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Prove that if $\alpha \in \mathbb{R}\backslash \mathbb{Q}$, there exists one (actually an infinity) of integers $p$ et $q$ such that $\big|\alpha -\frac{p}{q}\big| \leqslant \frac{1}{q^2}$ and $0<p<q$.

My attempt : I started by considering $\alpha\in [0,1[$ because if not we can just consider $\alpha - \lfloor \alpha \rfloor$ and then add $\lfloor \alpha \rfloor$ to the $p/q$ we find. But then I really don't know how to proceed. $p/q$ must be close to $\alpha$ but I don't see how to make the link with the denominator.

Could someone help me ?

math
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    Have a look at https://en.wikipedia.org/wiki/Dirichlet%27s_approximation_theorem – Tuvasbien Feb 24 '21 at 14:12
  • Surely $q = 2$ will always work, since every number is at most $\frac14$ away from some number of the form $\frac{p}{2}.$ For $q=3$ and above, it's maybe not so obvious. – Adam Rubinson Feb 24 '21 at 14:53
  • I forgot to say $0<p<q$. – math Feb 24 '21 at 15:34
  • $p < q$ is false for any $\alpha > 2$. If you want to avoid Adam Rubinson's trivial solution, you need to require $q > 2$. Or more generally, for any $N$ there is a pair with $q > N$ (which is why there are infinitely many). – Paul Sinclair Feb 24 '21 at 20:29

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