For an irrational number $a$ the fractional part of $na$ for $n\in\mathbb N$ is dense in $[0,1]$ [duplicate]

Let's start with along the lines of the standard proof.

Let us divide $[0,1]$ into $k$ intervals of length $1/k$; i.e. $[0,1/k]$, $[1/k,2/k]$, $[2/k,3/k]$, etc.

Now by Dirichlet principle there are two numbers $a\ne b$ such that $\{a\alpha\}$, $\{b\alpha\}$ which are in the same interval.

If $b>a$, then $(b-a)$ is a positive integer and either $\{(b-a)\alpha\}\in [0,1/k]$ or $\{(b-a)\alpha\}\in[1-1/k,1]$.

Since $\alpha$ is irrational, $\{(b-a)\alpha\}$ is non-zero. (The number $(b-a)\alpha$ cannot be an integer.)

Now if we take all multiples $n(b-a)\alpha$, $n\in\mathbb N$, then in each of the $k$ intervals must be at least one of the values $\{n(b-a)\alpha\}$. (We go either upwards from $[0,1/k]$, or downwards from the last interval, but we can never skip an interval.)

This implies that the set of all multiples is dense in $[0,1]$.

Pick any $k\in\mathbb{N}$. By the pigeonhole principle, there are two multiples of $\alpha$ whose fractional part lie within $1/k$ of each other. Taking the difference, there is a multiple of $\alpha$ with (positive) fractional part $<1/k$.

It follows that every $x\in [0,1]$ is within $1/k$ of some $\{n\alpha\}$, for any $k$.