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## Simultaneous diophantine approximation

Let $r(x)$ be the function $x$ mod $1$, i.e. $x$ minus its floor.

Now let $m$ be a given positive integer, and $c$ a vector in $\mathbb{R}^m$ whose components are linearly independent over $\mathbb{Q}$, where (without loss of generality) the first component is $c_1=1$. Is the set of points $(r(c_2n),\ldots,r(c_mn))$, for $n\in\mathbb{N}$, dense in the $(m-1)$-dimensional unit cube? (It is known that the origin is a limit point, under weaker assumptions.)

If not, is anything known about vectors $c$ for which this is the case?

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This is true, and known as the Kronecker Theorem on diophantine approximation.

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 Thanks for this - but your link doesn't work. Sources I have looked at suggest that Kronecker did just the case $m=2$. Do you have a reference for the general case? – cameroncounts Sep 10 at 15:37 Sorry, I corrected the link. – BS Sep 10 at 15:49 And according to Springer encyclopedia, he proved an even more general result in 1884, see encyclopediaofmath.org/index.php?title=k/k055910 – BS Sep 10 at 16:09

Let me share a simple proof I found during a childbirth class 8 years ago:

Let $x_1,\dots,x_d\in\mathbb{R}$ such that $1,x_1,...,x_d$ are linearly independent over $\mathbb{Q}$. Let $\epsilon>0$ and $a_1,\dots,a_d\in\mathbb{R}$ be arbitrary. We want to show that there are $n\in\mathbb{Z}$ and $y_1,\dots,y_d\in\mathbb{Z}$ such that $$|nx_i-y_i-a_i|<\epsilon,\quad 1\leq i\leq d.$$ We proceed by induction on $d$, the case of $d=0$ being trivial. The hypothesis is invariant under replacing $x_i$ with $nx_i-y_i$ for any nonzero $n\in\mathbb{Z}$ and any $y_1,\dots,y_d\in\mathbb{Z}$, while the conclusion only becomes stronger. Hence by Dirichlet's theorem on simultaneous diophantine approximation we can assume from the beginning that $$|x_i|<\epsilon,\quad 1\leq i\leq d.$$ By the induction hypothesis applied for $x_1/x_d,\dots,x_{d-1}/x_d$, there are $m\in\mathbb{Z}$ and $y_1,\dots,y_d\in\mathbb{Z}$ such that such that $r:=(m+a_d)/x_d$ satisfies $$|rx_i-y_i-a_i|<\epsilon/2,\quad 1\leq i\leq d.$$ Note that for $i=d$ this inequality is automatic with $y_d:=m$. Let $n$ be the closest integer to $r$, then $$|nx_i-y_i-a_i|\leq |rx_i-y_i-a_i|+|(n-r)x_i|<\epsilon/2+\epsilon/2=\epsilon,\quad 1\leq i\leq d.$$ The proof is complete.

Remark 1. I clarified the proof in response to some criticism.

Remark 2. Using Dirichlet's theorem again, there are infinitely many $n$'s with the required properties.

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"found during a childbirth class" that would be a good addition to an article of Günter Ziegler on where maths can be done :-) gegenworte.org/heft-16/ziegler16.html (text in German) – quid Sep 12 at 14:12
@quid: That's a great link, thank you! – GH Sep 12 at 14:56
Nice proof - much simpler than anything I've seen before. As a point of perfectionism: you may wish to remove extra "such that" and restate the sentence starting with "The statement does not change..." (the statement does change, what you really mean is that replacing the $x_i$ with $nx_i-y_i$ with a suitably chosen $n$ and $y_i$, one can assume without loss of generality that $|x_i|<\epsilon$). – Seva Jan 14 at 9:01
@Seva: Thank you. I agree with your comment. I updated the text, please check. – GH Jan 14 at 14:17

There is a flaw in the last line : ║(n-r)x║ is not lower than ║x║/2 because ║ ║, which means "distance to ℤ^d", is not positively homogeneous. Anyway, this proof cannot hold because it does not use the (necessary) hypothesis of linear independence but only that the x_i's are nonzero.

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I think my proof is ok. I really meant that $\| \|$ is the norm in $\mathbb{R}^d$, and from the very beginning we can assume that $\|x\|<\epsilon$ because we can replace $x$ with $mx+y$ for any $m\in\mathbb{Z}$ and $y\in\mathbb{Z}^d$. The independence hypothesis is used at two places: first we use that replacing $x$ with $mx+y$ as above leaves the hypothesis invariant, second we use it in the induction hypothesis in the form that $x_i/x_d$ for $i=1,\dots,d-1$ are linearly independent over $\mathbb{Q}$. – GH Jan 13 at 19:38
I clarified my proof above, please check. – GH Jan 13 at 20:06
Thank you very much, and shame on me !!! I put your proof on fr.wikipedia.org/wiki/…) – Anne Bauval Jan 14 at 0:07
Thanks for checking my proof carefully and putting it on Wikipedia. Please mention "GH from MO" and provide a link to this page. – GH Jan 14 at 0:14

Your problem is answered (positively) in the first two chapters of W. Schmidt, "Diophantine approximation." Lecture Notes in Mathematics, 785. 1980.

Very well written and not too long. The case of m=2 is treated separately, as it is especially elegant. More - in the case of m=2 - it is estimated how well you can approximate various numbers with growing n.

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