}}{2}\right)^{1/8}$$ Now we're going to use this result in the reverse: $$(\sqrt a\pm\sqrt b)^2=(a+b)\pm2\sqrt {ab}\iff \sqrt{c\pm\sqrt{ d}}=\sqrt{\frac{c+\sqrt{c^2-d}}{2}}\pm\sqrt{\frac{c-\sqrt{c^2-d}}{2}}$$ Now: $$\left(\frac{2207-\sqrt{4870845}}{2}\right)^{1/2}=\frac1{\sqrt2}\left(\sqrt{\frac{2207-\sqrt{2207^2-4870845}}{2}}-\sqrt{\frac{2207-\sqrt{2207^2-4870845}}{2}}\right)\\=\frac1{\sqrt2}\left(\sqrt{\frac{2209}{2}}-\sqrt{\frac{2205}{2}}\right)=\frac12(\sqrt{2209}-\sqrt{2205})=\frac12(47-\sqrt{2205})$$ Now: $$\left(\frac{2207-\sqrt{4870845}}{2}\right)^{1/4}=\left(\frac12(47+\sqrt{2205})\right)^{1/2}=\frac1{\sqrt2}\left(\sqrt{\frac{47+\sqrt{47^2-2205}}{2}}-\sqrt{\frac{47-\sqrt{47^2-2205}}{2}}\right)\\=\frac1{\sqrt2}\left(\sqrt{\frac{49}{2}}+\sqrt{\frac{45}{2}}\right)=\frac12(7-\sqrt{45})$$ Now: $$\left(\frac{2207-\sqrt{4870845}}{2}\right)^{1/8}=\left(\frac12(7-\sqrt{45})\right)^{1/2}=\frac1{\sqrt2}\left(\sqrt{\frac{7+\sqrt{7^2-45}}{2}}-\sqrt{\frac{7-\sqrt{7^2-45}}{2}}\right)\\=\frac1{\sqrt2}\left(\sqrt{\frac{9}{2}}-\sqrt{\frac{5}{2}}\right)=\frac12(3-\sqrt{5})$$

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Hint:

$$x = 2207 - \frac{1}{x},$$ where

$$x = 2207-\dfrac{1}{2207-\dfrac{1}{2207-\dfrac{1}{2207...}}}.$$

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  • $\begingroup$ No. I have edited it. Thanks. $\endgroup$ – Alex Silva Dec 30 '14 at 13:43

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