$\{ a + b\sqrt{2} \ : \ a, b \in \mathbb{Z} \}$ dense in $\mathbb{R}$?

Question

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.)

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 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
Related: Proving that $m+n\sqrt{2}$ 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

score 5 · Answer 1 · answered Dec 19 '14 at 17:27

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

score 4 · Answer 2 · answered Dec 19 '14 at 17:25

4

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 $\alpha^n \to 0$.

answered Dec 19 '14 at 17:25

lhf

216,483

Thanks to you and @Clement C for this. I had forgotten this result. – Simon S Dec 19 '14 at 17:27

MJD · Accepted Answer · 2019-04-25T10:19:22.717

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.

score 3 · Answer 4 · answered Dec 19 '14 at 17:24

3

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.

answered Dec 19 '14 at 17:24

Clement C.

67,323

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. – 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

$\{ a + b\sqrt{2} \ : \ a, b \in \mathbb{Z} \}$ dense in $\mathbb{R}$?

4 Answers4