In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 11 Dez., 19:53, Marshall <marshall.spi...@gmail.com> wrote: > > On Dec 11, 9:13 am, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > > > > > > > > On 11 Dez., 03:50, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > > > > > > > > > > > > If there are no infinite paths in that tree, 1/3 is not in that > > > > tree. > > > > > > > > > > 1/3 does not exist as a path. But everything you can ask for will be > > > > > found in the tree. > > > > > Everything of that kind is in the tree. > > > > > > This makes no sense. Every path in the tree (if all paths are finite) > > > > is > > > > a rational with a power of 2 as the denominator. So 1/3 does not exist > > > > as a path. In what way does it exist in the tree? > > > > > It exists in that fundamentally arithmetical way: You can find every > > > bit of it in my binary tree constructed from finite paths only. You > > > will fail to point to a digit of 1/3 that is missing in my tree. > > > Therefore I claim that every number that exists is in the tree. > > > > This argument can be inverted to "prove" the existence of > > a natural number whose decimal expansion is an infinite > > string of 3s. > > No. Why should it?
Why shouldn't it? It would make at least as much sense as your arguments.
> Of course the magnitude of natural numbers is not > limited. For every number with n digits 333...333 there is another > number with n^n digits. Nevertheless each one is finite.
But then, since all your paths are finite, they do not represent all reals, which can have non-finite representations. > > > The infinite-3 number exists in a fundamentally > > arithmetical way: you can find every digit of it in a preceding > > natural number constructed from finite string of 3s only. > > You will fail to point to a digit of ...333 that is missing in > > the natural numbers. > > Of course. That is because there is no digit missing in the natural > numbers. > > > Therefore you claim the naturals and > > the reals are just the same, but in the reverse direction, > > just as AP says they are. > > They *are* just the same, because your argument that above procedure > would prove an infinite string of 3's is wrong.
If the string of binary digits's in a path has a last position, then it does not represent 1/3. Then 1/3 can only be represented by an actually infinite set of such strings, at which point actual infiniteness is established and all of WM's objections to it fail.
> There is neither a > natural nor a rational with an infinite string of digits.
There is no finite string of binary digit representing 1/3. Or any rational not of form m/2^n for m an integer and n a natural.
> To be able > to determine every digit of a number you like does not imply that > there is a number with a never ending sequence of digits.
I like 1/3 which, in any base not divisible by 3, requires a never ending sequence of digits to represent exactly. > > Why should it??? Because Cantor believed that God knows infinite > strings? (He read it in civitate dei of St. Augustinus.)
Because it does. > > Or because Zermelo misunderstood Bolzana-Dedekind's definition of > infinity? Are you really thinking that infinity comes into being > because a Dr. Zermelo of Germany said so?
Does WM think he is more in touch with God than any of the many mathematical geniuses whom he trashes here?