Liouville's Theorem in Diophantine Approximation Liouville's Theorem states that for any algebraic $\alpha \in \mathbb{R}$ of degree $n$, there exists a positive constant $c:=c(\alpha)$ such that $$\left\lvert\alpha-\frac{p}{q}\right\rvert>\frac{c}{q^n}$$ for any $p \in \mathbb{Z}$ and $q \in \mathbb{N}.$
One can find an effective  lower bound for $c(\alpha).$ In the special case that $\alpha$ is a quadratic irrational, Exercise 27 in the following set of notes of Jorn Steuding 
http://www.math.uni-bremen.de/~bos/dioph.pdf
yields $$c(\alpha) \gg \frac{1}{(1+|\alpha|)H(\alpha)}  .$$
Here if $m_{\alpha}(x):=x^2+bx+c \in \mathbf{Q}[x]$ is the minimal polynomial of $\alpha$, the height $H(\alpha)$ is defined as the maximum of $|b|$ and $|c|.$ My question is whether one can find a better lower bound for $c(\alpha)$  when $\alpha$ is a quadratic irrational or if this is best possible.
 A: Let's assume $\alpha > 0$.  If $\alpha$ is a quadratic irrational, its simple continued fraction $a_0 + \dfrac{1}{a_1 + \frac{1}{a_2+\ldots}}$ is eventually periodic.
Every $p/q$ (in lowest terms) with $\left|\alpha - \dfrac{p}{q}\right| < \dfrac{1}{2q^2}$ is a convergent of $\alpha$,
and for the $n$'th convergent 
$$ \dfrac{1}{q_n^2 (a_{n+1}+2)} < \left| \alpha - \dfrac{p_n}{q_n} \right| \le \dfrac{1}{q_n^2 a_{n+1}}$$ 
Thus $\dfrac{1}{a_M+2} \le c(\alpha) \le \dfrac{1}{a_M}$ where $a_M$ is the largest element in the continued fraction of $\alpha$. 
A: The constant $c=1/\sqrt5$ (with $n=2$) works for any $\alpha$. If $\alpha$ is not, roughly speaking, the golden ratio, then $c$ can be improved to $1/2\sqrt2$, etc. If one removes a certain infinite sequence of quadratic irrationals, one can take $c=1/3$, but this is the best you can do in a general setting. 
A nice exposition can be found in [Cassels, An Introduction to Diophantine Approximation]. You may also start with a Wikipedia article. 
