On Tuesday, December 10, 2013 2:43:06 PM UTC-7, WM wrote: > Am Dienstag, 10. Dezember 2013 18:53:06 UTC+1 schrieb Zeit Geist: > > On Tuesday, December 10, 2013 4:52:18 AM UTC-7, WM wrote: > > > Am Montag, 9. Dezember 2013 22:01:18 UTC+1 schrieb Zeit Geist: > > > > > > I am not interested in your "proofs" but in defining an irrational number like the antidiagonal d by an arbitrarily large sequence of its digits d_n (including an infinite sequence). This is impossible because for every digits d_n there are infinitely many duplicates. > > > > > You not interested in any proofs at all. You shout Dogma. > > > > The above "proof" is invalid, but you are even too blind to that. > > > > I see that it is in agreement with mathematics: > > >For every finite or infinite sequence of digits of d there is a rational number including it. > > > That can and has been proven. > > > Where has that been proven? > > Proof: Take any digit. For every n: There are infinitely many digits following upon d_n, such that there are infinitely many rational numbers identical with d_1, ..., d_n. Since this holds for *all* infinitely many natural indices, there is no chance to identify 1/n! by any sequence of only finitely indexed digits.
So, all you proved was that each FIS of d has a Rational Equivalence. The conclusion does not follow. What's with the Factorial? How does this show a Rational includes d anyway?
> > What rational number q has the property that: q includes sqrt(2). > > This has been proven False for All rational numbers. > > q is not sqrt(2), but every digit that can be calculated by a sequence with limit sqrt(2) belongs to a rational approximation. >
You are saying the FIS of d, the approximations of d, are Rational Approximations. Well, d is Not an approximation, it is d. If d is Not Rational then no Finite sequence of Digits describes d.
> > > > The Logic in that above exposition resembles the flawed reasoning you would use. > > > > But it leads to the correct result: It is impossible to distinguish d by its digits from all entries of arationals-complete list. > > > First of all, I have admitted I was mistaken about the contents of that conjecture. > > Also, you have yet to prove this "impossibility to distinguish d by digits...". > > You haven't even defined what this means. > > To define what it means that two numbers differ by digits??? > Sorry, if you don't know then we should stop here. Perhaps that is the reason .... >
Two Real Numbers are Distinguishable by Digits iff There Exists a Digit such they Differ at that Digit. Is that what you mean.
Why not just say Unequal?
Now, tell me what makes this impossible for two Real Numbers.
> > I have argued that a Finite Formula does what is required, but you just disagree and don't give any support for you argument other than "but infinitely many follow". This is Not Sufficient to prove your claim. > > A finite formula does what is required. SUM 1/n! is uniquely defined. Every approximation can be calculated, that means every digit can be calculated. > > Formula ==> digits. But from the digits which all are rational numbers the formula cannot be obtained. >
But most Partial Sums of SUM 1/n! are less than the actual value of SUM 1/n!, and therefor Unequal to it.