Proving that an equation is solvable, Floor function

Let $x \in \mathbb{R} \backslash \mathbb{Q}, x>0$ and $q \in \mathbb{N}, q>0$, prove that there is an $r \in \mathbb{N}, r>0$ with: $r \cdot x - \left\lfloor r \cdot x \right\rfloor < \frac{1}{q}$ or $1-(r \cdot x - \left\lfloor r \cdot x \right\rfloor) < \frac{1}{q}.$

I was given the hint to divide the sequence $a_r := r \cdot x - \left\lfloor r \cdot x \right\rfloor \in [0,1)$ into q intervals $[0,\frac{1}{q}),[\frac{1}{q},\frac{2}{q}),\ldots,[\frac{q-1}{q},1)$ and use the pigeonhole principle, but I cannot see how this would help to the problem.

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The pigeonhole principle tells you that if you take $k+1$ elements of the sequence $a_r$, two of them will end up in the same interval. These correspond to two distinct multiples of $x$ that end up "close together". What can you do with them...? – Rahul Nov 14 '10 at 19:13
Currently (until the next edit), $[0,1)$ has been divided up into $q$ intervals and not $k$ intervals. Maybe this is causing you some confusion. – Derek Jennings Nov 14 '10 at 19:27
Yes thanks, I corrected this mistake – Listing Nov 14 '10 at 19:30
I think the problem is not right. You cannot have a positive number $y$ such that $y<\frac{1}{q}\vee 1 -y<\frac{1}{q}$. Try $q=4$ for example. – TCL Nov 15 '10 at 0:23
Ok. I guess the author meant $y<\frac{1}{q}$ or $1-y<\frac{1}{q}$. The way he wrote it, it looks like $y<(\frac{1}{q}\vee 1)-y<\frac{1}{q}$. – TCL Nov 15 '10 at 1:50

Write $\{ x\}$ for $x-\lfloor x\rfloor$.

1. Prove that $\{ rx\}$ for $r=0,1,\cdots,q$ are all distinct. (Here is where you use the condition that $x$ is irrational.)
2. Using pigeonhole principle, find integers $r,s, 0\le r<s,$ such that
$\{ sx \},\{ rx\}$ belong to the same subinterval.
3. Then prove that $s-r$ is your solution.
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Let's use the notation $\{x\} = x - \lfloor x \rfloor$. We can think of $\{\cdot\}$ as a function $\mathbb{R} \longrightarrow \mathbb{R}/\mathbb{Z}$, indeed a homomorphism of groups (with addition as the group operator). That means that $\{x + y\} = \{x\} + \{y\}$, or what matters more in your case, $\{x - y\} = \{x\} - \{y\}$.

Now suppose $\{r_1x\},\{r_2x\}$ are close, what does this say about $\{(r_1-r_2)x\}$?

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Well it tells us that $\{(r_1-r_2)x\} \approx 0$, right? – Listing Nov 14 '10 at 19:24
Right, and that's exactly what you want to prove. – Yuval Filmus Nov 14 '10 at 19:52
But I don't see where the case occurs that the result is $1-(r \cdot x - \left\lfloor r \cdot x \right\rfloor)$, nor do I think that $(a x-\lfloor a x\rfloor)-(b x-\lfloor b x\rfloor) = (a-b) x -\lfloor (a-b) x\rfloor$ – Listing Nov 14 '10 at 20:48
@user3123 All the computations are modulo 1. A small negative number is physically a number very close to 1. – Yuval Filmus Nov 15 '10 at 3:34