Simultaneous diophantine approximation - MathOverflow

Simultaneous diophantine approximation

2012-09-10T15:09:10Z

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?

Answer by BS for Simultaneous diophantine approximation

2012-09-10T15:23:08Z

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

Answer by GH for Simultaneous diophantine approximation

2012-09-12T14:08:03Z

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.

Answer by Anne Bauval for Simultaneous diophantine approximation

2013-01-13T15:10:32Z

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.

Answer by Adam Przezdziecki for Simultaneous diophantine approximation

2013-01-14T15:50:02Z

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.