"Multiplicative" Archimedean property in ordered fields

Question

There following are two equivalent formulations of the Archimedean property of an ordered field $F$:

1. $\mathbb N$ is unbounded in $F$.
2. (Formulation for the ordered group $(F, +)$). For any $x, \varepsilon > 0$ in $F$, there exists an $N\in\mathbb N$ such that $N\varepsilon > x$.

Now, (2) easily implies (using that $(1 + y)^N \ge 1 + Ny$), the following "multiplicative" version:

3. (Formulation for the ordered group $(F^+, \cdot)$). For any $x, \varepsilon > 1$ in $F$, there exists an $N\in\mathbb N$ such that $\varepsilon^N > x$.

Now I wonder if this multiplicative Archimedean property (3) is equivalent to the usual one (1) (or (2)).

The beauty of the potential equivalence makes me believe that it might be the case.

But I will not be shocked to see a counterexample for this, because addition and multiplication in a field are not "the same". They differ on how they distribute, for instance.

Any thoughts?

Answer 1

I’m turning my comment into an answer. Suppose (3) holds. Then, for all $x >1$, there is some $N \geq 1$ such that $2^N > x$ (taking $\epsilon=2$).

Thus $\mathbb{N}$ is unbounded in $F$, ie (1) holds.