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Topic: Matheology § 223: AC and AMS
Replies: 102   Last Post: Apr 18, 2013 12:26 AM

 Messages: [ Previous | Next ]
 mueckenh@rz.fh-augsburg.de Posts: 18,076 Registered: 1/29/05
Re: Matheology § 223: AC and AMS
Posted: Mar 19, 2013 5:40 PM

On 19 Mrz., 20:19, Virgil <vir...@ligriv.com> wrote:
> > You shoud try to get a course in mathematics, if ever possible.
>
> I have taught several different courses in mathematics at college level
> in the US of at least the level of the courses that you claim to teach.

And certainly you also have taught the chinese emporer.
Prove it, big mouth!

>
>  Of
>

> > course not only the rationals strictly between 0 and 1 have distinct
> > representations, but also, in decimal, 1.000... = 0.999... and
> > 1.1000... = 1.0999... and many, many more, namely every "terminating"
> > rational with no regard to its magnitude. Never heard of?

>
> The point is that when two different sequences produce onlyone  real, as
> they do here, that that one real is not then allowed to reproduce the
> two different sequences which are both needed to get all paths.

And your recognition of this point includes only the interval between
0 and 1? Poor boy.
>
> Thus your methodology crates a mapping which, besides being nonlinear,
> cannot be bijective.

Nonsense.
If the mapping is to be between binary sequences and paths, then ther
si no problem at all. And if it is to be between reals and paths, then
the remintaing path could be omitted.But as I said, nothing would
change with respect to cardinalities.
>
> > If you are disturbed by that fact, simply remove all rationals with
> > "terminating" representations.

>

Claimed are either all paths and all binary sequences OR all non-
terminating paths and all reals.
>

> > But again: This will not change anything with respect to cardinality.
> > I declare that there is an isomorphism between all paths of the usual
> > Binary Tree and the binary representations of all real numbers of the
> > unit interval.

>
> There is certainly a bijection between all binary sequences and all
> paths,

Of course.>
>
>

> > And if you are too stupid to understand that definition, then remain
> > what you are, a silly bigmouth claiming to have taught at a college
> > but incapable of proving that claim and most probably as incapable of
> > teaching as of learning.

>

You are even to silly to understand that also utside of the unit
interval every terminating rational has a periodic representation.

Regards, WM

Date Subject Author
3/14/13 Alan Smaill
3/14/13 mueckenh@rz.fh-augsburg.de
3/14/13 Virgil
3/14/13 fom
3/14/13 mueckenh@rz.fh-augsburg.de
3/14/13 fom
3/14/13 mueckenh@rz.fh-augsburg.de
3/14/13 fom
3/15/13 mueckenh@rz.fh-augsburg.de
3/15/13 fom
3/15/13 mueckenh@rz.fh-augsburg.de
3/15/13 Virgil
3/15/13 fom
3/16/13 mueckenh@rz.fh-augsburg.de
3/16/13 fom
3/16/13 mueckenh@rz.fh-augsburg.de
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3/16/13 mueckenh@rz.fh-augsburg.de
3/16/13 Virgil
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3/17/13 Virgil
3/16/13 mueckenh@rz.fh-augsburg.de
3/16/13 Virgil
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3/17/13 Virgil
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3/18/13 Virgil
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