Proving that the function $\frac{1}{q^2}$ is essential to Dirichlet's Approximation Theorem First post on math.stackexchange so I apologize in advance.
In my number theory class, we've been handed 3 things in order to do the following:
Prove using Dirichlet’s theorem and Liouville’s lemma that one cannot replace the function $\frac{1}{q2}$ in the statement of Dirichlet’s theorem by a faster decaying function by considering quadratic algebraic numbers (degree 2).
In this same assignment, we've been given three things - I believe as hints:

*

*Let $\phi, \psi : \mathbb{N} \rightarrow \mathbb{R}$. We say that:
a) $\phi$ is equivalent to $\psi$ if $lim_{q\rightarrow\infty}\frac{\phi(q)}{\psi q}=C$, where $C \neq 0$ is some real number.
b) $\phi$ decays faster than $\psi$ if $lim_{q\rightarrow\infty}\frac{\phi(q)}{\psi q}=0$
c) $\phi$ decays slower than $\psi$ if $lim_{q\rightarrow\infty}\frac{\phi(q)}{\psi q}=\infty$


*Recall Dirichlet's theorem: For every irrational $\alpha$, there exists infinitely distinct rationals $\frac{p}{q}$ such that $$|\alpha - \frac{p}{q}|<\frac{1}{q^2}$$


*Recall Liouville's Lemma: Let $\alpha$ be an irrational algebraic number of degree $n$. Then there exists $M\in \mathbb{N}$ such that for every $\frac{p}{q} \in \mathbb{Q}$, $$|\alpha-\frac{p}{q}|>\frac{1}{Mg^n}$$
Now I've already submitted my best attempt at this but I know for a fact it's wrong and I'd prefer not to embarrass myself on my first post on this website. I'm really just looking for an explanation that I can study so I don't miss it again.
 A: The main idea is just to note that Liousville's lemma tells you that you can only get so close to an irrational algebraic number. So if you could improve Dirichlet's theorem, you would eventually end up violating the lemma.
More rigorously, suppose $f(q)$ is a faster decaying function than $\frac1{q^2}$, so that $\lim_{q\to\infty}q^2f(q)=0$, and suppose for the sake of contradiction that we can replace $\frac1{q^2}$ with $f(q)$ in Dirichlet's theorem. Now let $\alpha$ be an irrational number of degree two, as hinted.
Liouville's lemma tells us that there exists some $M$ so that $\left\lvert\alpha-\frac pq\right\rvert>\frac1{Mq^2}$ for every $\frac pq$. By hypothesis, we know, however, that for infinitely many $\frac pq$, we must have $\left\lvert\alpha-\frac pq\right\rvert<f(q)$. In particular, it follows that there are infinitely many fractions $\frac pq$ such that
$$1=\frac{\left|\alpha-\frac pq\right|}{\left\lvert\alpha-\frac pq\right\rvert}<\frac{f(q)}{\frac1{Mq^2}}=Mq^2f(q).$$
But because $q^2f(q)\to0$, we know that there exists some $Q\in\mathbb N$ so that for all $q>Q$ we have $|q^2f(q)|<\frac1M$.
However, this would mean that $q\le Q$ for each of the infinitely many fractions $\frac pq$ with $\left|\alpha-\frac pq\right|<f(q)$. But since $f(q)$ is finite, there are only finitely many $p$ for any given $q$ which satisfy this. And since we just said that $q\in\{1,2,\dots,Q\}$, it follows that there are only finitely many pairs $(p,q)$ which satisfy that inequality. Obviously, this is impossible, so the $\frac1{q^2}$ in Dirichlet's theorem is the best possible.
