# Better approximation than convergent of continued fraction

Let $$\alpha = [a_0;a_1,a_2,\ldots] \in \mathbb{R}$$ and $$(p_n/q_n)$$ be the convergents to the continued fraction of $$\alpha$$. Prove that, if $$q_n \leq q < q_{n+1}$$, gdc($$p$$,$$q$$)$$=1$$ and $$p/q \not = p_n/q_n$$ then $$|\alpha - p/q| \leq |\alpha - p_n/q_n|$$ only if $$\frac{p}{q} = \frac{p_{n+1}-rp_n}{q_{n+1}-rq_n}$$, where $$r \in \mathbb{N}$$.

This is a problem I can't solve. I was only able to show that the continiued fraction of $$p/q$$ begins with $$a_0, a_1, \ldots, a_n$$. My guess is that the next step is showing that it actually ends at the $$(n+1)$$th term, and that, I think, would solve it.

• As a general comment, Alan Baker's "Concise introduction to the Theory of Numbers" is excellent on this and much more besides. – Richard Martin Nov 26 '18 at 17:56