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Re: Induction is Wrong
Posted:
Nov 30, 2012 12:44 PM


Dan Christensen wrote: > > 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 UTC8, 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 subtheory of Peano Arithmetic (PA). > > > > > > > > If we add an induction axiom (IND) to the axioms of Ring Theory (RT) > > > > > > > > then > > > > > > > > GA is also a subtheory 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 noncommutative rings. > > > > > > > > There are even finite noncommutative 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 subring 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, ...
For a ring you'll need 1, (1+1), (1+1+1), ... as well.
Do all your rings have a multiplicative neutral element?
 When a true genius appears in the world, you may know him by this sign, that the dunces are all in confederacy against him. Jonathan Swift: Thoughts on Various Subjects, Moral and Diverting



