Re: A question on GIT.

From: peter_douglass (baisly_at_gis.net)
Date: 09/09/04 

Date: Thu, 09 Sep 2004 13:05:35 GMT

"Herman Jurjus" wrote in message
news:2qan75Ftaq0lU1@uni-berlin.de...
> peter_douglass wrote:
> > "Herman Jurjus" wrote in message

> >>But what does it mean that Presburger Arithmetic is consistent?
> >>That FOL will not prove contradiction from it, in finitely many steps?
> >>OK. Please define 'finitely many steps', without somehow using the
> >>intuitive notion of N that we all currently use.

> > You may not like this, but use the notion of N that is in
> > Presburger Arithmetic. Do you need multiplication to do
> > induction?

You didn't answer this question. Do you need multiplication
to do induction?

> >>Now our intuitive N obviously satisfies PA. So i still say that, if PA
> >>is inconsistent (again, bizar, but...), then apparantly we have the
> >>wrong idea about N. So, imo, there would then be lots of theorems that
i
> >>would no longer trust, or that would become simply meaningless.
> >>Decidability and consistency of Presburger Arithmetic included.

> > If there is a sense in which we can speak of intuitive
> > notion "satisfying" a formalized theory, then I think
> > our intuitive N also satisfies Presburger Arithmetic.
> [snip]
> > Perhaps you should explain what it means for an
> > intuitive notion "satisfying" a theory.

> Well, apparantly you know what it means for your intuitive N to
> satisfy Presburger Arithmetic.

Not really. And I don't think the discussion will be fruitful
if we stay at the level of intuitive notions.

> Is there any axiom of PA that you think is false
> in 'your' picture of N, in the same sense?

There are axioms of PA that cannot be proven in Presburger Arithmetic:

(mult0) forall x . 0x = 0
(mult) forall x,y . x(y+1) = xy + x

These axioms can be found at
http://www.thoralf.uwaterloo.ca/htdocs/scav/fo_arith/fo_arith.html
but surprisingly are not found in
http://mathworld.wolfram.com/PeanosAxioms.html
or in
http://en.wikipedia.org/wiki/Peano_axioms#The_axioms
which gives the above "definition" of multiplication, but
does not describe it as axioms.

> Now then, if PA is inconsistent (i'm not saying that it is, i say
> _if_), wouldn't that mean that N, as you currently envision it,
> is an illusion?

How I envision things, and whether my intuitions are illusions
are not mathematically important.

> How would you deal with that? Would you throw away PA and keep
> Presburger Arithmetic? But then the only models of your theory
> would be non-standard, and would necessarily contain non-standard
> numbers.

What do you mean by "the only models of your theory would
be non-standard?" What do you mean by non-standard numbers?

>Because the standard picture, with all the finite numbers
> in them (no matter how astronomically large), but none of the
> infinite numbers, would then be no longer existing.

Um, there are integers in Presburger arithmetic. They are
all finite, but there is an infinite number of them. Would these
numbers "be no longer existing"?

> Now imagine what that would do to the _meaning_ of
> notions like 'fol-proof', 'consistent', 'recursively decidable'.
> All of these notions refer crucially to natural numbers,

Do they all involve multipication?

> and especially to our realistic picture of the
> 'intended model of PA'. (Only finite proofs are
> allowed, only calculations with finitely
> many steps are allowed. In all these cases,
> unfeasibly large numbers are included, but
> infinite numbers (non-standard numbers), are
> excluded.)

> The difficulty, of course, is that it is very hard to
> imagine any mathematics in which the natural
> number sequence does not exist, or is not unique.
> It's like assuming a contradiction and discussing
> the consequences.

What has all this to do with multiplication?

--PeterD

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