Let’s recall that an ordered field \(K\) is said to be Archimedean if for any \(x \in K\) there exists a natural number \(n\) such that \(n > x\).

The ordered fields \(\mathbb Q\) or \(\mathbb R\) are **Archimedean**. We introduce here the example of an ordered field which is not Archimedean. Let’s consider the field of rational functions

\[\mathbb R(x) = \left\{\frac{S(x)}{T(x)} \ | \ S, T \in \mathbb R[x] \right\}\] For \(f(x)=\frac{S(x)}{T(x)} \in \mathbb R(x)\) we can suppose that the polynomials have a constant polynomial greatest common divisor.

Now we define \(P\) as the set of elements \(f(x)=\frac{S(x)}{T(x)} \in \mathbb R(x)\) in which the leading coefficients of \(S\) and \(T\) have the same sign.

One can verify that the subset \(P \subset \mathbb R(x)\) satisfies following two conditions:

- ORD 1
- Given \(f(x) \in \mathbb R(x)\), we have either \(f(x) \in P\), or \(f(x)=0\), or \(-f(x) \in P\), and these three possibilities are mutually exclusive. In other words, \(\mathbb R(x)\) is the disjoint union of \(P\), \(\{0\}\) and \(-P\).
- ORD 2
- For \(f(x),g(x) \in P\), \(f(x)+g(x)\) and \(f(x)g(x)\) belong to \(P\).

This means that \(P\) is a positive cone of \(\mathbb R(x)\). Hence, \(\mathbb R(x)\) is ordered by the relation

\[f(x) > 0 \Leftrightarrow f(x) \in P.\]

Now let’s consider the rational fraction \(h(x)=\frac{1}{x} \in \mathbb R(x)\). \(h(x)\) is a positive element, i.e. belongs to \(P\). And for any \(n \in \mathbb N\), we have

\[h(x)-n=\frac{1}{x}-n=\frac{x-n}{x} \in P\] as the leading coefficients of \(x\) and \(x-n\) are both equal to \(1\). Therefore, we have \(h(x) > n\) for all \(n \in \mathbb N\), proving that \(\mathbb R(x)\) is not Archimedean.