Discreteness of $\{ a+b\sqrt2 \mid a,b \in \Bbb N \}$ in $\Bbb R$

Question

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.

What do you mean by discrete? Every point of it is open in itself? — pisco, 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, and at this question. — Dietrich Burde, Dec 30 '17 at 14:44
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
@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

score -1 · Answer 1 · answered Dec 30 '17 at 13:31

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.

Discreteness of $\{ a+b\sqrt2 \mid a,b \in \Bbb N \}$ in $\Bbb R$

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