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Topic: Matheology � 258
Replies: 53   Last Post: May 11, 2013 10:07 PM

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 DRMARJOHN Posts: 116 From: ROCHESTER NY Registered: 4/28/10
Re: Matheology � 258
Posted: May 10, 2013 10:36 AM

> In article
> <7130e63b-0e2e-4a6c-987c-0f249c3a9e9a@dl10g2000vbb.goo
> glegroups.com>,
> WM <mueckenh@rz.fh-augsburg.de> wrote:
>

> > On 6 Mai, 17:31, Dan <dan.ms.ch...@gmail.com>
> wrote:
> >
> > > The list is of non-terminating decimals (actual
> real numbers) .
> >
> > Their indices are natural numbers.
> > The indices of the digits 1 are finite initial

> sequences of real
> > numbers.
> > Do you agree?

>
> With such nonsense? Never!

> >
> > > What you need isn't what the argument is . The
> fact that the
> > > asymmetry  is blatant could be made evident to a
> >
> > The fact that both, reals and naturals, are in the

> list and can be
> > used according to my taste should be obvious to
>
> Except that WM's "Taste" only runs to misuse of them.
>
> Or

> > do you think that only men of genious can recognize
> that principle?
>
> It would take more than genius to find any prnciple
> in that.

> >
> > > And , for one
> > > who knows to walk them properly, it's those

>
> The levels of mathematics that WM can never reach.
> --
>
>

May I suggest comments that do not make people feel diminished? Fifth graders may read this, and think this is a model for reacting to other fifth graders. Another question can be: is this useful?
Neuroscientist Rex Jung, an expert in the psychology of creativity (Onbeing.org 5-3-13), makes a distinction between novel and novel and also useful. Novel ideas can be evaluated for their usefulness. One way to examine usefulness is to reduce an issue to a simple and basic principle, with an obvious example. As an illustration: consider, can A^n + B^n = 1 be possible, with A&B < 1, for prime n? A novel approach transforms the equation to A^n + (1-A^n) = 1, with A always rational. This demands an illustration: rational A (0,1), beginning with [.1, .9]. There is an A(1) of .8 and an A(2) of .6 such that .8^2 = .64, (1-A^2) = .36, and also .6^2 = .36, (1-A^2) =.64. There is an A that yields a (1-A^2) where the (1-A^2) is also the product of another A. It is possible for A^2 + B^2 = 1 to have a rational solution. This illustrates a novel principle: when there are two A's such that that A(2)^2= [1- A(1)^2], then there is a solution. For prime n>2, the novel question is: are there two rational A's such that A(2)^n= [1-A(1)^n]? If so, then there is a possible solution.