I am wondering about the possible best case approximation and worst case approximation of irrational numbers. I think the appropriate formulation is whether there are functions $\hat{b}(q)$ and $\check{b}(q)$ such that for any $\alpha\in [0,1)\setminus \mathbb{Q}$ we have for only finitely many $q's$

$$ \check{b}(q) \leq \Big\vert \alpha-\frac{p}{q} \Big\vert \quad \text{and} \quad \Big\vert \alpha-\frac{p}{q} \Big\vert \leq \hat{b}(q) \quad \text{for some}\;\; p\in \mathbb{Z}. $$

**Later edit:**
If I try to phrase it alternatively, for all irrational $\alpha$ we have functions $\check{b}(\alpha,q),\hat{b}(\alpha,q)$ satisfying for infinitely many $q$'s that
$$ \check{b}(\alpha,q) \leq \Big\vert \alpha-\frac{p}{q} \Big\vert \leq \hat{b}(\alpha,q). $$

Given $f:\mathbb{N}\to [1,\infty)$ satisfying that $f(n)\to \infty$, can we find $\alpha$ such that $\hat{b}(\alpha,q)=o\big(\frac{1}{f(q)})$? Can we find $\alpha$ irrational such that $\frac{1}{qf(q)}=O\big(\check{b}(\alpha,q)\big)$?

**End of later edit.**

By the Piegonhole principle, we know that $\check{b}(q)\geq \frac{1}{2q}$. Also, by Liouville's theorem we can see that $\hat{b}(q)=o(q^{-n})$ for all $n$. Are there more explicit asymptotic bounds ? Can we say something like $\hat{b}(q)=o(e^{-q})$ ? Can we say that $\check{b}(q)=\Theta(\frac{1}{q})$?

I tried to ask this questions on Math Stackexchange, but I had no response. I assume something like this is known and I have not found this in a short search online. I am unused to these notions so I apologize if this is a silly question.

I would appreciate any insights on this.

**Later edit:** By Oleg Eroshkin's comments, it seems like there is no effective bound on how small $\hat{b}(\alpha,q)$ can be.

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