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The Archimedean property for $\mathbb R$ follows from the axioms of complete ordered field.

$\mathbb Q$ is often characterized by stating it to be the minimal ordered field (with $0\ne 1$). But this (or any other equivalent axiomatic characterization, which pins it down to an isomorphism like the Peano axioms pin down $\mathbb N$) is rarely used for proving the Archimedean property for $\mathbb Q$. What is used (like here) is that they are formal quotients of integers, which is just a model of $\mathbb Q$.

I insist on proving Archimedean property for $\mathbb Q$ directly from its axiomatic characterization since particular models might in addition contains junk theorems (like the common set theoretic model of $\mathbb N$ requires $2\subset 3$).

Eric Wofsey
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Well, this depends on exactly what you mean by "the minimal ordered field". Let's say that we axiomatically define $\mathbb{Q}$ as an ordered field which contains no proper subfields.

Now, given such an ordered field $\mathbb{Q}$, let $K\subseteq\mathbb{Q}$ be the set of elements that can be written as a quotient of two integers (where we identify integers with elements of $\mathbb{Q}$ by taking sums of copies of $1$ or $-1$). Then it is easy to verify that $K$ is a subfield of $\mathbb{Q}$, and thus $K=\mathbb{Q}$. So for any $x\in\mathbb{Q}$, there are integers $a$ and $b$ such that $x=\frac{a}{b}$. Replacing $a$ and $b$ with their negatives if necessary, we may assume $b>0$. Since $b$ is an integer, this implies $b\geq 1$ (since $b$ is a sum of copies of $1$ and $1>0$). Thus $x=\frac{a}{b}\leq \frac{a}{1}=a$. Since $x\in\mathbb{Q}$ was arbitrary and $a$ is an integer, this proves $\mathbb{Q}$ is Archimedean.

Eric Wofsey
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