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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!

1 Answers 1

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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!

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    Please write the statement of Kronecker's theorem!2012-07-31
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    I've edited to give the link of Kronecker's Theorem. Hope this helps.2012-07-31
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    With $a=\pi$, I don't find a *single* integer $m$ so that $0$(0,1)$". – 2012-07-31
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    @celtschk The notation $\{ma\}$ is the decimal part of $ma$.2012-07-31
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    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.2012-07-31
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    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 $\mathbb{R}$).2012-07-31