A corollary of Dirichlet's approximation theorem is that for any irrational $\alpha$, there are infinitely many integer solutions of $$\left|\frac{p}{q}-\alpha\right|<\frac{1}{q^2}$$ Are there any similar results concerning the integer solutions of $$0<\frac{p}{q}-\alpha<\frac{1}{q^2}?$$
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Kronecker's approximation theorem allows to approximate $0$ with ${m\alpha +n \mid m, n \in \mathbb{Z}}$ (which is effectively dense) from any side of $\alpha$, although not as tight as $\frac{1}{m}$ – rtybase Mar 01 '18 at 11:15
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Yes. For every irrational $\alpha$ there are infinitely many integer solutions of $$0<\frac pq-\alpha<\frac1{q^2}. $$ (And by symmetry also infinitely many solutions of $0<\alpha-\frac pq<\frac1{q^2}$).
For irrational $\alpha\in(0,1)$, we can start with approximations $$ \frac 01<\alpha<\frac11$$ and step-wise improve by splitting at the Farey-sum, i.e. from $$\tag1 \frac ab<\alpha<\frac cd$$ we go to $$\tag2 \frac ab<\alpha<\frac {a+c}{b+d}\qquad\text{or}\qquad\frac {a+c}{b+d}<\alpha<\frac cd.$$ The difference between upper and lower bound in $(1)$ is $\frac1{bd}$, hence we have $0<\frac cd-\alpha<\frac1{d^2}$ at least whenever $b>d$. Hence the only way to not have infinitely many such approximations is to almost always choose the left interval in $(2)$. But that is impossible for irrational $\alpha$ because sooner or later the interval length becomes less than $\alpha-\frac ab$.
Hagen von Eitzen
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The approximations that come from the convergents to the continued fraction for $\alpha$ all satisfy the Dirichlet inequality and alternate between being bigger than and being smaller than $\alpha$.
Gerry Myerson
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