Approximation of a real number via a fraction of coprimes.

I'm reading a paper on number theory (which is not my field at all) stating, without any proof, a claim which can be rephrased as

Fix a positive integer $M$. Then, given any real number $\alpha$, there exist a positive integer $q\leq M$ and an integer $h$, such that:

$(q, h)=1;$
$|q\alpha - h|\leq \tfrac{1}{M}.$

I can see that this is true for $M=1$, with $q=1$.

So far I've found the following. First, it is enough to find two integers $q>0$ and $h$ which satisfy $|q\alpha - h|\leq \tfrac{1}{M}$, without imposing that they are coprime. Indeed, we can write $q=mq'$ and $h=mh'$ with $q'$ an $h'$ coprime, and $|q'\alpha - h'|\leq \tfrac{1}{Mm}\leq\tfrac{1}{m}$.

Then we can assume that $\alpha\in[0,1)$. As if this is not the case, we can write $\alpha = n + \alpha'$, with $\alpha'\in[0,1)$ and if $h', q'$ solve the problem for $\alpha'$, then $h=h'+q'n$ and $q=q'$ solve it for $\alpha$.

Now essentially I still need to show that the set $[\alpha, 2\alpha, \dots, M\alpha]$ has at least one element which is closer than $1/M$ to the closest integer. How can I proceed to show this?

The paper I'm reading is On Certain Sets of Integers by K. F. Roth (1953). And the claim I have rephrased above is equation (5).

Solution 1:

I think I have found a way to prove the last step. Assume that it is false and that there is no integer $j$ between $1$ and $M$ such that the decimal part of $j\alpha$ is in $[0, 1/M]\cup[1-1/M,1]$. Then I have $M$ elements ($[\alpha, 2\alpha, \dots, M\alpha]$) whose decimal parts are all in $(1/M, 1-1/M)$. It follows that at least two of this elements (let's say $u\alpha$ and $v\alpha$) which have decimal parts less than $1/M$ far apart. Let's say that $u>v$. We have that $(u-v)\alpha\in[\alpha, 2\alpha, \dots, M\alpha]$ and its decimal part must be in $[0, 1/M]\cup[1-1/M,1]$, which is a contradiction.