Date: Mar 26, 2013 5:52 PM
Author: Virgil
Subject: Re: Matheology � 224
In article

<03432ee8-3bbe-4c84-8179-490542b4bd53@r1g2000yql.googlegroups.com>,

WM <mueckenh@rz.fh-augsburg.de> wrote:

> 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 by

> induction runs as I said:

That WM said something is very nearly proof that it is false.

And induction requires only an inductive set, as I described above,

which need not be the set of positive naturals.

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