Re: CHAT: mathematics

John Cowan <cowan@...> wrote:
>
> On Sat, 18 Nov 2000, Dennis Paul Himes wrote:
>
> >     Goedel proved that a consistent theory's G is neither provable nor
> > refutable in that theory (and therefore, by Goedel's Completeness
> > Theorem, neither true nor false).
>
> Right, but that says nothing about the provability of ~G.
    Yes, it does.  "G is refutable" and "~G is provable" are just two
different ways of saying the same thing.

> > If G is refutable in, say, Peano Arithmetic, then
> > PA would be inconsistent, which is a possibility, but one which no
> > mathematician seriously entertains.
>
> No, it would only be omega-inconsistent, which is quite different.
    No, it would be outright inconsistent.  This is one of the things
Goedel proved.

> >     I don't understand this.  Are you saying that "nonstandard number
> > theory" proceeds from the assumption that PA is inconsistent?
>
> GT shows that G is independent of the other axioms of PA,
    Right (assuming GT means the Goedel Incompleteness Theorem which deals
with a Goedel statement (I can never remember which is the First and which
is the Second)), but this contradicts what you said above.  Specifically it
contradicts your statement that it "says nothing about the provability of
"~G".  "G is independent of the other axioms of PA" directly implies "~G is
not provable in PA".

> GT shows that G is independent of the other axioms of PA, so we can
> assume it (leading to standard number theory) or assume its negation
> (leading to nonstandard number theory).  In the latter case, omega
> inconsistency leads to the realization that 0,1,2,... are not all
> the integers there are, and so there are nonstandard numbers
> in such theories.
    I think you're confusing several results here.  It's certainly true that
there are models of PA which give us only the usual natural numbers, and
other models of PA which give us more, but this has nothing to do with
whether or not you add G to PA.  PA+G is consistent with both sorts of
models, as is PA+~G.

> But if ~G had a proof in PA (and this cannot be ruled out, because
> both G and ~G are consistent with PA),
    This statement directly contradicts itself, unless you are considering
the possibility that PA by itself is inconsistent.  Not being able to rule
out a proof of ~G in PA and G being consistent with PA are both only
possible if PA is inconsistent.

> >     It what sense is analysis or topology "nonstandard number theory"?
>
> No, I mean "nonstandard number theory, nonstandard analysis, nonstandard
> topology, etc.".  Nonstandard number theory carries with it a
> nonstandard theory of everything thing else mathematical, necessarily,
> since all mathematicses are grounded on PA, as you say:
    OK.  I didn't understand what you meant at first.

> > > But if there were an *independent* proof of ~G, ...
> >
> >     ... in what sense are you using "independent" ...
>
> By "independent" I mean a proof which does not depend on Goedel's Proof
> or meta-mathematical reasoning.
    If by "proof of ~G" you mean "proof, in some theory, of that theory's
~G", then we know no such proof is possible.  If by "proof of ~G" you mean
"proof, in some other theory, of some theory's ~G", then I think you're
pretty much committed to metamathematical reasoning.

 ===========================================================================

             Dennis Paul Himes    <>    dennis@himes.connix.com
                  http://www.connix.com/~dennis/dennis.htm

Disclaimer: "True, I talk of dreams; which are the children of an idle
brain, begot of nothing but vain fantasy; which is as thin of substance as
the air."                      - Romeo & Juliet, Act I Scene iv Verse 96-99