Proof that $\mathbb{Q}$ is dense in $\mathbb{R}$ I'm looking at a proof that $\mathbb{Q}$ is dense in $\mathbb{R}$, using only the Archimedean Property of $\mathbb{R}$ and basic properties of ordered fields.
One step asserts that for any $n \in \mathbb{N}$, $x \in \mathbb{R}$, there is an integer $m$ such that $m - 1 \leq nx < m$.  Why is this true?  (Ideally, this fact can be shown using only the Archimedean property of $\mathbb{R}$ and basic properties of ordered fields...)
 A: Assume first that $x>0$, so that $nx>0$. By the Archimedean property there is a $k\in\mathbb{N}$ such that $k>nx$; let $m$ be the least such $k$. Clearly $m-1\le nx<m$. If $x=0$, just take $m=1$. Finally, if $x<0$, then $-nx>0$, so by the first part of the argument there is an integer $k$ such that $k-1\le -nx<k$, and hence $-k<nx\le 1-k$. If $nx\ne 1-k$, you’re done: just take $m=1-k$. If $nx=1-k$, take $m=2-k$.
A: If $\mathbb{Q}$ is not dense in $\mathbb{R}$, then there are two members $x,y\in\mathbb{R}$ such that no member of $\mathbb{Q}$ is between them.  I claim that the distance $\varepsilon=|x-y|$ between $x$ and $y$ is an infinitesimal.  By the Archimedean property, this implies $\varepsilon=0$.
If
$$
\underbrace{\varepsilon+\cdots+\varepsilon}_{n\text{ terms}} > 1
$$
then $\varepsilon>1/n$, so some rational number of the form $k/n$ is between $x$ and $y$.
A: You mention in comments that your version of the Archimedean property is

For all $x,y\in\mathbb{R}$ with $x\gt 0$, there exists $n\in\mathbb{N}$ such that $nx\gt y$.

Let $z\in\mathbb{R}$. Assume first that $z\gt 0$. Now, using the Archimedean property with $y=z$ and $x=1$, it follows that there exists $n\in\mathbb{N}$ such that $n\gt z$. Thus, the set
$$\{ n\in\mathbb{N}\mid n\gt z\}$$
is nonempty. By the well-ordering principle, there is a least natural number $n_0$ such that $n_0\gt z$. Then $n_0\gt z$; if $n_0=1$, then $0\lt z\lt 1$ and $m=1$ works. If $n_0\gt 1$, then $n_0-1\in\mathbb{N}$, and minimality of $n_0$ means that $n_0-1\leq z$, Thus, $n_0-1\leq z\lt n_0$ and $m=n_0$ works again.
If $z=0$, take $m=1$.
If $z\lt 0$, then let $w=-z$. Then there exists, by the previous case, a nonnegative integer $k$ such that $k-1\leq w \lt k$. Therefore, $-k\lt z\leq 1-k$. If $z\lt 1-k$, then $m=1-k$ does the trick. If $z=1-k$, then $1-k\leq z\lt 2-k$, so $m=2-k$ works.
