$\{ a + b\sqrt{2} \ : \ a, b \in \mathbb{Z} \}$ dense in $\mathbb{R}$?

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 $|a+b\sqrt 2|<\epsilon$.

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 $\alpha^n \to 0$.

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.