$\{ a + b\sqrt{2} \ : \ a, b \in \mathbb{Z} \}$ dense in $\mathbb{R}$? 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.)
 A: 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 $|a+b\sqrt 2|<\epsilon$.
A: 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 $\alpha^n \to 0$.
A: 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.
A: 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.
