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2021-04-26, 20:44   #12
jyb

Aug 2005
Seattle, WA

24·3·37 Posts

Quote:
 Originally Posted by Happy5214 Note the difference. The axioms for the rational numbers (an ordered field) are the same as the real numbers (essentially the only Dedekind-complete ordered field) minus the least upper bound property/Dedekind completeness, so if your book has axioms for the real numbers, you can derive the axioms for the rational numbers from those.
This isn't quite correct. The axioms of the reals minus the least upper bound property gives you an ordered field, but it is not sufficient to characterize the rationals. Specifically, it need not be Archimedean, which the rationals are. E.g. see the surreal numbers, for which these axioms apply, but which are certainly not isomorphic to the rationals.

2021-04-26, 21:15   #13
Happy5214

"Alexander"
Nov 2008
The Alamo City

10111001012 Posts

Quote:
 Originally Posted by jyb This isn't quite correct. The axioms of the reals minus the least upper bound property gives you an ordered field, but it is not sufficient to characterize the rationals. Specifically, it need not be Archimedean, which the rationals are. E.g. see the surreal numbers, for which these axioms apply, but which are certainly not isomorphic to the rationals.
Correct. The rationals are the smallest ordered field, in the sense that they are embedded in every other ordered field, so that would be the missing axiom. I couldn't tell you how that implies the Archimedean property (I proved it for the reals in my analysis class, but that proof is based on completeness).

2021-04-27, 08:39   #14
Nick

Dec 2012
The Netherlands

32718 Posts

Quote:
 Originally Posted by Happy5214 I couldn't tell you how that implies the Archimedean property
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.

 2021-04-29, 15:35 #15 Dr Sardonicus     Feb 2017 Nowhere 22·3·401 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 "non-Archimedean valuation" (p-adic 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 p-adic rationals (and their extensions) into fields that are (Cauchy) complete WRT a non-Archimedean valuation. Last fiddled with by Dr Sardonicus on 2021-04-29 at 15:35 Reason: xifgin posty

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