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I have two questions regarding the Archimedean Property:

  1. I understand that for all $x \in \mathbb{C}$, there exists a positive $n \in \mathbb{N}$ such that $\vert nx \vert >1$ (Please correct me if I'm wrong). So does this mean that the complex number field satisfies the Archimedean property?

  2. This is regarding ordered fields. Can it be stated that, WLOG, all ordered fields have the Archimedean property if they do not contain infinitesimal (or infinite) elements, e.g. the Rationals?

Edited: Thanks.

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  • $\begingroup$ You have two distinctly different questions here! You realize that the complex number is NOT an ordered field, right? $\endgroup$ – user247327 Apr 9 '18 at 21:22
  • $\begingroup$ I have made an edit, thanks for your comment $\endgroup$ – T J. Kim Apr 9 '18 at 21:37
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    $\begingroup$ Part 2: yes, an ordered field is archimedean iff it contains no infinitesimals. $\endgroup$ – Rob Arthan Apr 9 '18 at 22:11
  • $\begingroup$ Thank you @RobArthan $\endgroup$ – T J. Kim Apr 9 '18 at 23:56
  • $\begingroup$ There are two distinct aspects of the word non-Archimedean: ordered fields and valued fields. For details see: math.stackexchange.com/q/349934 $\endgroup$ – Chilote Sep 26 '18 at 2:34
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By restricting for $z \in \mathbb C; z \ne 0$ that $|z|$ is such that there is an $n \in \mathbb N$ so that $n|z| > 1$ then you are only considering $|z| \in \mathbb R$. So this has absolutely nothing to do with complex numbers.

To have archimedian property have an sense it must be applied to an ordered set which $\mathbb C$ or $\mathbb R^n$ are not. But we can apply it to the norms of these spaces as the norms are all non-negative real numbers. But this is not considering anything other than $\mathbb R$.

"Can it be stated that, WLOG, all ordered fields have the Archimedean property"

As $\mathbb Q$ is a subfield of every ordered field and the rationals have the archemedian principal. [if $0< q < w < r$ and $q,r\in \mathbb Q$ but $w\not \in \mathbb Q$, then $n*q < n*w < n*r$ so if $1 < n*q$ then $1 < n*w$.

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  • $\begingroup$ Thanks. Is having the Archimedean principal the same as having the Archimedean property? $\endgroup$ – T J. Kim Apr 10 '18 at 23:16
  • $\begingroup$ Um, one or the other was probably me being casual and imprecise with language. I meant them to be the same but which, if either, term is the correct one, I'll have to look up. $\endgroup$ – fleablood Apr 11 '18 at 0:09
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    $\begingroup$ A breif google search: Archimedean Property is the more common and preferred term. However calling it the Archimedean Principal is not unheard of (Proof Wiki did so). However the Archimedes' Principal is a physical law that an object, is buoyed up by a force equal to the weight of the fluid displaced by the object. ... So... I should have been more careful. Archimedean property is the correct usage. $\endgroup$ – fleablood Apr 11 '18 at 0:14
  • $\begingroup$ I appreciate that, thanks. $\endgroup$ – T J. Kim Apr 11 '18 at 2:36
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It makes no sense to ask if a field satisfies the Archimedian property unless it is an ordered field. And $\mathbb C$ cannot became an ordered field (basically because $i^2=-1$).

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  • $\begingroup$ I see, does that mean the complex numbers do not have this property? $\endgroup$ – T J. Kim Apr 9 '18 at 21:33
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    $\begingroup$ @TJ.Kim No, it means that it makes no sense to ask whether or not it has this property. $\endgroup$ – José Carlos Santos Apr 9 '18 at 21:35
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    $\begingroup$ Notice the property you are refering is not about $z$ but about $|z|$. And $|z|$ is a real number. So, no, the Complex numbers do not have this property but the absolute values of the complex numbers do. $\endgroup$ – fleablood Apr 9 '18 at 21:42

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