On Dec 27, 9:57 am, WM <mueck...@rz.fh-augsburg.de> wrote: > On 22 Dez., 15:28, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > > > > Even a matheologian should understand that: If there is no digit at a > > > finite place up to that the sequence 0.333... identifies the number > > > 1/3, then there is no digit at a finite place up to that the number > > > 1/3 can be identified. > > > Right, there is no digit at a finite place up to that the number 1/3 can be > > identified. And as there are no digits at infinite places that appears to > > you to be a paradox. It is not. There is *no* finite sequence of digits > > that identifies 1/3. But there is an *infinite* sequence of digits that > > does so. > > Why not pronounce it a bit clearer? We know that every digit is the > last digit of a finite initial segment of digits. There is no other > digit, is it? > Therefore your statement can be expressed in the following way: There > is no finite initial segment that identifies 1/3. But there is an > infinite finite initial segment that identifies 1/3.
So you consider "infinite finite initial segment" as a clearer but equivalent form of "infinite sequence of digits". I guess you don't know what a sequence is. Or what infinite means. I'm going to give you the benefit of the doubt on "digits."
> If you read this sentence, perhaps you get an impression of what > matheology is?
Yes! It is your projection of your own inability to think clearly onto others.
> > > I said, if there is a sequence that identifies > > > 1/3, then the identifying digits must be at finite places. > > > Right, all identifying digits (there are infinitely many) are at finite > > places. > > Therefore your statement can be expressed in the following way: There > is no finite initial segment that identifies 1/3. But there is an > infinite finite initial segment that identifies 1/3.
Every time you use the phrase "infinite finite" you look stupid.
If only you were as dumb as you are stupid!
> > > Does the sequence of 1/3 not consist of a union of all finite > > > initial segments? > > > It is, but also (according to *your* definition) it is not a path. > > There can be no question: The union of all finite initial segments is > finite.
You have found another way to say "infinite finite" and hence another way to look stupid.