Archimedean ordered field

Let $F$ be an ordered field (Ordered field). For a natural number $n\ge 1$, write $n\cdot 1_F:=\underset{n}{\underbrace{1_F+\cdots +1_F}}$ 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

$x<n\cdot 1_F\mathrm{.}$

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 $1\mathrm{/}\epsilon$): for every $\epsilon >0$ in $F$ there is $n$ with $1\mathrm{/}(n\cdot 1_F)<\epsilon$, so the canonical fractions $1\mathrm{/}n$ 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).