What does $\tfrac{p}{q}$ represent in this continued fraction question? In this theorem for continued fractions
$\alpha \in \Bbb R /\Bbb Q $ suppose $n >1$ $, 0<q \leq q_n$
$\tfrac{p}{q}\neq\tfrac{p_n}{q_n}$, where $\tfrac{p_n}{q_n}$ is the $n^{th}$ convergent af $\alpha$ then $|\alpha-\tfrac{p_n}{q_n}|<|\alpha-\tfrac{p}{q}|$.
What does $\tfrac{p}{q}$ represent ? Originally I had thought it was $\alpha$ itself but obviously that would make no sense I realised after getting to the inequality at the end.
 A: $\alpha$ is an irrational and so it has a sequence of convergents $\frac{p_n}{q_n}$ (rational numbers) that approximate $\alpha$ closer and closer.
The statement you quote has as its conclusion that $|\alpha - \frac{p_n}{q_n}| < |\alpha-\frac{p}{q}|$, saying that $\frac{p_n}{q_n}$ is a better approximation of $\alpha$ than $\frac{p}{q}$ is (some other rational). But the condition on this other approximating rational $\frac{p}{q}$ makes no sense: what does it mean that "$\frac{p}{q}$ does not divide $\frac{p_n}{q_n}$"? We usually talk about divisibility in proper rings, but in fields (like $\Bbb Q$) every is divisible by any non-zero element). You might want to check what the lecturer means by that statement (or whether you copied it correctly). Check out the "best rational approximations" part of the Wikipedia page as well.
After changing the condition in the question:  After adding $\frac{p}{q} \neq \frac{p_n}{q_n}$ and $q \le q_n$ the world is right again and it's the statement in the Wikipedia page I linked to. So it's saying that the $n$-th convergent is the best rational approximation of $\alpha$ among all fractions of smaller or equal denominator. 
