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}
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!!