For an irrational number $\alpha$, prove that the set $\{a+b\alpha: a,b\in \mathbb{Z}\}$ is dense in $\mathbb R$

Question

I am not able to prove that this set is dense in $\mathbb{R}$. Will be pleased if you help in a easiest way. $A=\{a+b\alpha: a,b\in \mathbb{Z}\}$ where $\alpha$ is a fixed irrational number.

Answer 1

I will write $\{x\}$ to mean the fractional part of $x$, i.e. for $x$ minus the floor of $x$. What we need to show is that we can get arbitrarily close to $0$ by taking $\{m\alpha\}$ for varying integers $m$. Note that, because $\alpha$ is irrational, $\{m\alpha\} \neq \{m'\alpha\}$ for $m \neq m'$.

Let's show that we can get within $1/n$ of $0$ for an arbitrary positive integer $n$. Divide up the interval $[0, 1]$ into $n$ closed intervals of length $1/n$. We have $n+1$ distinct quantities $0, \{\alpha\}, \{2\alpha\}, \ldots, \{n\alpha\}$.

By the pigeonhole principle, two of these, say $\{i\alpha\}$ and $\{j\alpha\}$ with $i > j$, lie in the same closed interval $[k/n, (k+1)/n]$, and so their difference, which is $\{(i - j)\alpha\}$, is closer than $1/n$ to $0$; as $n$ was arbitrary, we're done.