optimality of 2 as a coefficient in a continued fraction theorem I'm giving some lectures on continued fractions to high school and college students, and I discussed the standard theorem that, for a real number $\alpha$ and integers $p$ and $q$ with $q \not= 0$, if $|\alpha-p/q| < 1/(2q^2)$ then $p/q$ is a convergent in the continued fraction expansion of $\alpha$. Someone in the audience asked if 2 is optimal: is there a positive number $c < 2$ such that, for every $\alpha$ (well, of course the case of real interest is irrational $\alpha$), when $|\alpha - p/q| < 1/(cq^2)$ it is guaranteed that $p/q$ is a convergent to the continued fraction expansion of $\alpha$?
Please note this is not answered by the theorem of Hurwitz, which says that an irrational $\alpha$ has $|\alpha - p_k/q_k| < 1/(\sqrt{5}q_k^2)$ for infinitely many convergents $p_k/q_k$, and that $\sqrt{5}$ is optimal: all $\alpha$ whose cont. frac. expansion ends with an infinite string of repeating 1's fail to satisfy such a property if $\sqrt{5}$ is replaced by any larger number. For the question the student at my lecture is asking, an optimal parameter is at most 2, not at least 2.
 A: 2 is optimal. Let $\alpha=[a,2,\beta]$, where $\beta$ is a (large) irrational, and let $p/q=[a,1]=(a+1)/1$. Then $p/q$ is not a convergent to $\alpha$, and $${p\over q}-\alpha={1\over2-{1\over \beta+1}}$$ which is ${1\over(2-\epsilon)q^2}$ since $q=1$. 
More complicated examples can be constructed.  I think this works, though I haven't done all the calculations. Let $\alpha=[a_0,a_1,\dots,a_n,m,2,\beta]$ with $m$ and $\beta$ large, let $p/q=[a_0,a_1,\dots,a_n,m,1]$. Then again $p/q$ is not a convergent to $\alpha$, while $$\left|{p\over q}-\alpha\right|={1\over(2-\epsilon)q^2}$$ for $m$ and $\beta$ sufficiently large. 
A: Roth's theorem, for which he won the 1958 Fields Medal, implies that for all algebraic numbers the exponent limit of 2 is optimal.

Theorem (Thue-Siegel-Roth). Suppose $x$ and $\alpha$ are real numbers, $\alpha >2$.
  If there are infinitely many reduced fractions $p,q$ satisfying 
  $$ |x − \frac{p}{q}|  < \frac{1}{q^{\alpha}}$$ then $x$ is
  transcendental.

As I understand it, this theorem was key in demonstrating the existence of many more (classes of) transcendental numbers (which were first proposed by Leibniz in the 1600s), and thus ending the work of approximating with rationals, that was begun by Louisville in 1844, and marking a major milestone in identifying classes of transcendental numbers that was begun by Euler in 1748.
