Roth's theorem says that for irrational algebraic number $\alpha$ and $\epsilon>0$, there are finitely many solutions to this: $$\displaystyle \left|\alpha-\frac pq\right|<\frac 1{q^{2+\epsilon}}$$ but from what I understand from continued fractions(theorem 5), there are infinitely many convergents of $\alpha$ $$\displaystyle \left|\alpha-\frac{p_n}{q_n}\right|<\frac {1}{q_nq_{n+1}}<\frac {1}{q_n^2}$$ Since $q_{n+1}>q_n$, we have $\frac {1}{q_nq_{n+1}}=\frac 1{q_n^{2+\epsilon_n}}$ for all $n$, contradicting Roth's theorem.
I probably missed something, but I am not sure what....