The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.math

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Proving archimedian property from "dedekind complete + orderer field"
Replies: 3   Last Post: Jul 28, 2013 12:55 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
David C. Ullrich

Posts: 3,555
Registered: 12/13/04
Re: Proving archimedian property from "dedekind complete + orderer field"
Posted: Jul 28, 2013 11:29 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On Sun, 28 Jul 2013 03:16:24 -0700 (PDT), Lax Clarke
<> wrote:

>I'm confused as to why the archimedian property (unbounded natural numbers in reals) needs to be proven from {least upper bound property + ordered field}"
>Aren't the naturals "obviously" infinite (meaning provable without all the other machinery)? Because n+1 is bigger than n which are different natural numbers (you can prove n+1 > n > 0 using the order properties of reals).
>Is it my understanding that this proof is intended to exclude any case where the naturals could wrap around on each other like 1+1+1...+1=0 in a finite field?

Decided to answer this in a separate post so as not to disrupt the
flow of the other one.

No, that's really not it. "Ordered field" rules out 1 + ... + 1 = 0,
but "ordered field" is not enough to show that the naturals
are not bounded. There _are_ ordered fields in which the
naturals _are_ bounded. (Of course they cannot be
least-upper-bound complete, by the theorem we're

I suspect that people will mention the "nonstandard
reals", or "hyperreals" as an example. There are much
simpler examples.

For example, let F be the set of all rational functions:
quotients of (real) polynomials. Given a rational function
f, define f > 0 to mean that

there exists a real number x such that f(t) > 0
for all t > x.

It's not hard to show that F is now an ordered field.
Define f(t) = t. Then f is an uppper bound for the
naturals in F.

(The multiplicative identity in F is the constant
function 1, so the "naturals" are the elements of
the form 1 + ... + 1, that is, the constant functions
that take a natural number for their value.)

>Why can't we prove that from the totality properly of the order < ?
>How would one prove that the naturals inside the rationals (without the sup property which is used in the proof I talk about above) is unbounded above?
>Thanks for the help.

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.