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How to prove that $\{ a+b\sqrt2 \mid a,b \in \Bbb N \}$ is discrete in $\Bbb R$?

If I sum over $\Bbb Z$ instead of over $\Bbb N$, it becomes dense, which is quite confusing to me.

Also, when I plot the points, they appear to become denser as I go to the right, which leads me to wonder if the set is really discrete.

Bill Dubuque
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Kenny Lau
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  • What do you mean by discrete? Every point of it is open in itself? – pisco Dec 30 '17 at 13:57
  • @pisco125 yes, that is what it means. – Kenny Lau Dec 30 '17 at 13:57
  • Then you basically answered the question below. Simply choose a small neighborhood which evades all those finite points. – pisco Dec 30 '17 at 14:00
  • @pisco125 I didn't know that the points are finite. Of course I answered the question below, or else I wouldn't have posted it as an answer. – Kenny Lau Dec 30 '17 at 14:01
  • See also the answers at [this question](https://math.stackexchange.com/questions/73262/proving-that-mn-sqrt2-is-dense-in-r), and at [this question](https://math.stackexchange.com/questions/1909/characterizing-dense-subgroups-of-the-reals). – Dietrich Burde Dec 30 '17 at 14:44
  • @DietrichBurde but my set is hardly a subgroup. – Kenny Lau Dec 31 '17 at 00:48
  • Ah, I see, I am sorry. But the part "If I sum over $\Bbb Z$ instead of over $\Bbb N$, it becomes dense, which is quite confusing to me." is covered by the duplicate at least. – Dietrich Burde Dec 31 '17 at 09:29
  • See more specifically [this answer](https://math.stackexchange.com/a/2130828/242) – Bill Dubuque Mar 11 '21 at 09:14
  • @BillDubuque Again, that question is about $\Bbb Z[\sqrt2]$ not $\Bbb N[\sqrt2]$. What is the criterion of duplicates? – Kenny Lau Mar 11 '21 at 14:04

1 Answers1

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It is discrete, as for every $N$ there are only finitely many elements from the set less than $N$, because for every $a+b\sqrt2 < N$, we know tha $a+b<N$, but there are only at most $N(N+1)/2$ possible pairs of $(a,b)$ with $a+b < N$, so the number of elements from the set that are less than $N$ is also finite.

Kenny Lau
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