Archimedean ordered field

Let $F$ be an ordered field (Ordered field). For a natural number $n \ge 1$, write $n \cdot 1_F := \underbrace{1_F + \cdots + 1_F}_{n}$ for the $n$-fold sum of the multiplicative identity, and $0 \cdot 1_F := 0$. These are the canonical natural numbers of $F$.

$F$ is Archimedean if for every $x \in F$ there is a natural number $n$ with

$x < n \cdot 1_F.$

Equivalently, the canonical naturals $(n \cdot 1_F)_{n \ge 1}$ are cofinal: no single element of $F$ is an upper bound for all of them.

• Equivalently (applying the definition to $1/\varepsilon$): for every $\varepsilon > 0$ in $F$ there is $n$ with $1/(n \cdot 1_F) < \varepsilon$, so the canonical fractions $1/n$ are arbitrarily small.
• That the canonical naturals are well-defined, positive, and strictly increasing is Canonical naturals are positive and strictly increasing. Every complete ordered field is Archimedean (Every complete ordered field is Archimedean); an ordered field need not be (Not every ordered field is Archimedean).