# Approximation of a random number with quadratic integers

Consider the following claim:

Claim: Let $$x$$ be a real random variable distributed according to the uniform distribution on the unit interval $$U(0,1)$$. Then for any quadratic irrational number $$\alpha$$, and real $$\epsilon>0$$, there exist finite real numbers $$c,\gamma,\Delta>0$$ such that $$\mathrm{Prob}\left( \min_{\substack{(p,q) \in \mathbb{Z}^2 \\ q \neq 0}}|\alpha q - p - x|q^{1+\epsilon} < \delta\right) \leq c \delta^\gamma$$ is true for all $$\delta < \Delta$$.

My Question: Does the claim hold?

It may be helpful to note that one can always find a $$p,q$$ independent constant $$C(\alpha,\epsilon)$$ such that the bound $$\left|\alpha - \frac{p}{q} \right| \geq \frac{C(\alpha,\epsilon)}{q^{2+\epsilon}}$$ is tight. This inequality holds for any algebraic $$\alpha$$ (Roth's theorem) and holds with probability one for $$\alpha$$ drawn from a piecewise continuous distribution (Gauss Kuzmin statistics). Thus for $$x = 0$$, we have that $$\min_{(p,q) \in \mathbb{Z}^2, \,q \neq 0}|\alpha q - p|q^{1+\epsilon} = C(\alpha,\epsilon),$$ and the question is essentially about how frequently the constant on the RHS becomes small when $$x$$ is varied away from zero.

My suspicion is that the claim holds, and that the bound is tight for $$\gamma = 1$$. Thus I would also be grateful for any insight regarding the follow up questions

• Can we choose $$\gamma = 1$$ always?
• Can $$c$$ be bounded?
• Does the claim hold with probability one for $$\alpha$$ itself drawn randomly from the unit interval (i.e. for $$\alpha$$ with Gauss-Kuzmin statistics)?
• are you sure it's $\alpha p + q$? I would have expected $\alpha q + p$ (occurring as the numerator of $\alpha + p/q$) Commented Mar 11, 2021 at 20:39
• @AndreaMarino why not $\alpha q - p$, since $\alpha-\frac{p}{q}$ is of interest? Commented Mar 11, 2021 at 21:53
• @AndreaMarino I think it doesn't change the answer, but in response to your comment I have changed it, as you are right, it is off-putting, and potentially needlessly confusing. Commented Mar 11, 2021 at 21:58
• @mathworker21 sure why not, I changed it, though I hope the audience is aware that $p \in \mathbb{Z}$ implies $(-p) \in \mathbb{Z}$. :) Commented Mar 11, 2021 at 21:59
• @ComptonScattering you need to rule out $q=0$ from the minimum. Commented Mar 11, 2021 at 22:05