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Essentially, my question is why $|\frac{1 + \sqrt{5}}{2} - \frac{a}{b}| < 1/b^c$ (for $c>2$) is satisfied by only a finite number of $\frac{a}{b}$. This is intrinsically related to Hurwitz's approximation for the golden ratio $= \alpha$ with $|\alpha - \frac{a}{b}| < \frac{1}{\sqrt{5}b^2}$. I realize that the $\frac{1}{\sqrt{5}}$ is the best we can do, I just don't really follow the proof.

My thought is that the best approximations will be the ratios of increasingly large Fibonacci numbers, and so since those ratios are reduced, you get $b$ increasingly large. E.g if we have $c = 3$, you quickly see that $\frac{1}{b*b^2} < \frac{1}{\sqrt{5}b^2}$. Is this the right idea? It's just not a formal proof.

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You get this because the golden ratio is a quadratic algebraic integer. Let $\tau=(1+\sqrt5)/2$ and $\overline{\tau}=(1-\sqrt5)/2$. These are the zeros of the polynomial $$ p(x)=(x-\tau)(x-\overline{\tau})=x^2-x-1. $$ Let then $a/b$ be a rational number (in reduced form). Because $\tau$ is irrational we get that $p(a/b)\neq0$. But $$ p(a/b)=(\frac ab)^2-\frac ab-1=\frac{a^2-ab-b^2}{b^2}. $$ Because $p(a/b)\neq0$ the numerator is non-zero. Because the numerator is an integer, it has absolute value $\ge1$. Therefore $$ |p(a/b)|\ge\frac1{b^2}. $$ But we also have $$ |p(a/b)|=|\frac ab-\tau|\cdot |\frac ab-\overline{\tau}|. $$ The following is the other key. If here $a/b$ is very close to $\tau$, then $|(a/b)-\overline{\tau}|$ is close to $|\tau-\overline{\tau}|=\sqrt5$. Therefore $$ |\frac ab-\tau|=\frac{|p(a/b)|}{|\frac ab-\overline{\tau}|}\ge \frac{1/b^2}K=\frac1{Kb^2}, $$ where $K$ is very close to $\sqrt5$.

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