In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> Am Donnerstag, 5. Dezember 2013 23:23:52 UTC+1 schrieb Zeit Geist: > > > > > > We need only show that for each Rational Number, r, there is One > > > > particular FIS(d_n) such that we have r ~= FIS(d_n). This is a > > > > Sufficient Condition to conclude that r =~ FIS(d_n). > > > > So your "logic" says. > > 1) There are all real numbers. All are accessible (can be taken). > 2) It is impossible to distinguish a real number d by its digits from all > other real numbers. (Proof by the fact that every FIS(d_n) = d_1, ..., d_n is > shared by uncountably many real numbers.)
As a proof, it stinks. One can, for example, distinguish the real number 0 from all other real numbers by its digits. A real number is not 0 if and only if it has a non-zero digit in its decimal ( or binary, or any other base) expansion.
That one exception to WM's claimed rule proves that his argument is false and his rule false.
> Can you please prove that you can "take" all r e Q?
Which members of Q does WM wish to claim one cannot take?
> This includes to be able > to take all n e N. In my opinion nobody has ever proven that but everybody > has been restricted to the case of very few numbers.
If one cites a general rule about natuals having no exceptions, WM is now claiming that there still must be exceptions.
If I state that every natural number has a unique successor natural number, WM must now prove that that does NOT hold for all of |N.
> Otherwise I must say: You are cheating.
WM should be a better judge of cheating than that since he is the foremost practitioner of it here. --