# Proof of continued fraction convergence theorem

How does one prove that $$|\alpha - \frac{p_n}{q_n}| < |\alpha - \frac{p_m}{q_m}|$$ for all $n>m$? I know that the left side is less than $\frac{1}{2q_n^2}$ and the right side is less than $\frac{1}{2q_m^2}$, but how do we know they are not equal? Thanks!

note: $\frac{p_n}{q_n}$ is the $n$th convergent of a continued fraction approximating some irrational $\alpha$.

All proofs of results of this type should be in Niven/Zuckerman/Montgomery, for example. They often follow from nice bounds for $\alpha-\frac{p_n}{q_n}$, such as Theorem 5 of this section.
Notice also that your desired inequality is a special case of: $|\alpha-\frac{p_n}{q_n}| < |\alpha-\frac cd|$ for all $d<q_n$.