Is the span of rationally independent real numbers dense in $\mathbb{R}$ Define the subset $S\subset\mathbb{R}$ to be equal to $\{a_1\mathbb{Z} + a_2\mathbb{Z} +\cdots+ a_n\mathbb{Z} \}$, where $a_1,a_2\cdots,a_n\in\mathbb{R}$ are rationally independent and $2\leq n<\infty$ is an arbitrary index. Is $S$ dense in $\mathbb{R}$?
 A: A subgroup of $\mathbb R$ is either dense or of the form $a\cdot\mathbb Z$ (this is a classical result):
If $S = \{0\}$, then $S = 0 \cdot \mathbb Z$ and we are done.
Otherwise, $S$ contains some $x \neq 0$ and so $\pm x > 0$ and so $S \cap(0, \infty) \neq \emptyset $. Set $a := \inf \left(S \cap (0, \infty)\right)$.
We have that $a \in S$ necessarily. Assume $a \notin S$, and so $a > 0$. There is then a sequence $x_n\in S$ converging to $a$, so $x_n-x_{n+1}$ converges to $0$ so $a=0$. Contradiction.
Now, if $a = 0$, you can see that $S$ is dense in $\mathbb R$.
Otherwise, the Eucledian division theorem says that if $x\in S$, $x=n\cdot a + r$ for some $n\in\mathbb Z$ and $0\leq r < a$. But $r=x-n\cdot a\in S$ so $r=0$ (because $a$ is the smallest positive element of $S$), and $x\in a\cdot \mathbb Z$, hence $S=a\cdot \mathbb Z$.
Coming back to the asked question, if $S$ is not dense, there is $a\in\mathbb R$ such that $S=a\cdot\mathbb Z$.
This implies that $a_1=na$ and $a_2=ma$ for some $n$ and $m$, hence $a_1/a_2=n/m$ so this contradicts the fact that they are rationnaly independent.
