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This question already has an answer here:

I'm guessing $\{ a + b\sqrt{2} \ : \ a, b \in \mathbb{Z} \}$ is dense in $\mathbb{R}$. I'm having a mental block. How do you show that?

(This is motivated by a different hypothesis: if $f$ is continuous with two periods $T_1$, $T_2$, then $f$ is constant if $T_1/T_2$ is not rational.)

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marked as duplicate by Jonas Meyer, jdoicj, MJD, Rick Decker, user147263 Dec 19 '14 at 17:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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There may be a simpler way to go about it, but I believe it follows from For an irrational number $a$ the fractional part of $na$ for $n\in\mathbb N$ is dense in $[0,1]$. The main argument used is the pigeonhole principle: Divide $[0,1]$ into $k$ subintervals; there are more than $k$ multiples of $a$, so two must be in the same subinterval; therefore there are two multiples of $a$ that differ by less than $\frac1k$.

Note that some of the fussy details in the answers there deal with the fact that the index set is $\Bbb N$ rather than $\Bbb Z$; since you want $\Bbb Z$ the arguments can be simplified.

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Try a kind of Euclid's algorithm to find $\gcd(1,\sqrt 2)$:

$$\sqrt 2=1+r_1$$ $$1=q_1r_1+r_2$$ $$\ldots$$

Obviously, you never end, precisely because of the irrationality of $\sqrt 2$. You should find that for any $\epsilon>0$, there exist $a,b$ such that $|a+b\sqrt 2|<\epsilon$.

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An additive subgroup $A$ of $\mathbb R$ is either cyclic or dense. This depends on whether $\inf A \cap \mathbb R^+ = 0$.

Your group contains $\alpha=-1+\sqrt 2$ and all its powers. Since $0 <\alpha <1$, the group cannot be cyclic, because $\alpha^n \to 0$.

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  • $\begingroup$ Thanks to you and @Clement C for this. I had forgotten this result. $\endgroup$ – Simon S Dec 19 '14 at 17:27
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Writing $G = \{ a + b\sqrt{2} \mid a, b \in \mathbb{Z} \}$, can you show that $(G,+)$ is a subgroup of $(\mathbb{R},+)$? Then, what do you know about the additive subgroups of $\mathbb{R}$? $(\dagger)$


$(\dagger)$ They are all of the form $c\mathbb{Z}$, or dense.

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  • $\begingroup$ This is just obscuring the question. Why are subgroups that are not of the form $c \mathbb{Z}$ dense? $\endgroup$ – Dzoooks Apr 21 at 2:27
  • $\begingroup$ @Dzoooks Because it is a standard exercise/fact, one very good to know, and not really hard to show. See, e.g., this. $\endgroup$ – Clement C. Apr 21 at 9:08
  • $\begingroup$ That is more words than and is equivalent to the pigeonhole principle answer on this page! $\endgroup$ – Dzoooks Apr 21 at 14:55
  • $\begingroup$ @Dzoooks and so? This is a valid answer, it works, and highlights another useful result worth knowing. Nothing asks you to use this approach; it is nonetheless a legitimate one. $\endgroup$ – Clement C. Apr 21 at 14:57
  • $\begingroup$ It is a way to overcome the "mental block" the OP had and solve the question. As such, it is an answer. I don't know what else to respond, nor what your issue is with this 5-year old post. Further, if you have any doubt as to whether this is useful, read the comment left by the OP on lhf's answer: you will see that the OP did, in fact, find this helpful. @Dzoooks $\endgroup$ – Clement C. Apr 21 at 15:03

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