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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.)

Simon S
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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 $[0,1]$](http://math.stackexchange.com/questions/903142/for-an-irrational-number-a-the-fractional-part-of-na-for-n-in-mathbb-n-is). The main argmuent used is the pigeonhole principle: Divide $[0,1]$ into $n$ subintervals; there are more than $n$ multiplies of $a$, so two must be in the same subinterval. – MJD Dec 19 '14 at 17:23
  • Yes it does. Looks like this is fairly easy to show, so I'm happy to run with that for now. Thanks – Simon S Dec 19 '14 at 17:25
  • Okay, I have posted that as an answer. – MJD Dec 19 '14 at 17:26
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    Related: [Proving that $m+n\sqrt{2}$ is dense in R](http://math.stackexchange.com/questions/73262/proving-that-mn-sqrt2-is-dense-in-r), http://math.stackexchange.com/q/889952/, http://math.stackexchange.com/questions/136665/for-every-irrational-alpha-the-set-ab-alpha-a-b-in-mathbbz-is-den?lq=1, http://math.stackexchange.com/questions/852210/why-does-the-additive-subgroup-of-mathbbr-generated-by-1-and-sqrt2-c?lq=1 – Jonas Meyer Dec 19 '14 at 17:27

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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 $|a+b\sqrt 2|<\epsilon$.

ajotatxe
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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 $\alpha^n \to 0$.

lhf
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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.

MJD
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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.

Clement C.
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  • This is just obscuring the question. Why are subgroups that are not of the form $c \mathbb{Z}$ dense? – Dzoooks Apr 21 '19 at 02:27
  • @Dzoooks Because it is a standard exercise/fact, one very good to know, and not really hard to show. See, e.g., [this](http://epsilon.2000.free.fr/Csup/ssgroupesdeR.pdf). – Clement C. Apr 21 '19 at 09:08
  • That is more words than and is equivalent to the pigeonhole principle answer on this page! – Dzoooks Apr 21 '19 at 14:55
  • @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. – Clement C. Apr 21 '19 at 14:57
  • 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 – Clement C. Apr 21 '19 at 15:03