Density of $\mathbb{Q}$ in $\mathbb{R}$ proof technicality I'm reading a proof of the density of $\mathbb{Q}$ in $\mathbb{R}$ and I was held up by a step in it (in Abbott's intro analysis book). Here's the proof up to the part in question:

Theorem: For every two real numbers $a$ and $b$ with $a < b$, there exists a rational number $r$ satisfying $a<r<b$.

Proof: We want to select $m \in \mathbb{Z}$ and $n \in \mathbb{N}$ such that $$a < m/n < b. \tag{1}$$ We may pick $n$ such that $1/n < b-a$ because of the Archimedean Property ($\mathbb{N}$ is unbounded in $\mathbb{R}$). Since $(1)$ is equivalent to $na < m < nb$, our goal is to find the smallest integer $m$ such that $na < m$; i.e., we may choose $m$ such that $m-1 \leq na < m$. [rest of proof shows this $m$ satisfies $(1)$]
The author seems to take for granted that we are allowed to choose this $m$. However, isn't that a major point of the proof we're supposed to show? Can someone explain why such a choice is allowed? 
 A: It is a fact of the natural numbers that if there is a natural satisfying some property, then there is a least such natural. This is known as the well-ordering theorem. Since the naturals are Archimidean, there is $m$ such that $n a < m$; so by the well-ordering theorem there is a least such $m$.
A: There is a property of the natural numbers, called the "well-ordering principle," which says any non-empty set of natural numbers has a least element.
It can be proven by induction. 
If $S$ is a set of natural numbers without a least element, then let $P(n)$ be the statement that $S$ does not contain any natural number less than $n$. Then $P(1)$ is true (or $P(0)$, depending on whether $0\in\mathbb N$,) and if $P(n)$ is true, then show $P(n+1)$ is true.
Implicitly, you'll need to show that there is no natural number between $n$ and $n+1$ to get the above step. That can also be proved by induction.
So any set of natural numbers without a least element is empty.
This is why it was important to first mention there is some $m$ with $m>na$ - you can't prove there is a least such natural number if you can't show there exists at least on such natural number.
