4

Possible Duplicate:
For every irrational $\alpha$, the set $\{a+b\alpha: a,b\in \mathbb{Z}\}$ is dense in $\mathbb R$

I want to show that

Given any irrational number $\alpha\in \mathbb{R}$, the set $\displaystyle S=\{ m+n\alpha : m,n\in Z \}$ is dense in $\mathbb{R}$.

Thanks in advance!

Community
  • 1
Kns
  • 3,165

1 Answers1

3

Kronecker's Theorem gives us that the set $\{ma\}_{m \in \mathbb{Z}}$ is dense in $(0,1)$ for irrational $a$, and hence, that the set $\{n+ma\}_{m,n \in \mathbb{Z}}$ is dense in $\mathbb{R}$.

If you want to prove Kronecker's Theorem, you may use Weyl's Criterion, which proves more strongly, that the said sequence is equidistributed too!

Rijul Saini
  • 2,544
  • Please write the statement of Kronecker's theorem! – Kns Jul 31 '12 at 16:12
  • I've edited to give the link of Kronecker's Theorem. Hope this helps. – Rijul Saini Jul 31 '12 at 16:15
  • With $a=\pi$, I don't find a single integer $m$ so that $0<ma<1$. That's quite the opposite of "dense in $(0,1)$". – celtschk Jul 31 '12 at 16:15
  • 1
    @celtschk The notation ${ma}$ is the decimal part of $ma$. – David Mitra Jul 31 '12 at 16:17
  • It may be noted that Kronecker's Theorem has a rather simple proof using the Pigeonhole Principle. See the answer to the question which is a duplicate of this. – David Mitra Jul 31 '12 at 16:20
  • 4
    OK, maybe when using ${}$ in any other way than to denote sets, it would be a good idea to explicitly say that :-) Especially since your second use obviously is using it just to denote a set (the set of fractional parts of $n+ma$ definitely is not dense in $\mathbb{R}$). – celtschk Jul 31 '12 at 16:24