Simultaneous diophantine approximation - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T17:55:46Z http://mathoverflow.net/feeds/question/106819 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106819/simultaneous-diophantine-approximation Simultaneous diophantine approximation cameroncounts 2012-09-10T15:09:10Z 2013-01-14T15:50:02Z <p>Let $r(x)$ be the function $x$ mod $1$, i.e. $x$ minus its floor.</p> <p>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.)</p> <p>If not, is anything known about vectors $c$ for which this is the case?</p> http://mathoverflow.net/questions/106819/simultaneous-diophantine-approximation/106821#106821 Answer by BS for Simultaneous diophantine approximation BS 2012-09-10T15:23:08Z 2012-09-10T15:47:06Z <p>This is true, and known as the <a href="http://en.wikipedia.org/wiki/Kronecker%27s_theorem" rel="nofollow">Kronecker Theorem</a> on diophantine approximation.</p> http://mathoverflow.net/questions/106819/simultaneous-diophantine-approximation/107010#107010 Answer by GH for Simultaneous diophantine approximation GH 2012-09-12T14:08:03Z 2013-01-14T14:19:08Z <p>Let me share a simple proof I found during a childbirth class 8 years ago:</p> <p>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 $$|nx_i-y_i-a_i|&lt;\epsilon,\quad 1\leq i\leq d.$$ 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 $$|x_i|&lt;\epsilon,\quad 1\leq i\leq d.$$ 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 $$|rx_i-y_i-a_i|&lt;\epsilon/2,\quad 1\leq i\leq d.$$ Note that for $i=d$ this inequality is automatic with $y_d:=m$. Let $n$ be the closest integer to $r$, then $$|nx_i-y_i-a_i|\leq |rx_i-y_i-a_i|+|(n-r)x_i|&lt;\epsilon/2+\epsilon/2=\epsilon,\quad 1\leq i\leq d.$$ The proof is complete.</p> <p><strong>Remark 1.</strong> I clarified the proof in response to some criticism.</p> <p><strong>Remark 2.</strong> Using Dirichlet's theorem again, there are infinitely many $n$'s with the required properties.</p> http://mathoverflow.net/questions/106819/simultaneous-diophantine-approximation/118814#118814 Answer by Anne Bauval for Simultaneous diophantine approximation Anne Bauval 2013-01-13T15:10:32Z 2013-01-13T15:10:32Z <p>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.</p> http://mathoverflow.net/questions/106819/simultaneous-diophantine-approximation/118889#118889 Answer by Adam Przezdziecki for Simultaneous diophantine approximation Adam Przezdziecki 2013-01-14T15:50:02Z 2013-01-14T15:50:02Z <p>Your problem is answered (positively) in the first two chapters of W. Schmidt, "Diophantine approximation." Lecture Notes in Mathematics, 785. 1980.</p> <p>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. </p>