20210426, 20:44  #12  
Aug 2005
Seattle, WA
2^{4}×107 Posts 
Quote:


20210426, 21:15  #13  
"Alexander"
Nov 2008
The Alamo City
5×11^{2} Posts 
Quote:


20210427, 08:39  #14 
Dec 2012
The Netherlands
2·839 Posts 
In that case, it's time for our 2nd important general principle of the thread:
given any property, always consider whether the set of all elements with that property is just a subset or a substructure (subgroup, subring, subfield or whatever). Here: all you have to do is show that the set of rational numbers that have the Archimedean property form a subfield. 
20210429, 15:35  #15 
Feb 2017
Nowhere
4544_{10} Posts 
The Archimedean property is that if x and y are in an ordered field, x > 0 and y > 0, there is a positive integer n such that n*x > y.
Assuming the ordered field is the field of rational numbers with the usual ordering, I note that if x and y are positive rational numbers, there is a positive integer M such that M*x and M*y are both positive integers. Then taking n = M*y + 1, we have n*(M*x) >= n*1 = M*y + 1 > M*y, so that n*x > y. If we ignore the ordering, and use instead a "nonArchimedean valuation" (padic valuation), anything dependent on ordering (like "upper bound" or "least upper bound," and therefore "Dedekind completeness") goes out the window. Luckily, "Cauchy completeness" (every Cauchy sequence in the field has a limit in the field) can still be used to embed the padic rationals (and their extensions) into fields that are (Cauchy) complete WRT a nonArchimedean valuation. Last fiddled with by Dr Sardonicus on 20210429 at 15:35 Reason: xifgin posty 
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