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Is there an understandable function $A(\epsilon)$ such that if $q < A(\epsilon)$ then $| q\pi - p| > \epsilon$ for all $p$?

I want to know how quickly $n\pi$ is getting close to integers, e.g., if $n\pi$ is within 0.0001 of an integer then $n>10^6$, if $n\pi$ is within 0.00000001 of an integer then $n>10^{10}$.

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    $\begingroup$ The irrationality measure of $\pi$ is known to be at most 7.6063, see iopscience.iop.org/0036-0279/63/3/L11/pdf/RMS_63_3_L11.pdf . $\endgroup$
    – Emil Jeřábek
    Commented Aug 11, 2011 at 15:54
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    $\begingroup$ Any irrational number $x$ has infinitely many approximations $p/q$ good to within $1/q^2$ (the numerator can be lowered but not indefinitely, try $x=\sqrt2$); thus $|q\pi−p|<1/q$ infinitely often. The closest approximations up to denominator $10^6$ and $10^{10}$ are respectively $1146408/364913$ and $21053343141/6701487259$ (courtesy of gp's "bestappr" function), and $6701487259\pi$ is within .0000000000002 of an integer. The result E.Jeřábek quotes means that $|\pi−p/q|<q^\theta$ has only finitely many solutions for each $\theta>7.6063$ . $\endgroup$
    – Noam D. Elkies
    Commented Aug 11, 2011 at 16:38
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    $\begingroup$ Moreover these "irrationality measures" (unlike those that arise in Thue-Siegel-Roth for algebraic numbers) are effective: the proof gives an explicit upper bound on $A(\epsilon)$, which at least in principle reduces the problem to a finite computation. $$ $$ And yes, I should have written $|\pi - p/q| < q^{-\theta}$, not $q^{+\theta}$; sorry for the typo. $\endgroup$
    – Noam D. Elkies
    Commented Aug 11, 2011 at 17:05
  • $\begingroup$ @Emil, given $\epsilon$ and having taken $q\lt A(\epsilon)$, surely $|q\pi-p|\gt\epsilon$ for all but at most one value of $p$. What did you mean to write? $\endgroup$
    – Gerry Myerson
    Commented Aug 12, 2011 at 0:53
  • $\begingroup$ @Gerry: You’re right, I somehow confused the quantifiers. It should be: for all but finitely many $q$, for all $\epsilon$ and $p$, if $q< A(\epsilon)$ then $|q\pi-p|>\epsilon$, where $A(\epsilon)=\epsilon^{1/(1-\theta)}$. In other words, there is an $\epsilon_0$ such that for all $\epsilon< \epsilon_0$, all $q< A(\epsilon)$, and all $p$, $|q\pi-p|>\epsilon$. What Noam writes means that one can (at least in principle) find $\epsilon_0$ explicitly. $\endgroup$
    – Emil Jeřábek
    Commented Aug 12, 2011 at 10:28

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