Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Matheology § 190
Replies: 15   Last Post: Jan 14, 2013 3:35 PM

 Messages: [ Previous | Next ]
 mueckenh@rz.fh-augsburg.de Posts: 18,076 Registered: 1/29/05
Re: Matheology § 190
Posted: Jan 12, 2013 4:05 AM

On 12 Jan., 09:45, Zuhair <zaljo...@gmail.com> wrote:
> On Jan 12, 11:07 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>

> > Matheology § 190
>
> > The Binary Tree can be constructed by aleph_0 finite paths.
>
> >         0
> >       1, 2
> >   3, 4, 5, 6
> > 7, ...

>
> > But wait! The Binary Tree has aleph_0 levels. At each level the number
> > of nodes doubles. We start with the (empty) finite path at level 0 and
> > get 2^(n+1) - 1 finite paths within the first n levels. The number of
> > all levels of the Binary Tree is called aleph_0. That results in
> > 2^(aleph_0 + 1) - 1 = 2^aleph_0 finite paths.

>
> NO. We get (2^(n+1)) -1 PATHS within the first n levels, This of
> course INCLUDES paths at level n itself.
>
> At level 0 the number of PATHS is 2^(1) -1 which is 1 that is the
> empty path
> At level 1 the number of PATHS is 2^2 -1 which is 3 this INCLUDES
> paths at level 0 and those at level 1 ITSELF! which are the empty path
> (at level 0), and 0-1, 0-1 paths at level 1.
>
> Now the number of ALL levels of the Binary Tree is called aleph_0,
> correct.
>
> But the result is The number of ALL *PATHS* up to Aleph_0

aleph_0 is not a natural number. Therefore I carefully excluded any
reference to last levels. And this is not required in mathematics.

> (and
> INCLUDING PATHS at level Aleph_0 OF COURSE: The later ones are the
> INFINITE paths)

You are in error. There are, according to set theory, aleph_0 finite
natural numbers. aleph_0 is not a natural number.

Instead of paths you could also say nodes, because there is a
bijection between finite paths and nodes. Do you defend the opinion
that there must be nodes at an infinite level in order to collect
infinitely many nodes? (This would be a sober position, but I doubt
that it is your position.)

> is (2^(Aleph_0 +1)) -1 which is 2^Aleph_0, this
> includes both FINITE as WELL as INFINITE paths of the complete
> infinite binary tree and NOT just finite paths as you are deceived by.

And you think that in the proof of Nicole d'Oresme also infinite
natural numbers play a role? You are in error. The sum of finite unit
fractions already surpasses any given number.

Regards, WM