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 by
induction 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