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I want to show this result starting from the property that $\phi = 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{...}}}$.

I presume that to get $\phi = \frac{1}{\phi} $ I need to find the limit of the equation above.

However, I'm unsure how to do this. It's easy to see where the "1" comes from, but I'm not sure about the $\frac{1}{\phi}$ part.

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  • $\begingroup$ If you define $\phi :=\frac{1+\sqrt{5}}{2}$ ($:=$ denotes "equal by definition"), then $\phi^2-\phi-1=0$, equivalently $\phi=1+\frac{1}{\phi}$. $\endgroup$ – user236182 Aug 6 '17 at 19:48
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    $\begingroup$ @dxiv please submit it as an aswer :) $\endgroup$ – user394255 Aug 6 '17 at 19:49
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By inspection: $\;\displaystyle\phi = \color{red}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{...}}}} = \color{blue}1 + \frac{1}{\color{red}{1 + \frac{1}{1 + \frac{1}{...}}}} = \color{blue}{1}+\frac{1}{\color{red}{\phi}}\,$.

For a formal proof, consider the recurrence $\;\displaystyle\phi_0=1, \;\phi_{n+1}=1+\frac{1}{\phi_n}\,$ and define $\,\phi\,$ as its limit.

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