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Topic: Answer to Dik T. Winter
Replies: 441   Last Post: Feb 5, 2013 6:25 AM

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Re: Answer to Dik T. Winter
Posted: Jun 12, 2009 10:33 AM

On Jun 12, 7:31 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> On 12 Jun., 09:17, Rainer Rosenthal <r.rosent...@web.de> wrote:
>
>
>
>
>

> > WM schrieb:
>
> > > On 11 Jun., 14:13, Rainer Rosenthal <r.rosent...@web.de> wrote:
> > >> WM wrote:
> > >>> My proof rests upon the fact that after the set of all finite paths of
> > >>> the form 0.1, 0.11, 0.111, ... has been constructed, there is no
> > >>> chance to construct the path 0.111... in addition.

> > >> 0.1, 0.11, 0.111, ...?

>
> > > [Many lines snipped, because I felt they didn't get to the point I
> > > was trying to ask]
> > > However, your contribution to this problem is of high value and I
> > > appreciate it very much.

>
> > Thank you, but it seems as if I wasn't clear in my question.
> > Didn't you say: "If I am going to enumerate the real numbers
> > in such and such a way then there will be some of them not
> > in my list"?

>
> It is not enumerating all real numbers.
> What I do is to use actually infinite paths (at the end we will see
> that they do not exist, but I assume their existence) for constructing
> the complete binary tree. As a simple example I use only such paths
> that have tails of the form 000..., i.e., "terminating" paths.
>
> The result is: They exhaust the complete infinite binary tree. There
> is no node that remains uncovered. And there remains no node and no
> set of nodes that could be used to construct a path that is not yet in
> the tree. The tree is complete at the end. With the tip of your finger
> you can follow every path, even 1/pi or 1/3.
>
>
>

> > But then, that doesn't prove all too much, because it's obvious
> > that there are many ways of not counting the reals.
> > The point is that there is no way of counting the reals.

>
> As I said, I do not count the reals. I cover all possible infinite
> sequences of bits by means of a countable set of paths = a countable
> set of binary sequences.
>
> not belong to the countable set (of "terminating" paths, i.e., such
> with tails 000...) that I use for construction, shows us, that the
> result is in no way different. But how can we explain that the result
> is not influenced by those paths? In my opinion by recognizing that
> such paths do not exist.
>
> Regards, WM- Hide quoted text -
>
> - Show quoted text -

Welcome to the "real" mathematics and not the fake "Matrix" movie
sorry excuse for chess you've been playing for the last 20 years.

Musatov

(Founder, Inverse 19 Mathematics) Not the only, but one without it
would not exist.

QED

Dear Simon,

Efforts by mathematicians to determine the value of pi since have led
to the conclusion that it ... Unlike rational numbers, pi cannot be
represented by a common fraction, ....

But what about an "uncommon fraction"?

Will you look at my drawing and in particular how the twelve points
are the outside points of the "sphere" are all "rationally" reached by
"decimal" numbers?

6.2035884*2 and doubled an infinite number of times should yield a
constant decimal binary expansion (what I mean by this is the numbers
to the right of the decimal point should always end in "4, then 8,
then 6, then 2, then 4, then 8, then 6, then 2...repeating for
infinity).

So in binary, by starting with this number we predict for n>2 the
infinite binary tree will end with decimal progression: 2, 4, 8, 6

For n<2 the infinite binary tree will begin and end (depends on if you
are increasing or decreasing the value) 1, 5, 5, 5 (repeating 5
infinitely) as we decrease to infinitely small.

Musatov Inverse Counting Chain of Infinite Proportions
(sample)
0.0000473372528076171875
0.000094674505615234375
0.00018934901123046875
0.0003786980224609375
0.000757396044921875
0.00151479208984375
0.0030295841796875
0.006059168359375
0.01211833671875
0.0242366734375
0.048473346875
0.09694669375
0.1938933875
0.387786775
0.77557355
1.5511471
3.1022942
6.2035884
12.4071768
24.8143536
49.6287072
99.2574144
198.5148288
397.0296576
794.0593152
1588.1186304
3176.2372608
6352.4745216
12704.9490432
25409.8980864
50819.7961728
101639.5923456
203279.1846912
406558.3693824
813116.7387648
1626233.4775296
3252466.9550592
6504933.9101184
13009867.8202368
26019735.6404736
52039471.2809472
104078942.5618944
208157885.1237888
416315770.2475776
832631540.4951552
1665263080.9903104
3330526161.9806208
6661052323.9612416
13322104647.9224832
26644209295.8449664
53288418591.6899328
106576837183.3798656
213153674366.7597312
426307348733.5194624
852614697467.0389248
1705229394934.0778496
3410458789868.1556992
6820917579736.3113984
13641835159472.6227968
27283670318945.2455936
54567340637890.4911872
109134681275780.9823744
218269362551561.9647488
436538725103123.9294976
873077450206247.8589952
1746154900412495.7179904
3492309800824991.4359808
6984619601649982.8719616
13969239203299965.7439232
27938478406599931.4878464
55876956813199862.9756928
111753913626399725.9513856
223507827252799451.9027712
447015654505598903.8055424
894031309011197807.6110848
1788062618022395615.2221696
3576125236044791230.4443392
7152250472089582460.8886784
14304500944179164921.7773568
28609001888358329843.5547136
57218003776716659687.1094272
114436007553433319374.2188544
228872015106866638748.4377088
457744030213733277496.8754176
915488060427466554993.7508352
1830976120854933109987.5016704
3661952241709866219975.0033408
7323904483419732439950.0066816
14647808966839464879900.0133632
29295617933678929759800.0267264
58591235867357859519600.0534528
117182471734715719039200.1069056
234364943469431438078400.2138112
468729886938862876156800.4276224
937459773877725752313600.8552448
1874919547755451504627201.7104896
3749839095510903009254403.4209792
7499678191021806018508806.8419584
(sample)

If you would like to do the honors Mr. Plouffe, I endear you to
calculate the angles and lengths of the number of ways, it is quite
obviously possible to "square a circle" based on the geometry and
numerical progression.

As is self-evident to anyone with even a modest understanding of
mathematics and geometry after visual inspection of the drawn proof
(http://www.flickr.com/photos/24869933@N07/3619373444/sizes/l/) it is
quite easily accomplished to connect and triangulate the outer points
of the rectangles to the outer points of the square and circle and
even by dimension sphere.

Yes, the drawing clearly by definition of inverse symmetry (take the
drawing cut it in half at the "equator" and turn the paper over and
rotate it and compare the top and bottom "mirrored symmetry" and you
will see it is an exact inverse of symmetry itself) the drawing
explains how to properly and mathematically soundly represent a fourth
dimension on a two dimensional plane.

In plane language if you take the picture and separate the top half of
the sphere by the bottom half of the sphere they are identical but
only if the sphere existed on a complete three dimensional plane
through expansion.

Since the paper is flat, this is an impossibility, nonetheless it is
completely expressed and retractable and countable infinitely.

By the definition of our standard we have established any k-digit
approximation to pi is rational.

Which could have many practical scientific applications. For instance,
search:

ON THE THEOREM OF IVA\$EV-MUSATOV IIIx e F(l) + F(2) + ... + F(p). This
remark is the basis of our proof of Theorem .... -y(M + 1), x' + y(M +
1)] = 0 it is easy to check that Conditions (i), ...... Now set \i =
pi x. *jj, v. Since p, x is positive,. (i)" fi is positive, ...
plms.oxfordjournals.org/cgi/reprint/s3-53/1/143.pdf
by TW Korner - 1986 - Cited by 7 - Related articles

REPRESENTATION OF FUNCTIONSBY SERIES AND CLASSES ?(1)with measure \E\>
2? -- ? and such that the trigonometric series for g(x) converges
uniformly on [0, 2?]. .... trigonometric series are due to
Ivashev-Musatov [40] who has shown that there are null series (in the
sense of ..... function ?(?) e C(0, 1) with ?(0) = ?(1) = 0 such that
...... <*i<Pi(0= ? ?;(0-4-a ...
www.iop.org/EJ/article/0036-0279/27/2/R01/RMS_27_2_R01.pdf
by PL Ul'yanov - 1972 - Cited by 24 - Related articles

Full text of "Bulletin of the British Museum (Natural History)"X 20.
FIG. 5. Well preserved large fragment showing the different aspect and
...... 0-1-0-5 P i n height) so giving the shell surface the
appearance of ...... 94-95, i pi. Maseru. BRINK, A. S. 1963. Two
cynodonts from the Ntawere ...... MUSATOV, D. L, NEMIROVSKAYA, V. N.,
SHIROKOVA, E. V. & ZHURAVLEVA, I. T. 1961. ...
www.archive.org/stream/bulletinofbritis17geollond/bulletinofbritis17geollond_djvu.txt

Trigonometric series with rapidly decreasing coefficients a i0, 5
...Nt,+i - Pi = n,. N tl+j. -pi =N tl+j. -i +pi +1. ..... number 8 6
(0,1) for the product (32) such that for every point x 6 E there are
...
www.iop.org/EJ/article/1064-5616/187/11/A01/MSB_187_11_A01.pdf
by SS Galstyan - 1996 - Related articles - All 7 versions

Plouffe's formula for pi - sci.math | Google Groups3 posts - 2 authors
- Last post: 41 minutes ago
x=pi e^(ipi)+1=0. Martin Musatov. Reply Forward. Martin Musatov to
Robert show details May 3 Reply. Thank you, Sir. No problem. ...
bf6d1ee13d87270b?lnk=raot
- 41 minutes ago

[PDF] ???? ?-?File Format: PDF/Adobe Acrobat
Gli storici ci dicono che, nel X secolo, il principe Vladi- ...... i
pi`u anziani e carismatici si evidenziavano i nomi di ...... tin
Somov, Viktor Borisov-Musatov, Aleksandr Benois e Lev Bakst ... 5/0;
Judovin 5/0; Korovin 2/1; Tukacov 1/0; Goli- cyn 1/1; Zacharov 1/1;
Jurkunas 5/0; Krasauskas 2/0;. Bulaka 3/0. ...
www.esamizdat.it/eSamizdat_2005_(III)_2-3.pdf

I am testing these identities:

(1/2) * ( - k ) * ( k - 1/2 ) * ( - 3 ) / k!

This equation produces

M = k! ( - 3 ) * ( k - 1/2) * ( - k ) * (1/2)

Sincerely,

Martin Musatov

---------- Forwarded message ----------
From: inverse 19 <hope9...@verizon.net>
Date: Apr 11, 7:41 pm
Subject: CHALLENGE TO MATHEMATICIANS OF THE WORLD
To: sci.math

On Apr 11, 11:40 am, "Harris Moran" <inva...@example.com> wrote:

> *plonk*

> "inverse 19" <hope9...@verizon.net> wrote in message

> > muh brane i needs one- Hide quoted text -

> - Show quoted text -

NOTE THE PROGRESSION OF NUMBERS PELLS PROPORTION X^2=3 or 4 = 3 as
the numbers progression, WOW if this does not proof this what will

1. Every odd number , number as Y value till infinity is exactly
solvable with a proportionate result using the base Pells equation i.e
3,5,7,9,etc etc till infinty and the answer is as in 2(2^2)+1=3^2 and
so on till infinity with a whole number integer.

2.Every even number , without exception, using Pells equation, the
integer always is a .75 till infinity consistant

as 63.75(2^4)=1=4^4 till infinity every integer will end with .75.
is that the same proportion as 3 over 4, Y=3 for the least with
numbers proportion

There is a definite pattern by pells to numbers, and I note that in
that pattern the number 19 proportion is strikingly proprtionate,
and I clearly surmise that that circluar progression is the same as
numbers for vector 19 progression.

Date Subject Author
5/27/09 mueckenh@rz.fh-augsburg.de
5/27/09 Dik T. Winter
5/27/09 mueckenh@rz.fh-augsburg.de
5/27/09 mueckenh@rz.fh-augsburg.de
5/27/09 Virgil
5/27/09 mueckenh@rz.fh-augsburg.de
5/27/09 Virgil
5/27/09 Virgil
5/28/09 mueckenh@rz.fh-augsburg.de
5/28/09 Virgil
5/28/09 Dik T. Winter
5/28/09 G. Frege
5/28/09 Jesse F. Hughes
5/29/09 Dik T. Winter
5/29/09 G. Frege
5/29/09 Dik T. Winter
5/30/09 G. Frege
5/30/09 Jesse F. Hughes
6/2/09 Dik T. Winter
5/30/09 Jesse F. Hughes
6/1/09 mueckenh@rz.fh-augsburg.de
6/1/09 Jesse F. Hughes
6/1/09 Virgil
6/2/09 george
6/2/09 Denis Feldmann
6/2/09 Dik T. Winter
6/11/09 mueckenh@rz.fh-augsburg.de
6/11/09 Virgil
6/11/09 William Hughes
6/11/09 mueckenh@rz.fh-augsburg.de
6/11/09 Virgil
6/11/09 William Hughes
6/11/09 Guest
6/11/09 mueckenh@rz.fh-augsburg.de
6/11/09 Virgil
6/11/09 William Hughes
6/12/09 mueckenh@rz.fh-augsburg.de
6/12/09 Virgil
6/12/09 William Hughes
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6/12/09 Virgil
6/12/09 William Hughes
6/12/09 mueckenh@rz.fh-augsburg.de
6/12/09 Virgil
6/12/09 William Hughes
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6/12/09 Virgil
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6/12/09 Virgil
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6/14/09 Virgil
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6/14/09 William Hughes
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6/15/09 Virgil
6/15/09 William Hughes
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6/15/09 Virgil
6/15/09 William Hughes
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6/15/09 Virgil
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6/16/09 Virgil
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6/16/09 Virgil
2/5/13
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6/16/09 Virgil
6/16/09 William Hughes
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6/17/09 Virgil
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6/17/09 Virgil
7/17/09 scriber77@yahoo.com
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6/19/09 William Hughes
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6/19/09 Virgil
6/19/09 William Hughes
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6/19/09 Virgil
6/19/09 William Hughes
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6/19/09 Virgil
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6/20/09 Virgil
6/20/09 William Hughes
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6/2/09 george
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6/3/09 george
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6/3/09 Guest
6/3/09 Marshall
6/3/09 Jack Markan
6/3/09 Dik T. Winter
6/3/09 Guest
6/7/09 mueckenh@rz.fh-augsburg.de
6/7/09 Virgil
6/7/09 mueckenh@rz.fh-augsburg.de
6/7/09 Virgil
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6/8/09 Virgil
6/9/09 mueckenh@rz.fh-augsburg.de
6/9/09 Virgil
6/9/09 William Hughes
6/9/09 Virgil
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6/10/09 Virgil
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6/11/09 Virgil
6/11/09 Dik T. Winter
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6/11/09 Virgil
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6/3/09 george
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5/28/09 William Hughes
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5/29/09 Virgil
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7/6/09 Dik T. Winter
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7/2/09 ross.finlayson@gmail.com
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7/2/09 ross.finlayson@gmail.com
7/2/09 Jack Markan
7/2/09 ross.finlayson@gmail.com
7/6/09 Jack Markan
7/6/09 ross.finlayson@gmail.com
7/6/09 Jack Markan
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