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We know that if $\alpha$ is irrational then set $\{\alpha n+m:(n,m)\in \mathbb{Z}^2\}$ is dense in $\mathbb{R}^1$.

Let's take a look at set $\{2\sqrt{2\pi n}+2\pi m+1: (n,m)\in \mathbb{Z^2}\}$. Is this set is dense in $\mathbb{R^1}$? If answer yes then can we prove it's density from density of $\{\alpha n+m:(n,m)\in \mathbb{Z}^2\}$?

Raheem Najib
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  • $\sqrt{2\pi n}$ is not defined for $n<0$. –  Nov 14 '15 at 09:25
  • @JohnMa, I have more general question: Is ${\alpha n+\beta m: (n,m)\in \mathbb{Z}^2}$ dense in reals? where $\alpha$ and $\beta$ are irrational – Raheem Najib Nov 14 '15 at 09:33
  • It depends on whether $\alpha/\beta$ is rational. –  Nov 14 '15 at 09:35
  • @RaheemNajib Yes, iff $\alpha/\beta$ is irrational (this is just a scaled version of ${\alpha n+m}$) – Mario Carneiro Nov 14 '15 at 09:35
  • @MarioCarneiro, Where I can find proof of this fact? Can you show the proof? – Raheem Najib Nov 14 '15 at 09:37
  • @RaheemNajib http://math.stackexchange.com/a/136685/50776 gives the proof for ${\alpha n+m}$, but you say you already know this fact. From that, it is simply a special case that if $A$ is dense then so is $rA$ for any positive real $r$, since any element $x\in A$ which is $\epsilon/r$-close to $y/r$ yields $rx\in rA$ which is $\epsilon$-close to $y$. – Mario Carneiro Nov 14 '15 at 09:44

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It is dense in $\Bbb R$, but for a much simpler reason: $2\sqrt {2\pi (n+1)}-2\sqrt {2\pi n}\to0$ as $n\to\infty$, while $2\sqrt {2\pi n}\to\infty$. Thus for any $\epsilon$ we eventually have $\epsilon$-close points above some $N$ (and in particular covering an interval $[N,N+2\pi]$ of length $2\pi$), and we can shift this to an interval covering any chosen point using the $2\pi m+1$ term.

Mario Carneiro
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  • Dear Mario! It's reaaly nice fact that if $a_{n+1}-a_{n}\to 0$ and $a_n\to \infty$ then ${a_n}$ is dense in real numbers. Right? Is proof of this is hard? – Raheem Najib Nov 14 '15 at 09:46
  • @RaheemNajib I've sketched the argument right here, but I'm afraid I don't know a good reference. You also need to assume that $a_n$ is monotonically increasing, and it is not ${a_n}$ but ${a_n+m:n\in\Bbb N,m\in\Bbb Z}$ that is dense. (The $2\pi$ and $+1$ factors can be easily removed by rescaling and shifting the set.) – Mario Carneiro Nov 14 '15 at 09:50
  • Dear Mario! Your statement seems really interesting and useful to me and I proved it here - http://math.stackexchange.com/questions/1528812/structure-of-dense-subset-of-mathbbr. Can you look at my proof? – RFZ Nov 14 '15 at 17:40