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

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
• @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, and at this question. – Dietrich Burde Dec 30 '17 at 14:44

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.