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...)

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.
