```Date: Mar 26, 2013 5:29 PM
Author: mueckenh@rz.fh-augsburg.de
Subject: Re: Matheology § 224

On 26 Mrz., 22:07, Virgil <vir...@ligriv.com> wrote:> > > One acceptable form of induction is:>> > > There exists a set of objects, N, and a zero object, 0, such that> > > 1. 0 is a member of N.> > > 2. Every member of N has a successor object in N.> > > 3. 0 is not the successor object of any object in N.> > > 4. If the successors of two objects in N are the same,> > > then the two original objects are the same.> > > 5. If a set, S, contains 0 and the successor object of every> > > object in S, then S contains N as a subset.>> > That is a definition of a sequence, not a proof by induction. It is> > not even a definition of the natural numbers, because even the ordered> > set> > N = (0, pi, pi^2, pi^3, ...)> > obeys your five points.>> Mathematical Induction does not require use of natural numbers, but only> of a set which is as inductive as (is order-isomorphic to) the set of> natural numbers, and my form satisfies that requirement.Your mess is not induction and is obviously not a proof. A proof byinduction runs as I said:> > > > Let P(1)> > > > and let P(x) ==> P(x+1)>> > > > Then P(n) at least for every natural number.Regards, WM
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