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Topic: Induction is Wrong
Replies: 4   Last Post: Nov 30, 2012 12:44 PM

 Messages: [ Previous | Next ]
 Dan Christensen Posts: 8,219 Registered: 7/9/08
Re: Induction is Wrong
Posted: Nov 30, 2012 10:50 AM

On Nov 30, 10:34 am, Dan Christensen <Dan_Christen...@sympatico.ca>
wrote:
> On Nov 30, 12:35 am, Dan Christensen <Dan_Christen...@sympatico.ca>
> wrote:
>
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>

> > On Nov 29, 11:45 pm, reaste...@gmail.com wrote:
>
> > > On Thursday, November 29, 2012 7:59:32 PM UTC-8, Dan Christensen wrote:
> > > > On Nov 29, 9:27 pm, RussellE <reaste...@gmail.com> wrote:
>
> > > > > Andrew Boucher has developed a theory called General Arithmetic (GA):http://www.andrewboucher.com/papers/ga.pdf
>
> > > > > GA is a sub-theory of Peano Arithmetic (PA).
>
> > > > > If we add an induction axiom (IND) to the axioms of Ring Theory (RT)
>
> > > > > then
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> > > > > GA is also a sub-theory of RT+IND. (We also need a weak successor
>
> > > > > axiom).
>
> > > > > Boucher proves Lagrange's four square theorem, every number is the sum
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> > > > > of four
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> > > > > squares, is a theorem of GA. Since the four square theorem is not true
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> > > > > in the
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> > > > > integers, the integers can not be a model for GA, PA, or RT+IND.
>
> > > > > GA also proves multiplication is commutative.
>
> > > > > It is well known there are non-commutative rings.
>
>
> > > > > Induction is wrong. It proves multiplication,
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> > > > > as defined by the axioms of ring theory,
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> > > > > must be commutative when this is not true.
>
> > > > Wrong. The set of integers, along with usual addition and
>
> > > > multiplication functions on the integers can be constructed starting
>
> > > > from Peano's Axioms (including induction) by using the axioms of logic
>
> > > > and set theory.
>
> > > > Dan
>
>
> > > Yes, you can construct the integers in PA,
> > > but the integers can not be the universe of a
> > > model of PA.

>
> > > My point is that induction proves multiplication
> > > must be commutative even when multiplication
> > > is defined with the axioms of ring theory.
> > > Yet, we know multiplication does not have to
> > > be commutative.

>
> > Induction cannot be applied to rings in general if that is what you
> > are getting at. That doesn't mean induction is "wrong."

>
> Hmmmm... I wonder if, within any infinite ring, there does exist an
> infinite sub-ring that is indeed commutative. There does exist within
> it, a smallest subset {1, 1+1, 1+1+1, ...}

That should be the smallest subset containing 1, 1+1, 1+1+1, ...

> on which induction holds
> (previous posting here), and + must therefore be commutative.
>
> Dan