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Hurwitz's Theorem states that for any irrational $\zeta$ there are infinitely many pairs of integers $p,q$ such that $$ |\zeta - p/q|<\frac{1}{\sqrt{5}q^2}. $$ A good Diophantine approximation $p/q$ to the number $\zeta$ satisfies $$ |\zeta - p/q| < |\zeta - p'/q'| $$ for all $q'<q$.

My question is, do all good Diophantine approximations satisfy the relation in Hurwitz's theorem?

I have looked at the cases I know how to calculate, examples where $\zeta$ is a quadratic integer, and it seems that they do. And based on the intuition these numbers are notably badly approximable, and that other number are generically more rational (that is they have rational approximations for which $|\zeta - p/q|$ falls faster than $q^{-2}$) one would expect that generic good Diophantine approximations satisfy the inequality in Hurwitz's theorem. However this is just hand waving.

The answer to this may be widely known but I wasn't able to find it clear statement.

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    $\begingroup$ You should have tried the notoriously worst of them, the Golden Ratio $\varphi=(1+\sqrt{5})/2$. $3/2$ is a good Diophantine approximation, but $2^2\sqrt{5}|\varphi-3/2|>1$. $\endgroup$ – Professor Vector Dec 12 '17 at 20:31
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Hurwitz' theorem states that if $\frac{p_{n-1}}{q_{n-1}},\frac{p_n}{q_n},\frac{p_{n+1}}{q_{n+1}}$ are three consecutive convergents of the continued fraction of $\alpha\in\mathbb{R}^+\setminus\mathbb{Q}$, at least one of them fulfills $\left|\alpha-\frac{p}{q}\right|\leq \frac{1}{q^2\sqrt{5}}$, but that is not granted to hold for any convergent, and it is pretty simple to construct counter-examples to such a claim, as Professor Vector points out in the comments.

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  • $\begingroup$ Thanks for the answer. Is it known if there exists a constant $c$ such that $|\alpha-p/q|\leq c q^{-2}$ for all irrational numbers? $\endgroup$ – ComptonScattering Dec 13 '17 at 0:20
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    $\begingroup$ @ComptonScattering: you question lacks some quantifiers. For any constant $c\geq \frac{1}{\sqrt{5}}$ and any $\alpha\in\mathbb{R}^+\setminus\mathbb{Q}$, there are an infinite number of rational approximations $p/q$ such that $|\alpha-p/q|\leq c q^{-2}$. Such constant is optimal in the following sense: for any $c<\frac{1}{\sqrt{5}}$ you may construct an irrational number $\alpha$ such that all the convergents of $\alpha$ fulfill $|\alpha-p/q|> c q^{-2}$. $\endgroup$ – Jack D'Aurizio Dec 13 '17 at 0:25
  • $\begingroup$ On the other hand, the inequality $|\alpha-p/q|\leq q^{-2}$ ($c=1$) holds for all the convergents of $\alpha$. $\endgroup$ – Jack D'Aurizio Dec 13 '17 at 0:27
  • $\begingroup$ Yes sorry I meant for all convergents of all irrational numbers. Which I understand you to have answered in the affirmative for $c=1$ in the second comment. Thank you! $\endgroup$ – ComptonScattering Dec 13 '17 at 2:25

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