DRMARJOHN
Posts:
116
From:
ROCHESTER NY
Registered:
4/28/10


Re: Matheology � 258
Posted:
May 10, 2013 10:36 AM


> In article > <7130e63b0e2e4a6c987c0f249c3a9e9a@dl10g2000vbb.goo > glegroups.com>, > WM <mueckenh@rz.fhaugsburg.de> wrote: > > > On 6 Mai, 17:31, Dan <dan.ms.ch...@gmail.com> > wrote: > > > > > The list is of nonterminating 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 > fifth grader . > > > > The fact that both, reals and naturals, are in the > list and can be > > used according to my taste should be obvious to > such a grader too. > > 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 > bridges that lead to > > 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 5313), 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 + (1A^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, (1A^2) = .36, and also .6^2 = .36, (1A^2) =.64. There is an A that yields a (1A^2) where the (1A^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= [1A(1)^n]? If so, then there is a possible solution.

