Is $\mathbb{Z}+\mathbb{Z}\alpha$ dense in $\mathbb{R}$ if $\alpha\in\mathbb{R}\setminus\mathbb{Q}$? I know this should be easy, but I am at a loss right now.
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1It might be easier to show that the $\mathbb Z \alpha \pmod 1$ is dense in $(0,1)$. – Quang Hoang Nov 30 '20 at 14:53

@QuangHoang those are equivalent statements. – The Thin Whistler Nov 30 '20 at 14:58

Yes they exactly equivalent. What I meant is that attending to $(0,1)$ only might be a little easier. – Quang Hoang Nov 30 '20 at 15:00

Take a look at [this](https://math.stackexchange.com/q/73262/11619). My answer there is specific to $\alpha=\sqrt2$, but Mariano SuárezÁlvarez posted an answer that works for all irrationals. – Jyrki Lahtonen Nov 30 '20 at 15:52

Use [KAT](https://mathworld.wolfram.com/KroneckersApproximationTheorem.html). – rtybase Nov 30 '20 at 16:43

@TheThinWhistler: In [this answer](https://math.stackexchange.com/questions/189397/whyisthisquotientspacenothausdorff/189402#189402) I gave an elementary proof that $\{nr\bmod1:n\in\Bbb Z\}$ is dense in $[0,1)$. – Brian M. Scott Nov 30 '20 at 20:45
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$G=\mathbb{Z}+\mathbb{Z}\alpha$ is an additive subgroup of $\mathbb R$. And the additive subgroups of the real numbers are either discrete or dense.
If $G$ was discrete, it would exist $0 \neq \beta \in \mathbb R$ and $a,b \in \mathbb Z$ such that
$$\begin{cases} 1 &= b \beta\\ \alpha &= a \beta \end{cases}$$
Therefore, $\alpha = a/b$ would be rational. A contradiction.
Hence $G$ is dense.
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2What if $a=1,b=0$? Since this answer is downvoted , please note that I didn't downvote. – PNDas Nov 30 '20 at 14:47

Aren't you assuming exactly what OP has to prove  that there are only those two options – Empy2 Nov 30 '20 at 14:51

how do I know the additive subgroups are either discrete or dense? – The Thin Whistler Nov 30 '20 at 14:58


@TheThinWhistler See [Subgroup of Real Numbers is Discrete or Dense](https://proofwiki.org/wiki/Subgroup_of_Real_Numbers_is_Discrete_or_Dense). – mathcounterexamples.net Nov 30 '20 at 15:09

1@TheThinWhistler if the group has a limit point $x$, subtract everything by $x$ then $0$ is an limit point. It should be easy to show then that the group is dense in a neighborhood of $0$, hence dense everywhere. – Quang Hoang Nov 30 '20 at 15:09

Still the updated version doesn't seem to clarify the situation. Why does the expression follow from the discreteness? – kesa Nov 30 '20 at 15:16

If $G$ is discrete, $G= \beta \mathbb Z$ for $0 \neq \beta \in \mathbb R$. And $1=1+0 \alpha$, $\alpha = 0 + 1 \alpha$ both belong to $G$. – mathcounterexamples.net Nov 30 '20 at 15:17

Still, these are assumptions that one, who asks this question, does not necessarily know – kesa Nov 30 '20 at 15:19

@kesa Sure. Proves of a result may depend on the theorems that you apply. This result about additive subgroups of the reals is pretty *well known* though. – mathcounterexamples.net Nov 30 '20 at 15:21

I'm am aware of that, my point is that you seem to ignore that someone who asks this question might not – kesa Nov 30 '20 at 15:26
