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

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David C. Ullrich

Posts: 3,555
Registered: 12/13/04
Re: Proving archimedian property from "dedekind complete + orderer field"
Posted: Jul 28, 2013 11:13 AM
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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}"

We need to prove this for the same reason as we need to prove anything
else. One typically defines the reals to be "the" least-upper-bound
complete ordered field, and we need to know that the natural
numbers are not bounded above in the reals.

>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).

Yes, this is obvious without all the rest of that. But this doesn't
say that the natural numbers are unbounded _in the reals_!

It's obvious that

(1) there is no natural number N such that
n<= N for every natural number n,

for exactly the reason you give. But what we
want to show is

(2) there is no real number x such that
n<=x for every natural number n.

Possibly it seems to you that (2) is clear from
(1). If not, skip the next paragraph or so. If
you _do_ think that (2) is obvious from (1),
that's because when you dealing with things
you've dealt with all your life and you're
suddenly trying to prove things from axioms,
it's hard to forget all the obvious things you

Anyway, here's the obvious proof of (2) from
(1): Suppose that x is as in (2). Choose a
natural number N with N >= x. Then
N is as in (1). So if there is no N as in (1)
then there is no x as in (2), qed.

That's totally bogus. Given x, how do you know
there exists a natural N with N >= x? If you
think about it, you see this is precisely because
(2) holds!

So. It's obvious that (2) follows from (1), except
that the proof already _assumes_ (2).

>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?
>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?

I take it you know the actual proof of (2): Assuming x is a real upper
bound for the naturals, it follows that x-1 is also a real upper
bound for the naturals. So the naturals cannot have a least
upper bound, hence they cannot have a real upper bound.

The proof of uboundedness of the naturals in the rationals
is totally different, purely algebraic. Say r is a rational
uppper bound for the naturals. Say r = p/q, where p
and q are natural numbers. Then

p+1 < p/q

p + 1 <= (p+1)q < p,


>Thanks for the help.

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