How does this proof of density of $\mathbb{Q}$ in $\mathbb{R}$ require $a \geq 0$? My Real Analysis text offers the following proof of "given two real numbers a and b, with $a < b$, there exists a rational number r strictly in between of those two", conditional on $0 \leq a < b$. It also says that the case of $a < 0$ follows easily from this proof after small rearrangement. My problem is that I don't see where $a \geq 0$ comes into play in this proof, as I can't find where any of the proof's arguments are dependent on a's sign. Here's the proof:
We need to produce m, n $\in \mathbb{N}$ such that $a < \frac{m}{n} < b$.
By Archimedean Property we know that there exists $n$ such that $\frac{1}{n}<b-a$. 
Proceed to $na < m < nb$. Now we need to pick such an m so that it's the smallest natural number greater than na; in other words $m-1 \leq na < m$. From here we get $m > na$, which is half way. Going back to our Archimedean Property equation, we can rewrite it as $a < b-\frac{1}{n}$, and thus write 
$$m \leq na+1$$
$$< n(b-\frac{1}{n})+1$$
$$=nb.$$
Thus, conditions for $a < \frac{m}{n} < b$ are satisfied. Once again, I don't see where a's positivity plays a role here. Thanks!!
 A: (As requested, posted as an answer). 
For some reason, the author wants to find a positive rational between $a$ and $b$ (note he is looking for natural numbers $n$ and $m$); of course, that will not happen in general. Technically, because he is requiring $m$ to be a natural number (hence nonnegative [or possibly positive, depending on whether the author's version of $\mathbb{N}$ includes $0$ or not]), it may be impossible to find such a number that satisfies $m-1 \leq na \lt m$ (e.g., if $na\lt -1$). So he is asking that $a$ be nonnegative to ensure that one can pick $m$ nonnegative that satisfies the condition.
That said, there is absolutely no need to divide the argument like this. The exact same argument works if we pick $m$ to be the unique integer such that $m-1\leq na\lt m$. The argument carries through then to produce a rational between $a$ and $b$, regardless of the signs of $a$ and of $b$. So you are right that the essence of the argument does not require $a$ to be nonnegative, and a minor tweak to the argument would eliminate the need for that assumption. 
A: Let us begin from the point where we have established that $na < m < nb$. Let
  $$
  S = \{k\in\mathbb{Z}: na < k < nb\}.
  $$
Since $m\in S$, we know that $S$ is nonempty. Since $a\ge 0$, we know that $S\subset \mathbb{N}$. Hence, by the well-ordering principle (every nonempty set of positive integers has a smallest element), we know that $S$ has a smallest element. Call it $m_0$.
Since $m_0\in S$, we know that $na < m_0 < nb$. But since $m_0$ is the smallest element in $S$, we know that $m_0-1\notin S$, and this implies $m_0-1\le na$. Thus, $m_0-1\le na<m_0$, and we may proceed with the rest of the proof.
I would guess that this is what the author(s) meant.
