```Date: Nov 30, 2012 10:50 AM
Author: Dan Christensen
Subject: Re: Induction is Wrong

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:>>>>>>>>>> > 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>> > > > > 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>> > > > > of four>> > > > > squares, is a theorem of GA. Since the four square theorem is not true>> > > > > in the>> > > > > 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.>> > > > > There are even finite non-commutative rings:http://answers.yahoo.com/question/index?qid=20090827201012AAD7qJg>> > > > > Induction is wrong. It proves multiplication,>> > > > > as defined by the axioms of ring theory,>> > > > > 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>> > > > Download my DC Proof 2.0 software athttp://www.dcproof.com>> > > 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> Download my DC Proof 2.0 software athttp://www.dcproof.com
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