Let $p\in 1^*$ with $0 < p < 1$. There exists an $n\in \mathbb N$ such that
$$
p < 1 - \frac 1 {m + 1} = \frac m {m + 1} \tag{1}
$$
for each $m\in \mathbb N$, $m \geq n$.
Let $r\in \alpha, r >0$ and $0 < q < r/n$. There exists an $m$ such that $m q\in \alpha$ and $(m + 1)q\notin \alpha$. Evidently we have $m \geq n$.
Inequality (1) implies
$$
\frac p {mq} < \frac m {m + 1}\cdot \frac 1 {mq} = \frac 1 {(m + 1) q}
$$
so $\frac p {mq} \in \alpha^{-1}$ and
$$
p = (mq)\cdot \frac p {mq}.
$$
The natural numbers can be defined very rigorously using either Zermelo-Fraenkel or Von Neumann-Bernays-Godel Set Theory. Both are axiomatic constructions of almost all of mathematics as we know it.
http://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory
http://en.wikipedia.org/wiki/Von_Neumann%E2%80%93Bernays%E2%80%93G%C3%B6del_set_theory
The natural numbers are usually defined as $\emptyset=0, \{\emptyset\}=\{0\}=1, \{\{\emptyset\},\emptyset\}=\{1,0\}=2$ and so on. This has the useful property that $n\in n+1$ for any $n\in\mathbb{N}$ and $2=0\cup 1$. We can then define addition inductively in terms of the successor function $0^+=0+1=1$, and $n^+=n+1=(n-1)\cup n=\{0,1,2,\ldots,n-1\}\cup n$.
It takes a bit of work, but you can then prove some things like the successor function (which is basically $n^+=n+1$) is injective.
We then define an equivalence relation ~, as $(a,b)$~$(c,d))$ iff $a+d=b+c$, so given $a,b\in \mathbb{N}$, define $a-b:=\{(c,d)\in\mathbb{N}^2|a+d=c+b\}=:[(a,b)]$. This basically says that if you have something like $5-3$, will will consider that as being equal to $6-4$, since $5+4=6+3$. We can now construct $\mathbb{Z}$.
$\mathbb{Z}:=\{a-b|(a,b)\in\mathbb{N}^2\}$
This says any integer is just the difference of two natural numbers. Now make the equivalence relation ~, as $(a,b)$~$(c,d)$ iff $ad=bc$ and define division, for any $a,b\in\mathbb{Z}, b\neq 0$ as,
$\frac{a}{b}:=\{(c,d)\in\mathbb{Z}|ad=bc \mbox{ and }d\neq 0\}=[(a,b)]$
Again, because we don't have division, we have to define this equivalence class in terms of multiplication (the brother of division) to our definition of division. You'll find as well, that these definitions of division, subtraction, multiplication and addition all behave as you'd expect, and you can prove everything you'd expect you could like commutivity, associtativity, distribution of addition and multiplication, inverses etc.
Finally we'll define $\mathbb{Q}$ as,
$\mathbb{Q}:=\{\frac{a}{b}|a,b\in\mathbb{Z} \mbox{ and } b\neq0\}$
This definition assures that numbers like $\frac{2}{4}=\frac{1}{2}$ by using these equivalence relations and classes.
The reals are then often defined as the completion of $\mathbb{Q}$.
Best Answer
By definition there is some rational $q$ that is not in $\alpha$. Let $p=|q|+1$; then $p>q$, so $p\notin\alpha$, and $p>0$. The rational $w$ is also positive, so by the Archimedean property of the rationals there is a positive integer $m$ such that $mw>p$, and therefore $mw\notin\alpha$.
We also know that there is some rational $r\in\alpha$. Let $s=-|r|-1$; then $s<r$, so $s\in\alpha$, and $-s>0$, so again by the Archimedean property there is a positive integer $k$ such that $kw>-s$. But then $(-k)w<s$, so $(-k)w\in\alpha$. We now have positive integers $k$ and $m$ such that $-kw\in\alpha$ and $mw\notin\alpha$.
Let $S=\{i\in\Bbb N:(i-k)w\notin\alpha\}$; $m+k\in S$, so $S\ne\varnothing$. Since $\Bbb N$ is well-ordered, $S$ has a smallest element, $j$. Note that $0\notin S$, since $-kw\in\alpha$, so $j>0$. Let $n=(j-1)-k$. Then $nw\in\alpha$, and $(n+1)w\notin\alpha$.