There is an integer multiple of each real to get close to an integer

Question

Let $\alpha\in\mathbb{R}\setminus \mathbb{Q}$. I want to show that for any $\varepsilon>0$, there exists integers $m$ and $n$ such that \begin{equation} n\in(m\alpha-\varepsilon, m\alpha+\varepsilon) \end{equation} Obviously, $\alpha$ doesn't have any integer multiple (integer multiple). I tried a proof by contradiction, but didn't work and don't have any other ideas.

Answer 1

Let $f=\inf \{|m\alpha -n|: m,n \in \Bbb Z\}\backslash \{0\}.$ Suppose (by contradiction) that $f>0.$ We have $f< 1/2.$ (Let $m=1$ and let $n$ be the nearest integer to $\alpha.$ )

Take $m,n \in \Bbb Z$ such that $f\leq |m\alpha-n|<3f/2.$

For brevity let $g=m\alpha - n.$ We have $0<|g|<1.$

$\bullet \;$ There exists $k\in \Bbb N$ such that $k|g|<1<(k+1)|g|.$ ( We cannot have $k|g|=1$ or $(k+1)|g|=1 $ for any $k\in \Bbb N,$ else $g,$ and hence $x$ also, is rational.)

The positive numbers $1-k|g|$ and $(k+1)|g|-1$ add to $|g|$ so their minimum is at most $|g|/2.$

$\bullet \;$ So $0<\min (1-k|g|, (k+1)|g|-1)< |g|/2.$ (We cannot have a min of $|g|/2,$ else $g$ is rational.)

Let $k'=k$ if $1-k|g|< |g|/2.$ Otherwise let $k'=k+1.$ In both cases we have $0<|\;k'|g|-1|< |g|/2.$

Let $m'=mk'$ and $n'=nk'.$

Now if $g>0$ then $0<|m'\alpha - (n'+1)|=|\;k'|g|-1|<|g|/2<(3f/2)/2=3f/4<f,$ contrary to the definition of $f.$

Or if $g<0$ then $0<|m'\alpha- (n'-1)|=|\;k'|g|-1|< |g|/2<(3f/2)/2=3f/4<f,$ again a contradiction.

Answer 2

Hint:   by Dirichlet's approximation theorem the inequality $\left|\alpha - \dfrac{n}{m}\right| \lt \dfrac{1}{m^2}$ with irrational $\alpha$ is satisfied for infinitely many integers $n, m$. Note that this is the same as $-\dfrac{1}{m}\lt m \alpha - n \lt \dfrac{1}{m}$.