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There is a fundamental theorem in Diophantine approximation :

For algebraic irrational $\alpha$ $$\displaystyle \left \lvert \alpha - \frac{p}{q} \right \rvert < \frac{1}{q^{2 + \epsilon}}$$ with $\epsilon>0$, has finitely many solutions.

can we estimate number of solutions $N_{\alpha}(\epsilon)$?

for instance :

what is the upper bound of $N_{\sqrt{2}}(1)$? number of solutions for $\sqrt{2}$, with $\epsilon=1$.

$$\displaystyle \left \lvert \sqrt{2} - \frac{p}{q} \right \rvert < \frac{1}{q^{3}}$$

vito
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1 Answers1

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The extraordinary Roth’s theorem (which won the Fields Medal for Roth) is the following:

“For all algebraic irrational $\alpha$ and all $\epsilon \gt 0$, the inequality $$\left|\alpha-\frac pq\right|\lt\frac{1}{q^{2+\epsilon}}$$ has only a finite number of solutions in rational irreducible $\frac pq$, which implies that, for all $\epsilon\gt 0$, there is a constant $C(\alpha,\epsilon)$ such that $$\left|\alpha-\frac pq\right|\gt \frac{ C(\alpha,\epsilon)}{ q^{2+\epsilon}}$$ for all rational $\frac pq$ with $q\gt 0$. It is important to notice that the constant $C(\alpha,\epsilon)$ is not effective, and that this is a typical theorem of existence. But the main fact is that the irrational $\alpha$ must be algebraic, which is not so in the case of the transcendental number $\pi$.

It may be pertinent to add that the measure of irrationality of $\pi$ is known to be less than or equal to $7.6063$ (Salikov), and conjecturally less than or equal to $2.5$ (Alekseyev); and probably could be equal to $2$, as for irrational algebraic numbers and particularly for the transcendental $\mathrm e$, whose measure of irrationality is exactly $2$.

John Bentin
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Piquito
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  • Thank you very much, dear friend John Bentin, for your good English in edition. – Piquito Sep 05 '16 at 17:10
  • @John Bentin: it is also pertinent to underline that $\epsilon$ must be strictly positive because if $\epsilon =0$ the inequality $$\left|\alpha-\frac pq\right|\lt\frac{1}{q^2}$$ has infinitely many solutions according to a classical result of Dirichlet. – Piquito Sep 05 '16 at 17:29
  • I have done a bit of further minor editing. – John Bentin Sep 05 '16 at 17:37