best rational approximation to an irrational number is the one with the largest denominator

Question

This comes as a common sense to me but don't know where to start for a formal proof.

Lets denote the closest rational approximation to a a positive irrational number $\alpha$, for a given denominator $y \in \mathbb Z^{+}$, by $best(\alpha,y)$. My claim is that for larger $y$ we should find a better approximation to $\alpha$.

OR $$ |best(\alpha,z)-\alpha| \gt |best(\alpha,y)-\alpha| $$ for some $z>y$

Answer 1

A hint: consider the integer $n=\lceil\dfrac1{|best(\alpha,y)-\alpha|}\rceil$. Suppose $best(\alpha,y)=\frac my$, and consider $\frac my-\frac1n = \frac{nm-y}{ny}$. Can you show that $best(\alpha,ny)\lt best(\alpha,y)$?