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