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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.

The Thin Whistler
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    It 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 Suárez-Álvarez 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/why-is-this-quotient-space-not-hausdorff/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.

mathcounterexamples.net
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