Dense sequence in $[0,1]$

Question

There is the theorem proved by Liouville which states that if $x$ is irrational then there are infinitely many fractions $\frac{p}{q}$ such that $|x-\frac{p}{q}|<\frac{1}{q^2}$, i.e. $|qx-p|<\frac{1}{q}$. It means that the fractional part of $qx$ (let's denote it by $\{qx\}$) is either close to $0$ or close to $1$. I guess it can help me to prove that the set $\{\{mx\}:m\in \mathbb{N}\}$ for irrational $x$ is dense in $[0,1]$. Could you help me to prove that?

Answer 1

You can use the Euclidean algorithm to prove that. Run it on $x$ and $1$.

If $x$ is irrational, it never stops and produces arbitrarily small positive values $ax+b$, where $a,b\in\mathbb Z$.

So say $1>ax+b=\varepsilon>0$ (so $a\ne0)$.

If $a>0$, then $\{kax\}=\{kax+kb\}=\{k\varepsilon\}=k\varepsilon$ as long as $k\varepsilon<1$.

If $a<0$, then $\{-kax\}=1-\{kax\}=1-\{kax+kb\}=1-\{k\varepsilon\}=1-k\varepsilon$ as long as $k\varepsilon<1$.

This gives you subsets of $[0,1]$ such that every number is at most $\varepsilon$ away from some $\{mx\}$, where $m\ge1$ is an integer.