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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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3 Answers 3

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

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  • $\begingroup$ 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? $\endgroup$ Sep 10, 2012 at 15:37
  • $\begingroup$ Sorry, I corrected the link. $\endgroup$
    – BS.
    Sep 10, 2012 at 15:49
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    $\begingroup$ And according to Springer encyclopedia, he proved an even more general result in 1884, see encyclopediaofmath.org/index.php?title=k/k055910 $\endgroup$
    – BS.
    Sep 10, 2012 at 16:09
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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 $$|nx_i-y_i-a_i|<\epsilon,\qquad 1\leq i\leq d.$$ We proceed by induction on $d$, the case of $d=0$ being trivial. So let us assume that $d\geq 1$ and the statement holds for $d-1$ in place of $d$. The initial hypothesis is invariant under replacing $x_i$ by $nx_i-y_i$ 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 $$|x_i|<\epsilon,\qquad 1\leq i\leq d.$$ By the induction hypothesis applied to the $d-1$ numbers $x_1/x_d,\dots,x_{d-1}/x_d\in\mathbb{R}$, there are $m\in\mathbb{Z}$ and $y_1,\dots,y_{d-1}\in\mathbb{Z}$ such that such that $r:=(m+a_d)/x_d$ satisfies $$|rx_i-y_i-a_i|<\epsilon/2,\qquad 1\leq i\leq d-1.$$ Moreover, this inequality also holds for $i=d$ if we set $y_d:=m$. Now let $n$ be the closest integer to $r$. Then $$|nx_i-y_i-a_i|\leq |rx_i-y_i-a_i|+|(n-r)x_i|<\epsilon/2+\epsilon/2=\epsilon,\qquad 1\leq i\leq d.$$ The proof is complete.

Added on 24 November 2023. In fact the proof above is essentially the same as the one in Estermann: A proof of Kronecker's theorem by induction, J. London Math. Soc. 8 (1933), 18-20. It also appears in Section 23.8 of Hardy-Wright: An introduction to the theory of numbers.

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    $\begingroup$ "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) $\endgroup$
    – user9072
    Sep 12, 2012 at 14:12
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    $\begingroup$ 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 $|x_i|<\epsilon$). $\endgroup$
    – Seva
    Jan 14, 2013 at 9:01
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    $\begingroup$ Nice argument, thanks for sharing! I have one suggestion: in the induction step explicitly state that you choose $a_i - \frac{x_i}{x_d} a_d$, for $i=1,\ldots,d-1$. I also have one correction: since you induction argument uses $d-1 \geq 1$, it supposes that $d \geq 2$. Thus, you must start the induction proving $d=1$, the one-dimensional Kronecker Theorem, which you must assume since your induction argument does not prove it. $\endgroup$
    – Lucas Seco
    May 30, 2023 at 14:16
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    $\begingroup$ Thanks for your reply! I do see that $d=1$ is the statement of one-dimensional Kronecker Theorem, but I do not understand what is the statement of the induction step for $d=0$: could you please clarify? $\endgroup$
    – Lucas Seco
    May 30, 2023 at 19:51
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    $\begingroup$ The proof of the induction step is pristine clear, but the proof of the $d=1$ step is not clear, unless you are assuming the one-dimensional Kronecker Theorem in which case there is nothing to be proved. $\endgroup$
    – Lucas Seco
    May 30, 2023 at 20:26
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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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