Reactions to/against the Binary Tree
WM
implications on set theory have been on topic frequently. Here is the
collection of counterarguments that have come to my attention.
If you think there is another counter-argument (or pro-argument), or
if you think to have an argument that improves one of the presented
arguments, please feel free to append it. I would be glad however, if
only serious contributions were added and if all unqualified comments
and statements of pure opinions could be suppressed.
Regards, WM
Theorem: The complete infinite binary tree has only countably many
infinite paths.
0.
/ \
0 1
/ \ / \
0 10 1
...
Proof A) Construct the binary tree starting from a "tree" that has
only one path, say
p_0 = 0.000...
0.
|
0
|
0
|
0
...
Add all paths that end by infinitely many zeros. Every path that you
add must start at a node of p_0 or at a node of a path already
constructed. The number of paths in the tree grows by not more and not
less than 1 when 1 path is added. However, after completing that
procedure all nodes and every infinite sequence of bits (including the
path 0.111...) is represented in the infinite binary tree.
Proof B) The complete infinite binary tree can be constructed by an
infinite agglomeration of basic elements of the form
|
o
/ \
The number of distinct lines is increased by 1 by 1 node.
lines going out - lines coming in - nodes =
2 - 1 - 1 = 0 .
This procedure, even when applied infinitely often, cannot but yield
the result 0 implying a countable number of lines. The set of all
lines limits the set of all distinct paths. The construction is
possible in (B) as well as in (A) because the numbers of elements used
for construction is countably infinite.
Proof C) Consider the edges of the complete binary tree (an edge
connects two subsequent nodes of a path). Take all edges and put them
on one and the same level of the tree, side by side, such that the
"tree" now is an array of parallel edges:
|||||||...
This array limits the number of possible paths of the tree. It is an
upper limit, because every path there has only one edge. And there is
no further edge remaining to distinguish any further paths.
Remark: Of course a set of n edges can be put in n! different
sequences. But the edges of the binary tree are not subject to
arbitrary ordering. Each one has one and only one fixed place in the
tree. Therefore the number of edges limits the number of paths. Proof
C gives an upper limit.
Remark: The tree contains all possible sequences of bits including
0.111... .
Nevertheless the tree contains only a countable number of paths, as we
see by each of the arguments (A, B, and C). This shows that "most" of
the real numbers cannot exist as independent bit sequences. It shows
further that most of the real numbers are not subject to being put in
a list or resulting from a list as an anti-diagonal number.
For visualising the construction of the binary tree see:
http://www.hs-augsburg.de/~mueckenh/GU/GU12.PPT#335,24,Folie 24
========Counter Arguments============
Vaughan Pratt:
http://www.cs.nyu.edu/pipermail/fom/2009-March/013493.html
In the finite case, each node at depth d from the root of a tree of
height h participates in 2^(h-d) paths. As h increases that node
participates in an exponentially growing number of paths. Therefore
in the limit each node participates in 2^N paths. Now you might say,
oh, but there are twice as many new nodes at every height increment.
So then we have to ask whether this is enough node growth at depth h
to offset the exponentially growing number of paths shared by one node
at depth d. In the limit each path is shared between N nodes, but
that still leaves 2^N/N paths per node, which cannot be N because N^2
is countable and 2^N is not.
My comment:
It is just under investigation whether 2^N is larger than the number
of nodes which, by the way, is 2^N, if the tree has N levels.
===================================
Vaughan Pratt:
http://www.cs.nyu.edu/pipermail/fom/2009-March/013493.html
Since the infinite case does not count the leaves (there being none),
it would be misleading to count them in the finite case if we want
insight into what happens at infinity. For trees of height h we find
a cardinality gap: 2^h - 1 interior nodes vs. 2^h paths. Why should
passing to the limit close this gap?
My answer:
A good argument. Let us start with a pre-root node
o
|
o
/ \
Then there is no gap from the beginning. Why should passing to the
limit open a gap?
====================================
Vaughan Pratt:
http://www.cs.nyu.edu/pipermail/fom/2009-March/013493.html
Every pair of nodes is connected by a finite path, whereas every pair
of paths has in common only finitely many nodes and they differ by
infinitely many nodes. Therefore any comparison of nodes to paths is
an apples-to-oranges comparison for which it would very surprising to
find a bijection between them.
My answer:
There is no bijection, but as proof A shows, there are not more paths
than nodes.
=====================================
David C. Ullrich (alluding to proof B):
http://groups.google.com/group/sci.logic/msg/82cd07114d4e6d7e?hl=de&dmode=source
You say
"This procedure, even when applied infinitely often, cannot but yield
the result 0 respectively a countable number of lines."
over and over, but it's simply not true. Your asserting it does not
make it true. The fact that it seems clearly true to you does not make
it true. You need to give a _proof_ of this, and you've never done so
- in each of your attempts there's always a similar unjustified
assertion.
My answer:
For the infinite sum over all (countably many) basic elements it
should be clear that
SUM[n --> oo] 2 - 1 - 1 = SUM[n --> oo] 0 = 0.
Without proof. But even if we had only
SUM[n --> oo] 2 - 1 - 1 = SUM[n --> oo] 0 < 2^aleph_0
there would be less than 2^aleph_0 paths.
Proof of the latter:
SUM[n --> oo] 0 =< SUM[n --> oo] 1 = oo
< 2^aleph_0 (if the disproved result were correct.)
===================================
Owen Jacobson:
http://groups.google.com/group/sci.logic/msg/e86f55016eac8d8c?hl=de&dmode=source
Every infinitely-long binary string constructed this way has only
finitely many 1s. There infinitely-long binary strings which contain
more than finitely many 1s, but your construction will not create them
at any step, so you cannot inductively prove things from the
properties of the strings your construction creates to strings that
your construction omits.
More surprising, though, is the fact that your construction
demonstrably constructs at least one path through every node in the
tree! For any node, we can construct an initial path leading to it
using a finite number of 1s and 0s, and we can then proceed using
finitely many more 1s to any of its descendent nodes, but at some
point on each path, your construction "gives up" and generates
infinitely many zeros.
My answer:
I don't give up before Cantor gives up in his diagonal argument.
Construction according to A or B supplies the complete set of paths.
======================================
Virgil
http://groups.google.com/group/sci.logic/msg/5839bd6427920b33?hl=de&dmode=source
All those with infinitely many 1's in them [are missing]
My answer:
I cannot find any missing path. Construction according to A or B
supplies the complete set of paths.
=========================================
Calvin Ostrum:
http://groups.google.com/group/sci.logic/msg/52c1b3a5e2fbde58?hl=de&dmode=source
Here, I think, is your problem. You are not noticing that when you
add a single path, you are actually adding many paths.
My answer:
According to construction A this is false.
========================================
Calvin Ostrum :
http://groups.google.com/group/sci.logic/msg/52c1b3a5e2fbde58?hl=de&dmode=source
The net result in the end is that after all your paths have been
added, there are many other paths that exist, because of the edges
that have been added to make the paths, which you have not noticed.
My answer:
According to construction A this is false.
=========================================
Calvin Ostrum:
http://groups.google.com/group/sci.logic/msg/52c1b3a5e2fbde58?hl=de&dmode=source
Every path you add is of the form a finite sequence of 0's and 1's,
following by infinite sequence of 0's. However, there are paths in
the final tree that are not like this. That is because the paths
interact with one another. You must look at the individual edges and
how then can be made into paths in the final tree. You do not add
paths, you add edges. Paths come FOR FREE. You cannot pick and choose
the paths you add. They sneak in as sets of joined edges and there is
nothing you can do about it.
My answer:
I cannot believe that an uncountable set of different paths can be
individually constructed by a countable number of steps.
=======================================
Christopher Creutzig:
http://groups.google.com/group/de.sci.mathematik/msg/a35c36864fd75c82?dmode=source
My translation (of the German original, cp. also for some other
arguments, related to some already discussed above): A path is not
defined by a single node but by an infinite number of them. Therefore
the number of nodes is not an upper limit for the number of paths, at
least not automatically.
My answer:
Every path can be distinguished by one single node from every
"selected" set of paths. Of course first the selected set must be
defined by individually "picking" the paths belonging to that set, as
is usual in logic. If infinitely many nodes were needed to define a
single path, then you should be able to present at least one of the
set of needed nodes.
====================================
Christopher Creutzig:
http://groups.google.com/group/de.sci.mathematik/msg/a35c36864fd75c82?dmode=source
All attempts to construct an explicit surjection of the set of nodes
onto the set of paths have been contradicted by the explicit
presentation of paths that are not in this surjection.
My answer:
I do not present a surjection. Only countably many paths can be
counted. But in order to construct the whole tree countably many paths
are sufficient.
Further, according to proof A, the number of nodes occupied by
constructed paths grows by aleph_0 with every _single_ path appended.
====================================
Georg Cantor
Letter to Vivanti of Dec. 3 1885
Cantor discusses what now is known as the construction of the binary
tree.
Consider all "finite paths":
z_n = (a_1)/2 + (a_2)/2^2 + ... + (a_(n-1))/2^(n-1) + 1/2^n
with a_k having value 0 oder 1. The number of them is z_n is 2^
(n-1).
For n = 1, 2, 3, ... in inf. then
1+ 2 + 4 + 8 + ... in inf. z == {z_1, z_2, z_3, ...}
But these numbers are merely a very small subset of all numbers that
have the form
z_n = (a_1)/2 + (a_2)/2^2 + ... + (a_n)/2^n + ... in inf.
and make up a set of second cardinality (the infinite paths).
This apparent difficulty is solved as follows: The set of all numbers
of the unit interval is not the limit of all "finite path" z_n for n =
oo.
My answer:
If the set of real numbers of the unit interval (infinite paths)
cannot be considered as the limit of the set of finite paths, why then
should any irrational number be considered as the limit of the
sequence of its finite initial segments (i.e., as an infinite path in
the binary tree)? This would be extremely inconsequent. The question
would arise: How many real numbers can be considered as the limits of
the sequences of their finite initial segments? Only one at a time, or
two or ten or aleph_0 or some number depending on the brain capacity
of the observer? But if consequently denying the limit process, then
also 0.111... is not the limit of
0,1
0,11
0,111
...
Then Cantor's diagonal argument fails, because then there is not a
diagonal number 0.111... that is different from all its finite
initial segments. Then also Cantor's construction of irrational
numbers fails, as given by himself in ["Bemerkungen mit Bezug auf den
Aufsatz: Zur Weierstraß-Cantorschen Theorie der
Irrationalzahlen" (Werke p. 114)] :
sqrt(3) = (1,7, 1,73, 1,732, ...)
Marshall
>
> If you think there is another counter-argument (or pro-argument), or
> if you think to have an argument that improves one of the presented
> arguments, please feel free to append it. I would be glad however, if
> only serious contributions were added and if all unqualified comments
> and statements of pure opinions could be suppressed.
This is not an argument either pro- or anti-, but it is a serious
comment nevertheless.
You have put enormous effort into tearing down parts of existing
mathematics. You find fault with set theory, with the real numbers,
the existence of very large natural numbers, etc.
Leaving entirely aside what I think of your arguments, I notice
that you never advance any theory of your own. You have never
provided anything constructive. Always with you it is trying
to destroy, never to create.
Don't you think your time would be better spent, don't you think
your arguments would be more persuasive, if you provided an
actual alternative?
I think you should put your time into formalizing your ideas.
As it stands, and again even without any judgment of your
arguments, you are no pioneer, merely a naysayer.
Marshall
WM
On 24 Mai, 18:38, Marshall <marshall.spi...@gmail.com> wrote:
> On May 24, 8:01 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
>
>
> > If you think there is another counter-argument (or pro-argument), or
> > if you think to have an argument that improves one of the presented
> > arguments, please feel free to append it. I would be glad however, if
> > only serious contributions were added and if all unqualified comments
> > and statements of pure opinions could be suppressed.
>
> This is not an argument either pro- or anti-, but it is a serious
> comment nevertheless.
I see and appreciate that. Therefore I answer.
>
> You have put enormous effort into tearing down parts of existing
> mathematics. You find fault with set theory, with the real numbers,
Both are unresolvably connected. An entangled state. Cantor knew it.
> the existence of very large natural numbers, etc.
I do not deny the existence of very large numbers, but only the
existence of numbers that require more information for identification
(or definition) than is available.
>
> Leaving entirely aside what I think of your arguments,
That's deplorable. I think that in particular Cantor's text and my
answer shows the problem in clearest possible way.
> I notice
> that you never advance any theory of your own. You have never
> provided anything constructive. Always with you it is trying
> to destroy, never to create.
Recently I have written a text book on elementary mathematics. I did
not find anything to change with current mathematics. The reason is
that set theory has not really influenced mathematics. So I simply
explained in the preface that there are no actual infinities --- and
everything runs its way. What should I change?
*That* mathematics does not allow quantifier exchange.
>
> Don't you think your time would be better spent, don't you think
> your arguments would be more persuasive, if you provided an
> actual alternative?
>
> I think you should put your time into formalizing your ideas.
I agree with Doron Zeilberger: "The Human Obsession With "Formal
Proofs" is a Waste of the Computer's Time, and, Even More Regretfully,
of Humans' Time." (Opinion 94)
Formalization has not protected many extremely intelligent minds from
set theory.
> As it stands, and again even without any judgment of your
> arguments, you are no pioneer, merely a naysayer.
Without us evolution would sometimes run out of control
Regards, WM
Martin Musatov
Very good: Musatov proved P=NP with witnesses:
http://groups.google.com/group/sci.math/browse_thread/thread/ec7e95813d6886f2/f1469447927935b1?lnk=raot#f1469447927935b1
Window of opportunity
Marshall
> On 24 Mai, 18:38, Marshall <marshall.spi...@gmail.com> wrote:
>
> > You have put enormous effort into tearing down parts of existing
> > mathematics. You find fault with set theory, with the real numbers,
>
> Both are unresolvably connected. An entangled state. Cantor knew it.
I agree.
> > the existence of very large natural numbers, etc.
>
> I do not deny the existence of very large numbers, but only the
> existence of numbers that require more information for identification
> (or definition) than is available.
Yes; that is what I should have said. Sometimes this excessive
information requirement results in a very large number, but in
other cases it could be between 0 and 1.
> > Leaving entirely aside what I think of your arguments,
>
> That's deplorable. I think that in particular Cantor's text and my
> answer shows the problem in clearest possible way.
Well, I would hope to be able to address one issue at a
time, rather than be required to discuss everything all at
once. Would it work for you if I said "Leaving aside
*temporarily* what I think of your arguments ..."
> > I notice
> > that you never advance any theory of your own. You have never
> > provided anything constructive. Always with you it is trying
> > to destroy, never to create.
>
> Recently I have written a text book on elementary mathematics. I did
> not find anything to change with current mathematics. The reason is
> that set theory has not really influenced mathematics. So I simply
> explained in the preface that there are no actual infinities --- and
> everything runs its way. What should I change?
My immediate reaction is, if nothing changes, then why
is it such an issue for you? Why bother arguing over a
difference that makes no difference? It seems pointless.
> *That* mathematics does not allow quantifier exchange.
>
> http://www.oldenbourg-wissenschaftsverlag.de/olb/de/1.c.1598342.de?ha...
My congratulations to you on the occasion of the publishing of
your book.
> > Don't you think your time would be better spent, don't you think
> > your arguments would be more persuasive, if you provided an
> > actual alternative?
>
> > I think you should put your time into formalizing your ideas.
>
> I agree with Doron Zeilberger: "The Human Obsession With "Formal
> Proofs" is a Waste of the Computer's Time, and, Even More Regretfully,
> of Humans' Time." (Opinion 94)
I disagree. Machine-checked anything is useful because machines
are so much more fastidious than humans. It is true that machine
checked proof is no defense against "garbage in, garbage out"
but it is certainly an excellent defense against mistaken
deduction, and that is something all humans are subject to.
I also have my example of the affirmative conclusion to the
Robbins Conjecture, which was pursued "manually" by humans
for decades to no avail, but was solved instead by the EQN
software. This software was itself written by humans of course.
Also: mechanical proof software is fun. (Perhaps not universally,
though; I suppose I could *imagine* someone somewhere
might not find it as interesting as I do. My wife, say.)
> Formalization has not protected many extremely intelligent minds from
> set theory.
>
> > As it stands, and again even without any judgment of your
> > arguments, you are no pioneer, merely a naysayer.
>
> Without us evolution would sometimes run out of control
I am curious, can you name any event in math history that
would be an example of a success of the type you seek:
finding fault with established theory? This is not a rhetorical
question; I have some modest interest in math history, and
I can't think of an example. But if there is one, I'd be interested
to hear of it.
Marshall
Martin Musatov
Marshall,
Here is an op-ed piece I wrote:
An Open Letter to the Moral and Scientific Community:A Quote:"If smart
people all had Ph.D.'s we would not have light bulbs." --Martin
Musatov speaking on American Entrepreneur and Innovator Thomas
EdisonPreface: "Computational Complexity"So much of what I have seen
since I have began studying computational complexity simply amazes me.
I have come from an outsiders perspective peering into this vast new
world where obvious things hide themselves and complex things take
center stage to be studied like pellets of sand beneath a microscope.
I will say this one thing: I have never been treated with more disdain
in an academic setting. I have had M.I.T. Assistant Professor Scott
Aaronson publicly threaten to contact my Internet Service Provider and
call me a "goon" for disproving his theorem "P does not Equal NP"
publicly. I have been called a "troll" and "couch boy" the latter I
have no idea what the colloquial means. I have had my I.P. address
blocked from contributing to Wikipedia and have been sent threatening
emails from Wikipedia administrators saying, "Wikipedia doesn't need
you." Since I began pursuing my proof of computational complexity my
Wikipedia profile for my work as a screenwriter (which had remained
untouched for the better part of three years) was immediately flagged
as "non-notable" and deleted. And all because the mathematics and code
I was inputting was too advanced for wiki language to swallow without
causing system problems and offending apparently some very sensitive
people. And all over a tiny little problem in theoretical computer
science called P=NP.Basically, as the case may certainly be there seem
to be a lot of people out there absolutely insistent that "P" does not
equal "NP". But I have to wonder, if it is only theory we are debating
here, what is so vested by this people that they defend an insistent
of an impossibility as if it were the holy grail? It just does not
make sense to me. I will say this, especially, it does not make sense
to argue that something such as P equals NP has to be impossible. If
it were true there are well documented published articles such as one
in the Boston Globe which blatantly list all the potential benefits we
might experience if the scientific community would accept P=NP. The
list includes advances in "Protein Folding" which could spur
unprecedented growth and advances in biological research which may
well include cures for diseases like cancer and H.I.V. So dare I say,
why are noted professors at top universities such as Scott Aaronson at
M.I.T. and Stephen Arthur Cook at the University of Toronto so
insistent of its impossibility? What could be so motivating as one
would defend such a contrary position to which being contrariety holds
no obvious benefit for society at large. The elephant in the room
seems to be that this argument has been raging and churning for years
ever since Stephen Cook invented the class "NP-Complete" back in
1970.My goal, my dream, in pursuing a proof that P=NP was not to win a
million dollars and notoriety, but to help the people in the world use
the technology to better take care of themselves and their families.
My goals personally are to help my young niece who just had an implant
put in her ear so she could hear better and to spur advances in cancer
research as my uncle and Godfather Michael Schultz was in the last
month diagnosed with kidney and bone cancer. So still, I continue on,
every morning pursuing the solution despite the animosity and
ignorance. My dreams are simply bigger than theirs. My dreams are not
to predict the S&P 500 and compromise the security of banks by
collapsing known elements of cryptography. My dreams are that a young
researcher in Tibet working by himself may uncover a cure for cancer
that no one had seen. My goal is that a hobby mechanic in rural Russia
with access to the Internet will invent a hybrid computer driven
engine which will best all the struggling automakers who we continue
to float financially like giant sick whales out to sea. My dream is
that the academic community would allow open access to citizens at
large and not simply the ones who can afford the prestigious school
tuition. The basis of my plea: history has shown it to be the best
path. With only three months of formal education he became one of the
greatest inventors and industrial leaders in history. Edison obtained
1,093 United States patents, the most issued to any individual.Call
this my prayer or call it my plea it is my cry to the scientific
community and to God in heaven can we please work together here and
accomplish some good in the world instead of warbled disagreement? My
last thought is to ask yourself why would anyone insist on the
absolute impossibility of something that could bring so much good to
the world?Quotes by Thomas Edison:"Hell, there are no rules here we're
trying to accomplish something.""I didn't fail ten thousand times. I
successfully eliminated, ten thousand times, materials and combination
which wouldn't work.""I never perfected an invention that I did not
think about in terms of the service it might give others.""I am more
of a sponge than an inventor. I absorb ideas from every source. My
principal business is giving commercial value to the brilliant but
misdirected ideas of others.""Time is really the only capital that any
human being has, and the one thing that he can't afford to lose.""I
find out what the world needs. Then I go ahead and try to invent
it.""I have more respect for the fellow with a single idea who gets
there than for the fellow with a thousand ideas who does
nothing."Thank you for reading this letter. If you have any comments
or suggestions please feel free to contact me through the publisher.
Thank you,
Martin Musatov
P.S. P=NP, Call me the Charlton Heston if math, but they can have my
theorem when they pry it...actually, no. It's truth. Truth belongs to
the people.
WM
> On May 24, 10:57 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> Yes; that is what I should have said. Sometimes this excessive
> information requirement results in a very large number, but in
> other cases it could be between 0 and 1.
So it is.
Pi is defined by many finite formulas.
But if a finite formula is lacking, as unavoidably must be the case
for most reals, then it is impossible to define or to use them as
individuals.
The decimal expansion of pi or even that of 1/33 = 0.030303... does
not exist _in the real world_, because not all natural numbers
required as indices (counting the places) are available, be it in
written, thought or what ever form.
>
> > > Leaving entirely aside what I think of your arguments,
>
> > That's deplorable. I think that in particular Cantor's text and my
> > answer shows the problem in clearest possible way.
>
> Well, I would hope to be able to address one issue at a
> time, rather than be required to discuss everything all at
> once. Would it work for you if I said "Leaving aside
> *temporarily* what I think of your arguments ..."
>
> > > I notice
> > > that you never advance any theory of your own. You have never
> > > provided anything constructive. Always with you it is trying
> > > to destroy, never to create.
>
> > Recently I have written a text book on elementary mathematics. I did
> > not find anything to change with current mathematics. The reason is
> > that set theory has not really influenced mathematics. So I simply
> > explained in the preface that there are no actual infinities --- and
> > everything runs its way. What should I change?
>
> My immediate reaction is, if nothing changes, then why
> is it such an issue for you? Why bother arguing over a
> difference that makes no difference? It seems pointless.
There are two reasons: First set theorists have conquered mathematics
claming that only mathematics based upon set theory (and only in
formalized form) is real mathematics. Everything else is unclear
prose. I do not believe that. I am sure that later times will find
that automatized proofs are not automatically true.
Second, there are some scholars who try to prove theorems about finite
numbers by means of infinite cardinals. Such attempts are in vain.
But in fact, I see that finished infinity is such a hocus pocus that I
cannot be silent.
>
> I also have my example of the affirmative conclusion to the
> Robbins Conjecture, which was pursued "manually" by humans
> for decades to no avail, but was solved instead by the EQN
> software. This software was itself written by humans of course.
That is another issue. Of course a machine may be used to find a
solution of an involved or difficult combinatorical problem. I use
calculators and Mathematica for many purposes. But I do not believe,
like some, that a proof of Cantor's theorem, once checked by a
machine, establishes absolute truth. The machine uses some basics that
can be false.
> I am curious, can you name any event in math history that
> would be an example of a success of the type you seek:
> finding fault with established theory? This is not a rhetorical
> question; I have some modest interest in math history, and
> I can't think of an example. But if there is one, I'd be interested
> to hear of it.
Naysayer Hippasos von Metapont showed that the pythagorean theorem
"everything is number" (or ratio) was wrong (without introducing
irrational numbers, he left that to Eudoxos).
Naysayer Nicole d'Oresme denied established knowledge, then the Bible,
and proposed the heliocentric system, contradicting the Bible.
(Fortunately for him, nobody cared at that time.)
In 1833 Legendre presented to the French academy six proofs of the
parallel axiom, three of them working with infinite angular areas.
Naysayer Gauss did not dare to publish his results concerning
geometry.
The other way round: Christian Doppler stated his theory by 1850.
Though being extremely simple, it was rejected by all great Austrian
(and other) mathematicians for more than 30 years. As late as in 1900
a physics handbook appeared, saying that Doppler's theory "is probably
correct".
Regards, WM
WM
> Thank you,
> Martin Musatov
>
May I ask what is your opinion with respect to the binary tree?
Regards, WM
Martin Musatov
Marshall
That guy is just a prankster. His only interest is to wind
people up.
Marshall
Ralf Bader
Probably. And what does it tell about Mueckenheim that he didn't
notice that but asked the question above?
Ralf
LudovicoVan
> During the last years the complete infinite binary tree and its
> implications on set theory have been on topic frequently. Here is the
> collection of counterarguments that have come to my attention.
>
> If you think there is another counter-argument (or pro-argument), or
> if you think to have an argument that improves one of the presented
> arguments, please feel free to append it. I would be glad however, if
> only serious contributions were added and if all unqualified comments
> and statements of pure opinions could be suppressed.
A simple counter-argument (I suppose this might be an informal version
of your proof A):
[tree height] < [num. of paths] < [num. of nodes]
0 < 0 < 0
1 < 2 < 2
2 < 4 < 6
3 < 8 < 14
4 < 16 < 30
...
n < 2^n < 2^(n+1)-2 (for n > 0)
...
Extending to the infinite case (here the notation might need fixing
and I'm not sure if the inequalities should still be strict, but this
is irrelevant to the conclusion):
w <= 2^w <= 2^(w+1)-2
Since the number of nodes is countable, i.e.:
|w| = |2^(w+1)-2|
Follows:
|w| = |2^w| = |2^(w+1)-2|
IOW, that is:
The number of nodes is countable;
The number of paths is dominated by the number of nodes;
Therefore:
The number of paths is countable.
-LV
Marshall
You can't just blithely extend to the infinite case. Infinite is not
just a bigger kind of natural number.
Marshall
William Elliot
> Theorem: The complete infinite binary tree has only countably many
> infinite paths.
>
> 0.
> / \
> 0 1
> / \ / \
> 0 10 1
> ...
You are trying to prove the impossible as the conjectured theorem is false.
If you want to prove the reals are countable, then consider the Skolem
paradox that shows that if ZF is consistent, then it has a countable
model.
WM
Regards, WM
WM
> On Sun, 24 May 2009, WM wrote:
> > Theorem: The complete infinite binary tree has only countably many
> > infinite paths.
>
> > 0.
> > / \
> > 0 1
> > / \ / \
> > 0 10 1
> > ...
>
> You are trying to prove the impossible as the conjectured theorem is false.
If so, then we have to tolerate the follwing situation:
The real number sqrt(3) is the limit of a rational sequence.
Every real number is the imit of a rational sequence.
Not every number is the limit of a rational sequence.
Because not all real numbers are the limits of rational sequences.
I do not tolerate that situation.
>
> If you want to prove the reals are countable, then consider the Skolem
> paradox that shows that if ZF is consistent, then it has a countable
> model.
Skolem said: "Erstens habe ich eine genauere Begründung des
allgemeinen mengentheoretischen Relativismus gegeben, der besonders
die Konsequenz hat, daß das Absolut-nicht-abzählbare auf axiomatischer
Grundlage keine Existenzberechtigung hat. ... "
The absolute meaning of" uncountable" is not justified on an axiomatic
basis.
Regards, WM
WM
> On 24 May, 16:01, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > During the last years the complete infinite binary tree and its
> > implications on set theory have been on topic frequently. Here is the
> > collection of counterarguments that have come to my attention.
>
> > If you think there is another counter-argument (or pro-argument), or
> > if you think to have an argument that improves one of the presented
> > arguments, please feel free to append it. I would be glad however, if
> > only serious contributions were added and if all unqualified comments
> > and statements of pure opinions could be suppressed.
>
> A simple counter-argument (I suppose this might be an informal version
> of your proof A):
>
> [tree height] < [num. of paths] < [num. of nodes]
>
> 0 < 0 < 0
> 1 < 2 < 2
> 2 < 4 < 6
> 3 < 8 < 14
> 4 < 16 < 30
> ...
> n < 2^n < 2^(n+1)-2 (for n > 0)
> ...
>
> Extending to the infinite case (here the notation might need fixing
> and I'm not sure if the inequalities should still be strict, but this
> is irrelevant to the conclusion):
So it is.
>
> w <= 2^w <= 2^(w+1)-2
>
> Since the number of nodes is countable, i.e.:
>
> |w| = |2^(w+1)-2|
>
> Follows:
>
> |w| = |2^w| = |2^(w+1)-2|
>
> IOW, that is:
>
> The number of nodes is countable;
> The number of paths is dominated by the number of nodes;
> Therefore:
> The number of paths is countable.
>
I agree. There is another argument. For every appended path we inrease
the number of nodes by aleph_0, the number of paths by 1. This should
suddenly switch when finishing the infinite?
Ad if it cannot be finished, then the whole discussion about
cardinalities is in vain.
Regards, WM
William Elliot
> On 25 Mai, 08:57, William Elliot <ma...@rdrop.remove.com> wrote:
>> On Sun, 24 May 2009, WM wrote:
>>> Theorem: The complete infinite binary tree has only countably many
>>> infinite paths.
>>
>>> � 0.
>>> � / \
>>> �0 1
>>> / \ / \
>>> 0 10 1
>>> ...
>>
>> You are trying to prove the impossible as the conjectured theorem is false.
>
> The real number sqrt(3) is the limit of a rational sequence.
> Not every number is the limit of a rational sequence.
> Because not all real numbers are the limits of rational sequences.
>
reducio ad absurdum that your proposition is false.
> I do not tolerate that situation.
It's extremely obvious to all of us, except to you, that
you can't tolerate that situation where you're wrong.
>> If you want to prove the reals are countable, then consider the Skolem
>> paradox that shows that if ZF is consistent, then it has a countable
>> model.
>
> Skolem said: "Erstens habe ich eine genauere Begr�ndung des
> allgemeinen mengentheoretischen Relativismus gegeben, der besonders
> die Konsequenz hat, da� das Absolut-nicht-abz�hlbare auf axiomatischer
> Grundlage keine Existenzberechtigung hat. ... "
>
> The absolute meaning of" uncountable" is not justified on an axiomatic
> basis.
>
Then stop proving they're uncountable by using the usual axiomatic basis.
David C. Ullrich
<ju...@diegidio.name> wrote:
>[...]
>
>Extending to the infinite case
Oops. You can't just "extend" an inequality valid for natural
numbers to the infinite case (at least not without justification).
Things simply don't work that way.
Look. Here's Ullrich's amazing theorem:
Theorem: w is finite.
Proof: For every natural number n, n is finite.
Extending to the infinite case, w is finite. QED.
This really doesn't have all that much to do with
infinity. You're assuming that if every natural number
n has a certain property then w must also have that
property. That's exactly as valid as assuming that
if every natural number n < 10 has a certain property
then 10 must also have the same property. Which is
to say, not valid at all.
(Exercise: Show that 10^2 < 100.
Hint: First prove an inequality valid for all n < 10.
Then "extend" to the case n = 10.)
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
WM
> On Mon, 25 May 2009, WM wrote:
> > On 25 Mai, 08:57, William Elliot <ma...@rdrop.remove.com> wrote:
> >> On Sun, 24 May 2009, WM wrote:
> >>> Theorem: The complete infinite binary tree has only countably many
> >>> infinite paths.
>
> >>> 0.
> >>> / \
> >>> 0 1
> >>> / \ / \
> >>> 0 10 1
> >>> ...
>
> >> You are trying to prove the impossible as the conjectured theorem is false.
>
> > If so, then we have to tolerate the following situation:
> > The real number sqrt(3) is the limit of a rational sequence.
> > Every real number is the limit of a rational sequence.
> > Not every number is the limit of a rational sequence.
> > Because not all real numbers are the limits of rational sequences.
>
> You've contradicted your self. Thus you've shown by
> reducio ad absurdum that your proposition is false.
Bare nonsense should be avoided in this thread.
>
> > I do not tolerate that situation.
>
> It's extremely obvious to all of us,
That is not a proof of anything.
>
> >> If you want to prove the reals are countable, then consider the Skolem
> >> paradox that shows that if ZF is consistent, then it has a countable
> >> model.
>
> > Skolem said: "Erstens habe ich eine genauere Begründung des
> > allgemeinen mengentheoretischen Relativismus gegeben, der besonders
> > die Konsequenz hat, daß das Absolut-nicht-abzählbare auf axiomatischer
> > Grundlage keine Existenzberechtigung hat. ... "
>
> > The absolute meaning of" uncountable" is not justified on an axiomatic
> > basis.
>
> Then stop proving they're uncountable by using the usual axiomatic basis.-
I do just the contrary. I show they are countable by not using the
axiomatic basis. Are you an alias of Musatov?
Regards, WM
LudovicoVan
> On Sun, 24 May 2009 22:28:27 -0700 (PDT), LudovicoVan
>
> <ju...@diegidio.name> wrote:
> >[...]
>
> >Extending to the infinite case
>
> Oops. You can't just "extend" an inequality valid for natural
> numbers to the infinite case (at least not without justification).
There is no need to justify that inequality in my argument. You might
enjoy exercising yourself against the final syllogism.
> Look. Here's Ullrich's amazing theorem:
Lovely.
-LV
WM
> On Sun, 24 May 2009 22:28:27 -0700 (PDT), LudovicoVan
>
> <ju...@diegidio.name> wrote:
> >[...]
>
> >Extending to the infinite case
>
> Oops. You can't just "extend" an inequality valid for natural
> numbers to the infinite case (at least not without justification).
> Things simply don't work that way.
They work that way if irrational numbers are limits, e.g., in case of
Cantor's "Fundamentalfolgen".
>
> Look. Here's Ullrich's amazing theorem:
>
> Theorem: w is finite.
>
> Proof: For every natural number n, n is finite.
> Extending to the infinite case, w is finite. QED.
That is not amazing, it is the simple truth.
>
> This really doesn't have all that much to do with
> infinity.
It is the only form of infinity (outside of matheology). It is called
potential infinity.
> You're assuming that if every natural number
> n has a certain property then w must also have that
> property. That's exactly as valid as assuming that
> if every natural number n < 10 has a certain property
> then 10 must also have the same property. Which is
> to say, not valid at all.
That is purest nonsense.
A valid comparison would be: Showing that every initial segment of N
is finite shows that N cannot be actually infinite.
>
> (Exercise: Show that 10^2 < 100.
>
> Hint: First prove an inequality valid for all n < 10.
> Then "extend" to the case n = 10.)
>
Hint: Try to understand and then to explain, such that normal
mathematicians can understand, what you may have meant by
David C. Ullrich (alluding to proof B):
http://groups.google.com/group/sci.logic/msg/82cd07114d4e6d7e?hl=de&dmode=source
You say
"This procedure, even when applied infinitely often, cannot but yield
the result 0 respectively a countable number of lines."
over and over, but it's simply not true. Your asserting it does not
make it true. The fact that it seems clearly true to you does not make
it true. You need to give a _proof_ of this, and you've never done so
- in each of your attempts there's always a similar unjustified
assertion.
Regards, WM
William Hughes
> Since the number of nodes is countable, i.e.:
what you want here is the statment.
|w| = aleph_0
The false statement
> |w| = |2^(w+1)-2|
is not a statement that the number of nodes is countable.
Recall
|2^(w+1)-2| = |2^w|
and
|x| < |2^x|
for any x. So The cardinality of the paths
is strictly greater than that of the nodes.
- William Hughes
LudovicoVan
> On May 25, 1:28 am, LudovicoVan <ju...@diegidio.name> wrote:
>
> > Since the number of nodes is countable, i.e.:
>
> what you want here is the statment.
>
> |w| = aleph_0
>
> The false statement
>
> > |w| = |2^(w+1)-2|
>
> is not a statement that the number of nodes is countable.
Then, give the proper statement.
> Recall
> |2^(w+1)-2| = |2^w|
> and
> |x| < |2^x|
> for any x.
What is x now, and what that *means*?
> So The cardinality of the paths
> is strictly greater than that of the nodes.
Surely m notation is flawed in many ways, but your conclusion is a
nonsequitur -- if it makes any sense at all.
As you'd love to say, try again Sam.
-LV
William Hughes
> On 25 May, 16:55, William Hughes <wpihug...@hotmail.com> wrote:
>
<snip>
> > The false statement
>
> > > |w| = |2^(w+1)-2|
>
> > is not a statement that the number of nodes is countable.
>
> Then, give the proper statement.
Let the set of nodes for the infinite tree be N_inf
|N_inf| = aleph_0
Note
Let the set of nodes for the tree of height n be N_n
We can show lim(n->inf) |N_n| = |N_inf|
However this is not always true
Let the set of paths for the infinite tree be P_inf
|et the set of paths for the tree of height n be P_n
It is not true that lim(n->inf) |P_n| = |P_inf|
(the difference being that there are no "infinite nodes"
but there are infinite paths)
Hence your inequality does not hold in the infinite
case
- William Hughes
LudovicoVan
Why is it not true? How does the fact that the paths are infinite (an
"internal" property of a path object) make any difference here?
Moreover, you have infinite paths only in the limit case, otherwise a
path is finite. (But, again, I don't see how the specific "internal
constitution" of a path object makes any difference.)
To attack it from an other side: given that in the finite (over the
height of the tree) the number of paths is lesser than the number of
nodes, how can we have that the opposite relation holds in the
infinite?
-LV
William Hughes
> To attack it from an other side: given that in the finite (over the
> height of the tree) the number of paths is lesser than the number of
> nodes, how can we have that the opposite relation holds in the
> infinite?
>
Simple, in the infinite tree there are no nodes that cannot
be found in some finite tree. However. in the infinite tree there
are many (uncountably many) paths that are not found in any
finite tree. Or put another way, the finite paths, the paths found in
finite
trees, are countable. It is the infinite paths that are uncountable,
and as you note, you only get infinite paths in the infinite tree.
- William Hughes
LudovicoVan
> On May 25, 2:04 pm, LudovicoVan <ju...@diegidio.name> wrote:
>
> > To attack it from an other side: given that in the finite (over the
> > height of the tree) the number of paths is lesser than the number of
> > nodes, how can we have that the opposite relation holds in the
> > infinite?
>
> Simple, in the infinite tree there are no nodes that cannot
> be found in some finite tree. However. in the infinite tree there
> are many (uncountably many) paths that are not found in any
> finite tree.
To be sure: isn't a "path" that thing that develops from the root
along a parent-child trajectory? If so, what are you talking about
William?
> Or put another way, the finite paths, the paths found in finite
> trees, are countable. It is the infinite paths that are uncountable,
> and as you note, you only get infinite paths in the infinite tree.
You are have restated the thesis: your thesis, for which I still see
no support.
-LV
Gus Gassmann
> Bare nonsense should be avoided in this thread.
Then why the fuck do you keep posting?
William Hughes
> To be sure: isn't a "path" that thing that develops from the root
> along a parent-child trajectory? If so, what are you talking about
> William?
>
The trivial observation that there is no infinite thing that develops
from the root along a parent child trajectory in any finite tree.
So there are paths in the infinite tree that are not in any finite
tree.
Recall your question
... in the finite (over the height of the tree) the number of paths
is
lesser than the number of nodes, how can we have that
the opposite relation holds in the infinite?
For the finite trees we have nodes and paths
that are in finite trees, for the infinite tree we still
have nodes that are in finite trees, but we have paths
that are not in finite trees.
- William Hughes
calvin
of 0 and 1, both repeating and non-repeating, all assumed to
follow the binary point. Thus it contains all real numbers
between zero and one, and that set of real numbers is
uncountable.
William Elliot
>>>>> Theorem: The complete infinite binary tree has only countably many
>>>>> infinite paths.
>>
>>>> You are trying to prove the impossible as the conjectured theorem is false.
>>
>>> If so, then we have to tolerate the following situation:
>>> The real number sqrt(3) is the limit of a rational sequence.
>>> Every real number is the limit of a rational sequence.
>>> Not every number is the limit of a rational sequence.
>>> Because not all real numbers are the limits of rational sequences.
>>
>> You've contradicted your self. �Thus you've shown by
>> reducio ad absurdum that your proposition is false.
>
> Bare nonsense should be avoided in this thread.
Then shut up.
>>> I do not tolerate that situation.
>>
>> It's extremely obvious to all of us,
>
> That is not a proof of anything.
On the contrary, having removed my full statement
is proof that you can't tolerate being wrong.
>>>> If you want to prove the reals are countable, then consider the Skolem
>>>> paradox that shows that if ZF is consistent, then it has a countable
>>>> model.
>>
>>> Skolem said: �"Erstens habe ich eine genauere Begr�ndung des
>>
>>> The absolute meaning of" uncountable" is not justified on an axiomatic
>>> basis.
>>
>> Then stop proving they're uncountable by using the usual axiomatic basis.-
>
> I do just the contrary. I show they are countable by not using the
> axiomatic basis. Are you an alias of Musatov?
>
That's already been done by Skolem. As for you not using axiomatic
methods, it's quite obvious to all of us but you, that you're rule
of derivation is hand waving non-sense, therefore I'm right. Indeed, that
is not the axiomatic method. Now remember, as according to you, there's to
be no bare nonsense in this thread. Very well then, you can shut up your
oracle of bare nonsense and this thread will no longer blather bare
nonsense.
David C. Ullrich
<ju...@diegidio.name> wrote:
>On 25 May, 12:57, David C. Ullrich <dullr...@sprynet.com> wrote:
>> On Sun, 24 May 2009 22:28:27 -0700 (PDT), LudovicoVan
>>
>> <ju...@diegidio.name> wrote:
>> >[...]
>>
>> >Extending to the infinite case
>>
>> Oops. You can't just "extend" an inequality valid for natural
>> numbers to the infinite case (at least not without justification).
>
>There is no need to justify that inequality in my argument.
Very convenient.
> You might
>enjoy exercising yourself against the final syllogism.
Why? You're claiming to prove something that is easily
seen to be false. Finding one error is sufficient - when
you say "extending to the infinite case" it's clear we've
found at least one error.
And probably "the" error - assuming that we can
simply extend this or that to the infinite case is
precsiely the error in many "proofs" that various
uncountable sets are countable.
>> Look. Here's Ullrich's amazing theorem:
>
>Lovely.
>
>-LV
David C. Ullrich
WM
> On Mon, 25 May 2009 06:00:18 -0700 (PDT), LudovicoVan
>
> <ju...@diegidio.name> wrote:
> >On 25 May, 12:57, David C. Ullrich <dullr...@sprynet.com> wrote:
> >> On Sun, 24 May 2009 22:28:27 -0700 (PDT), LudovicoVan
>
> >> <ju...@diegidio.name> wrote:
> >> >[...]
>
> >> >Extending to the infinite case
>
> >> Oops. You can't just "extend" an inequality valid for natural
> >> numbers to the infinite case (at least not without justification).
>
> >There is no need to justify that inequality in my argument.
>
> Very convenient.
Is tere any need, in your opinon, to justify the followig inequality?
SUM[n --> oo] 2 - 1 - 1 = SUM[n --> oo] 0 =< SUM[n --> oo] 2^-n <
2^aleph_0
> And probably "the" error - assuming that we can
> simply extend this or that to the infinite case is
> precisely the error in many "proofs" that various
> uncountable sets are countable.
Among these errors one of the first class, seems to me, to be Cantor's
theroem and its basic assumption, namely that sqrt(3) = (1,7, 1,73,
1,732, ...). Withou this error, there are no uncountable sets, and
with this error they don't exist either, and there is no need to show
the latter by committing the same error.
Regards, WM
Regards, WM
WM
> The complete binary tree contains all possible infinite sequences
> of 0 and 1, both repeating and non-repeating, all assumed to
> follow the binary point.
If such sequences exist at all, then that statement is certainly true.
> Thus it contains all real numbers
> between zero and one, and that set of real numbers is
> uncountable.
The paths of the binary tree are not more abundant than the countable
set of nodes. This theorem stands firm as you may see from the
helpless attempts to contradict it collected in the original post.
Regards, WM
calvin
> On 26 Mai, 06:38, calvin <cri...@windstream.net> wrote:
>
> > The complete binary tree contains all possible infinite sequences
> > of 0 and 1, both repeating and non-repeating, all assumed to
> > follow the binary point.
>
> If such sequences exist at all, then that statement is certainly true.
We know that many such sequences exist. The repeating
sequences, representing all of the rational numbers
betwen zero and one certainly exist.
And we know that certain non-repeating, but well defined,
sequences exist, such as .01001000100001000001..., so
we know that sequences for certain irrational numbers
betwen zero and one also exist.
What remains are the sequences that are neither repeating
nor well defined (to the extent of our knowledge about
them).
Of these, some are computable (in theory) to any position,
for example the part of pi that exceeds 3. So I think you
have to admit that those sequences exist also.
Finally, what remains are those sequences that are neither
repeating, nor well defined, nor computable (yet). I take
it that those are the sequences whose existence you doubt.
Am I right?
>
> > Thus it contains all real numbers
> > between zero and one, and that set of real numbers is
> > uncountable.
>
> The paths of the binary tree are not more abundant than the countable
> set of nodes. This theorem stands firm as you may see from the
> helpless attempts to contradict it collected in the original post.
That is not a theorem; it is an assertion, which does not
stand firm because it has not been proven. Nor will it
ever be proven, because it is not true.
A Nony Mouse
<95d34fcc-0210-45e2...@g19g2000vbi.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 26 Mai, 12:51, David C. Ullrich <dullr...@sprynet.com> wrote:
> > On Mon, 25 May 2009 06:00:18 -0700 (PDT), LudovicoVan
> >
> > <ju...@diegidio.name> wrote:
> > >On 25 May, 12:57, David C. Ullrich <dullr...@sprynet.com> wrote:
> > >> On Sun, 24 May 2009 22:28:27 -0700 (PDT), LudovicoVan
> >
> > >> <ju...@diegidio.name> wrote:
> > >> >[...]
> >
> > >> >Extending to the infinite case
> >
> > >> Oops. You can't just "extend" an inequality valid for natural
> > >> numbers to the infinite case (at least not without justification).
> >
> > >There is no need to justify that inequality in my argument.
> >
> > Very convenient.
>
> Is tere any need, in your opinon, to justify the followig inequality?
>
> SUM[n --> oo] 2 - 1 - 1 = SUM[n --> oo] 0 =< SUM[n --> oo] 2^-n <
> 2^aleph_0
As that inequality is irrelevant to any property of a complete infinite
binary tree, at least outside of WM's dingbat world, there is no need to
justify that particular irrelevancy.
>
>
> > And probably "the" error - assuming that we can
> > simply extend this or that to the infinite case is
> > precisely the error in many "proofs" that various
> > uncountable sets are countable.
>
> Among these errors one of the first class, seems to me, to be Cantor's
> theroem and its basic assumption, namely that sqrt(3) = (1,7, 1,73,
> 1,732, ...).
For the reals of Cantor's first uncountability theorem, a real is either
an equivalence class of Cauchy sequences or a Dedekind cut, and not
merely a sequence.
> Withou this error
Outside of WM's we wee world. it is not an error, and in that wee wee
world everything is error.
> there are no uncountable sets
In WM's wee wee world, there aren't any sets at all, since they are not
defined.
--
Virgil
A Nony Mouse
<bd757eb7-71e6-423a...@t11g2000vbc.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 26 Mai, 06:38, calvin <cri...@windstream.net> wrote:
> > The complete binary tree contains all possible infinite sequences
> > of 0 and 1, both repeating and non-repeating, all assumed to
> > follow the binary point.
>
> If such sequences exist at all, then that statement is certainly true.
What is not true in WM's wee wee world is quite often true everywhere
else. And what is false everywhere else is whatever WM wants it to be in
his wee wee world.
But those two worlds have very little overlap.
>
> > �Thus it contains all real numbers
> > between zero and one, and that set of real numbers is
> > uncountable.
>
> The paths of the binary tree are not more abundant than the countable
> set of nodes.
Perhaps not WM's trees, but for everyone else's complete infinite binary
trees, there is no sequence counting all paths of such trees.
As Cantor proved, and WM cannot fault that proof.
> This theorem stands firm as you may see from the
> helpless attempts to contradict it collected in the original post.
Then let WM explicitly enumerate ALL the paths of a complete infinite
binary tree, or, equivalently, enumerate ALL infinite binary sequences.
Since he cannot do either, the falseness of his claim is revealed.
WM's error here is to accept the existence of a complete infinite binary
tree, which requires accepting some actually infinite sets.
Once that is accepted, there is no legitimate way to refute the
uncountability of the set of (infinitely long) paths in such a tree.
If WM were consistent, he would simply deny that any complete infinite
binary tree exists at all, along with any other infinite sets.
One can develop a perfectly consistent, though limited, set theory that
way.
If WM were at all logical, he would also not object to stating
any axioms or proper definitions the way he does. His wandering around
in a perpetual fog of ambiguity convinces no one of the virtues of his
position.
--
Virgil
calvin
> ...
> For the reals of Cantor's first uncountability theorem, a real is either
> an equivalence class of Cauchy sequences or a Dedekind cut, and not
> merely a sequence.
That's fine, but once it has been proven that
all the reals can be represented by repeating
or non-repeating decimals (or binaries),
and that all such decimals (or binaries)
represent reals, then it is no longer necessary to
refer to the definitions of the reals when
discussing the binary tree.
Virgil
<f8af62ae-bec2-47a8...@g1g2000yqh.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 24 Mai, 21:01, Marshall <marshall.spi...@gmail.com> wrote:
> > On May 24, 10:57 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > Yes; that is what I should have said. Sometimes this excessive
> > information requirement results in a very large number, but in
> > other cases it could be between 0 and 1.
>
> So it is.
> Pi is defined by many finite formulas.
> But if a finite formula is lacking, as unavoidably must be the case
> for most reals, then it is impossible to define or to use them as
> individuals.
> The decimal expansion of pi or even that of 1/33 = 0.030303... does
> not exist _in the real world_, because not all natural numbers
> required as indices (counting the places) are available, be it in
> written, thought or what ever form.
WM seems to believe that without a decimal expansion, a number cannot
exist, whereas it is only the decimal expansion (a special form of name
for the number) which does not exist.
WM has long been confused about the difference between a name and the
thing named. He could do with a good dose of Korzybski
http://en.wikipedia.org/wiki/Alfred_Korzybski
> >
> > > > Leaving entirely aside what I think of your arguments,
> >
> > > That's deplorable. I think that in particular Cantor's text and my
> > > answer shows the problem in clearest possible way.
> >
> > Well, I would hope to be able to address one issue at a
> > time, rather than be required to discuss everything all at
> > once. Would it work for you if I said "Leaving aside
> > *temporarily* what I think of your arguments ..."
> >
> > > > I notice
> > > > that you never advance any theory of your own. You have never
> > > > provided anything constructive. Always with you it is trying
> > > > to destroy, never to create.
> >
> > > Recently I have written a text book on elementary mathematics.
It is only applied math, which, from the point of view of pure
mathematicians, is mere engineering, and not proper creative mathematics
at all.
> There are two reasons: First set theorists have conquered mathematics
> claming that only mathematics based upon set theory (and only in
> formalized form) is real mathematics.
Category theorists claim all mathematics is based on categories.
Logicians claim all mathematics is based solely on logic.
And so on.
> Second, there are some scholars who try to prove theorems about finite
> numbers by means of infinite cardinals.
Name one.
>
> But in fact, I see that finished infinity is such a hocus pocus that I
> cannot be silent.
WM's alleged potentialy infinite sets are such a hocus pocus that many
of us cannot keep silent.
WM's lack of clear definitions are such a hocus pocus that many of us
cannot keep silent.
WM's lack of any clear statement of the foundation of his "system" is
such a hocus pocus that many of us cannot keep silent.
WM's multitudinous claims without proofs of things which can be
disproved are such a hocus pocus that many of us cannot keep silent.
>
> That is another issue. Of course a machine may be used to find a
> solution of an involved or difficult combinatorical problem. I use
> calculators and Mathematica for many purposes. But I do not believe,
> like some, that a proof of Cantor's theorem, once checked by a
> machine, establishes absolute truth. The machine uses some basics that
> can be false.
No one needs computers to check either proof of Cantor's uncountability
theorems (plural). The "diagonal" proof, in particular, of the
uncountability of the collection of all binary sequences is already too
pellucid to need computerization. And WM hates it because it is so
simple that he cannot find any flaw in it.
>
> > I am curious, can you name any event in math history that
> > would be an example of a success of the type you seek:
> > finding fault with established theory? This is not a rhetorical
> > question; I have some modest interest in math history, and
> > I can't think of an example. But if there is one, I'd be interested
> > to hear of it.
>
> Naysayer Hippasos von Metapont showed that the pythagorean theorem
> "everything is number" (or ratio) was wrong (without introducing
> irrational numbers, he left that to Eudoxos).
The Pythagoreans took "everything is number" as an axion, not as a
theorem provable only derived from other axioms. And finding that axiom
systems have problems occurs several times.
> Naysayer Nicole d'Oresme denied established knowledge, then the Bible,
> and proposed the heliocentric system, contradicting the Bible.
> (Fortunately for him, nobody cared at that time.)
Denying physical assumptions is irrelevant.
> In 1833 Legendre presented to the French academy six proofs of the
> parallel axiom, three of them working with infinite angular areas.
> Naysayer Gauss did not dare to publish his results concerning
> geometry.
At least he chose not to. But a lot of his workt he chose not to publish.
> The other way round: Christian Doppler stated his theory by 1850.
Physics again, irrelevant again.
--
Virgil
Virgil
<7a889015-f4e0-496d...@q16g2000yqg.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 24 Mai, 21:57, Martin Musatov <marty.musa...@gmail.com> wrote:
>
> > Thank you,
> > Martin Musatov
> >
> May I ask what is your opinion with respect to the binary tree?
>
> Regards, WM
MM and WM, two of a kind!!
--
Virgil
Virgil
<f1062213-9d3a-4d0c...@c9g2000yqm.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 25 Mai, 00:42, Marshall <marshall.spi...@gmail.com> wrote:
> > On May 24, 1:54�pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > On 24 Mai, 21:57, Martin Musatov <marty.musa...@gmail.com> wrote:
> >
> > > > Thank you,
> > > > Martin Musatov
> >
> > > May I ask what is your opinion with respect to the binary tree?
> >
> > That guy is just a prankster. His only interest is to wind
> > people up.
> >
> In this case my interest was the same.
>
> Regards, WM
That admission makes WM kookhood undeniable!
--
Virgil
Virgil
<b21cf8a7-1f96-42ef...@h2g2000yqg.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 25 Mai, 08:57, William Elliot <ma...@rdrop.remove.com> wrote:
> > On Sun, 24 May 2009, WM wrote:
> > > Theorem: The complete infinite binary tree has only countably many
> > > infinite paths.
> >
> > > � 0.
> > > � / \
> > > �0 1
> > > / \ / \
> > > 0 10 1
> > > ...
> >
> > You are trying to prove the impossible as the conjectured theorem is false.
>
> If so, then we have to tolerate the follwing situation:
> The real number sqrt(3) is the limit of a rational sequence.
That is only one of the properties of sqrt(3).
> Every real number is the imit of a rational sequence.
Including the rational numbers themselves, but that is only one property
of a number which has other properties.
> Not every number is the limit of a rational sequence.
Because not all numbers are real numbers.
Numbers which are not real need not be limits of rational sequences.
> Because not all real numbers are the limits of rational sequences.
Name one which is not!
>
> I do not tolerate that situation.
Since it is a phony "situation" of your own making, why not?
--
Virgil
Virgil
<22165fae-c1ca-475b...@u8g2000yqn.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 25 Mai, 07:28, LudovicoVan <ju...@diegidio.name> wrote:
> > On 24 May, 16:01, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > During the last years the complete infinite binary tree and its
> > > implications on set theory have been on topic frequently. Here is the
> > > collection of counterarguments that have come to my attention.
> >
> > > If you think there is another counter-argument (or pro-argument), or
> > > if you think to have an argument that improves one of the presented
> > > arguments, please feel free to append it. I would be glad however, if
> > > only serious contributions were added and if all unqualified comments
> > > and statements of pure opinions could be suppressed.
> >
> > A simple counter-argument (I suppose this might be an informal version
> > of your proof A):
> >
> > [tree height] < [num. of paths] < [num. of nodes]
> >
> > � 0 < 0 < 0
> > � 1 < 2 < 2
> > � 2 < 4 < 6
> > � 3 < 8 < 14
> > � 4 < 16 < 30
> > � ...
> > � n < 2^n < 2^(n+1)-2 �(for n > 0)
> > � ...
> >
> > Extending to the infinite case (here the notation might need fixing
> > and I'm not sure if the inequalities should still be strict, but this
> > is irrelevant to the conclusion):
>
> So it is.
> >
> > � w <= 2^w <= 2^(w+1)-2
> >
> > Since the number of nodes is countable, i.e.:
> >
> > � |w| = |2^(w+1)-2|
> >
> > Follows:
> >
> > � |w| = |2^w| = |2^(w+1)-2|
> >
> > IOW, that is:
> >
> > � � The number of nodes is countable;
> > � � The number of paths is dominated by the number of nodes;
> > � � Therefore:
> > � � � � The number of paths is countable.
> >
> I agree. There is another argument. For every appended path we inrease
> the number of nodes by aleph_0, the number of paths by 1. This should
> suddenly switch when finishing the infinite?
To "finish" the job of constructing a complete infinite binary tree in
this way, by appending new paths one at a time, one must a priori assume
that the set of paths being appended is countable, so your argument is
circular and therefore invalid.
> Ad if it cannot be finished, then the whole discussion about
> cardinalities is in vain.
Why? WM misreads the evidence again!
Being unable to finish by adding one path at a time is evidence that the
set of paths cannot be counted out one at a time like the naturals can
be counted.
--
Virgil
Virgil
<f5b171b3-3270-4594...@3g2000yqk.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 25 Mai, 10:58, William Elliot <ma...@rdrop.remove.com> wrote:
> > On Mon, 25 May 2009, WM wrote:
> > > On 25 Mai, 08:57, William Elliot <ma...@rdrop.remove.com> wrote:
> > >> On Sun, 24 May 2009, WM wrote:
> > >>> Theorem: The complete infinite binary tree has only countably many
> > >>> infinite paths.
> >
> > >>> � 0.
> > >>> � / \
> > >>> �0 1
> > >>> / \ / \
> > >>> 0 10 1
> > >>> ...
> >
> > >> You are trying to prove the impossible as the conjectured theorem is
> > >> false.
> >
> > > If so, then we have to tolerate the following situation:
> > > The real number sqrt(3) is the limit of a rational sequence.
> > > Every real number is the limit of a rational sequence.
> > > Not every number is the limit of a rational sequence.
> > > Because not all real numbers are the limits of rational sequences.
> >
> > You've contradicted your self. �Thus you've shown by
> > reducio ad absurdum that your proposition is false.
>
> Bare nonsense should be avoided in this thread.
Then WM should cease posting in it, as that is apparently the only way
to eliminate his nonsense.
> >
> > > I do not tolerate that situation.
> >
> > It's extremely obvious to all of us,
>
> That is not a proof of anything.
Except WM's foolishness.
> >
> > > The absolute meaning of" uncountable" is not justified on an axiomatic
> > > basis.
> >
> > Then stop proving they're uncountable by using the usual axiomatic basis.-
>
> I do just the contrary. I show they are countable by not using the
> axiomatic basis. Are you an alias of Musatov?
WM deludes himself if he thinks that he has established any of his
claims to the satisfaction of anyone but himself. Or perhaps MM.
WM is more like MM than anyone else posting in this thread, he's just
like MM but half upside down.
--
Virgil
Virgil
<7ed54433-7894-45f2...@o30g2000vbc.googlegroups.com>,
LudovicoVan <ju...@diegidio.name> wrote:
Since 1/n > 0 for every n in N, LV seems to be be arguing that
Lim_{n -> oo} 1/n > 0.
--
Virgil
Virgil
<0b5f514a-c010-4b2a...@h28g2000yqd.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 25 Mai, 13:57, David C. Ullrich <dullr...@sprynet.com> wrote:
> > On Sun, 24 May 2009 22:28:27 -0700 (PDT), LudovicoVan
> >
> > <ju...@diegidio.name> wrote:
> > >[...]
> >
> > >Extending to the infinite case
> >
> > Oops. You can't just "extend" an inequality valid for natural
> > numbers to the infinite case (at least not without justification).
> > Things simply don't work that way.
>
> They work that way if irrational numbers are limits, e.g., in case of
> Cantor's "Fundamentalfolgen".
Not to mathematicians, but then we know that WM is mathematically
incompetent.
> >
> > Look. Here's Ullrich's amazing theorem:
> >
> > Theorem: w is finite.
> >
> > Proof: For every natural number n, n is finite.
> > Extending to the infinite case, w is finite. QED.
>
> That is not amazing, it is the simple truth.
Only in WM's weird weird world.
> >
> > This really doesn't have all that much to do with
> > infinity.
>
> It is the only form of infinity (outside of matheology). It is called
> potential infinity.
"Potentially infinite sets do not exist in mathematics, however much WM
may dream of them.
>
> > You're assuming that if every natural number
> > n has a certain property then w must also have that
> > property. That's exactly as valid as assuming that
> > if every natural number n < 10 has a certain property
> > then 10 must also have the same property. Which is
> > to say, not valid at all.
>
> That is purest nonsense.
The nonsense is all WM's.
Consider this: Every natural is smaller that some other natural so that,
according to WM, N must also be smaller than some other natural.
> A valid comparison would be: Showing that every initial segment of N
> is finite shows that N cannot be actually infinite.
Wrong!!!
First of all, not every initial segment of N is finite, since N is an
initial segment of itself which is not finite.
And if one limits oneself to finite initial segments being finite, then
one has to beg the question to conclude that N is finite.
But WM begs most of his questions.
--
Virgil
Virgil
<ca2e1d14-ef47-4839...@r34g2000vba.googlegroups.com>,
LudovicoVan <ju...@diegidio.name> wrote:
> On 25 May, 20:51, William Hughes <wpihug...@hotmail.com> wrote:
> > On May 25, 2:04 pm, LudovicoVan <ju...@diegidio.name> wrote:
> >
> > > To attack it from an other side: given that in the finite (over the
> > > height of the tree) the number of paths is lesser than the number of
> > > nodes, how can we have that the opposite relation holds in the
> > > infinite?
> >
> > Simple, in the infinite tree there are no nodes that cannot
> > be found in some finite tree. �However. in the infinite tree there
> > are many (uncountably many) paths that are not found in any
> > finite tree.
>
> To be sure: isn't a "path" that thing that develops from the root
> along a parent-child trajectory? If so, what are you talking about
> William?
If you can't figure it out, perhaps you should recuse yourself from this
thread.
>
> > Or put another way, the finite paths, the paths found in finite
> > trees, are countable. �It is the infinite paths that are uncountable,
> > and as you note, you only get infinite paths in the infinite tree.
>
> You are have restated the thesis: your thesis, for which I still see
> no support.
One can easily biject the set of paths of the complete infinite binary
tree with the set of infinite binary strings, and Cantors original
diagonal proof clearly shows that any attempt to count those binary
strings must fail.
>
> -LV
--
Virgil
lwa...@lausd.net
> In article
> <f5b171b3-3270-4594-8894-2ef663df0...@3g2000yqk.googlegroups.com>,
With the old thread having passed the Google limit of a thousand
posts WM has started a new thread. But instead of continuing with
the discussion of Potential Infinity, which I believe could be an
interesting theory, WM has gone back to the Binary Tree. The
theory that a binary tree could be countable could also be an
interesting theory, but it's already been pointed out that WM
doesn't believe in actually infinite sets, including the infinite
binary tree, and this is actually intended as an attempt to prove
that ZFC is inconsistent.
Since I don't accept that the inconsistency of ZFC, if it's ever
proved, will be proved using binary trees, I have no further
comment on binary trees.
But I've noticed that a new so-called "crank" has appeared in
this thread, namely Martin Musatov. Most of Musatov's previous
posts attempt to prove either FLT or P=NP, and thus his posts are
similar in tone to JSH's. I defend neither JSH nor Musatov.
And so it seems interesting that WM would accuse one of the
standard theorists as being an alias of Musatov, which is
followed by:
> WM is more like MM than anyone else posting in this thread, he's just
> like MM but half upside down.
So this Musatov is obviously someone so "cranky" that not even
the other "cranks" want to be associated with him, and WM
associates Musatov with his other opponents. Musatov is likely
an example of what Dudley, the authority on "cranks," would
consider to be "crazy," which is beyond being a "crank."
Jesse F. Hughes
> So this Musatov is obviously someone so "cranky" that not even
> the other "cranks" want to be associated with him, and WM
> associates Musatov with his other opponents. Musatov is likely
> an example of what Dudley, the authority on "cranks," would
> consider to be "crazy," which is beyond being a "crank."
I doubt it. Martin seems to be a troll, just fooling around to see
how much attention he can generate. One hint: his first P=NP post was
a recycled years-old April Fools joke by a different poster. That's
not the behavior of a crank. That's the behavior of a practical
joker.
(His second joke, involving a link to an unreadable text which he
pretended contained a coherent argument, was much more original and
hence funnier. He had me fooled until I realized he had earlier
recycled that April Fool's post.)
--
Jesse F. Hughes
"The people who made up the words could have said 'newspaper' is
'trees'." -- Quincy P. Hughes, five-year-old Wittgensteinian
(This comment came out of the blue at breakfast.)
Martin Musatov
Marshall wrote:
> On May 24, 1:54 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 24 Mai, 21:57, Martin Musatov <marty.musa...@gmail.com> wrote:
> >
> > > Thank you,
> > > Martin Musatov
> >
> > May I ask what is your opinion with respect to the binary tree?
>
> That guy is just a prankster. His only interest is to wind
> people up.
>
>
> Marshall
Contrary to Marshall's supposition I am not a prankster and no, my
only goal is not to wind people up.
Appreciatively with great conviction, Musatov
___________________
P=NP has become a metaphor for equality.
Martin Musatov
Why do you call me that? I am merely testing the limits of a theory. --
MMM
lwa...@lausd.net
> lwal...@lausd.net wrote:
> > So this Musatov is obviously someone so "cranky" that not even
> > the other "cranks" want to be associated with him, and WM
> > associates Musatov with his other opponents. Musatov is likely
> > an example of what Dudley, the authority on "cranks," would
> > consider to be "crazy," which is beyond being a "crank."
> Why do you call me that? I am merely testing the limits of a theory. --
Welcome to sci.math. Since you are new here, let me explain
a few things to you.
Lately, there has been a war taking place in many different
threads here on sci.math. There are two sides to this debate,
namely the "bullies" and the "cranks." The so-called "cranks"
are those who reject some part of standard theory, while the
"bullies" are those who call the "cranks" out for it.
You joined sci.math right in the middle of this war. Now
where do you fit in this conflict?
Virgil is on the side of the "bullies." Naturally, he has
already belittled your ideas. But what I found surprising was
that WM, on the side of the "cranks," is also making fun of
your ideas. WM criticized William Hughes, on the "bully" side
of the great debate, as being you in disguise. Then Virgil
compared WM to you.
And so both Virgil and WM consider you to be on their
opponent's side of the debate. Neither one of them considers
you to be their ally. And so I wondered whether you were too
"crankish" even for the "cranks."
Whose side am I on in this debate? I've been referred to as a
"meta-crank," because I've expressed disagreement with the
categorization of certain posters as "cranks."
I admit that I was wrong to call you "crazy." Indeed, Jesse
Hughes (on the "bully" side of the debate) has stated that
the word "troll" describes you better. Indeed, this isn't the
first time that I've gotten in trouble for initiating the use
of one of those five-letter words in a thread ("crank,"
"troll" and now "crazy"). In the future, I must be careful to
let the "bullies" be the ones to initiate the use of those
five-letter words -- after all, use of those words is why I
call them "bullies" in the first place. So "bully" is the only
five-letter word that I'll initiate in a thread (although the
first to use the word was actually galathaea).
WM
> (the difference being that there are no "infinite nodes"
> but there are infinite paths)
Oh yeah. There are paths stretching even into that domain where no
nodes exist.
And, oh wonder, they can even branch out in that vacum.
Matheology at its greatest!
Regards, WM
WM
> On May 25, 2:04 pm, LudovicoVan <ju...@diegidio.name> wrote:
>
> > To attack it from an other side: given that in the finite (over the
> > height of the tree) the number of paths is lesser than the number of
> > nodes, how can we have that the opposite relation holds in the
> > infinite?
>
> Simple, in the infinite tree there are no nodes that cannot
> be found in some finite tree. However. in the infinite tree there
> are many (uncountably many) paths that are not found in any
> finite
> trees, are countable. It is the infinite paths that are uncountable,
In consequence you are arguing that the cardinal number of the path
(infinite binary sequence) 1/pi, i.e., |{1/pi}|, is larger than the
cardinal number of the set of its finite initial segments (because
they all can be found in finite trees). So we have
|{1/pi}| = 1 > aleph_0
Again a fine result! And so counterintuitive!! Logicians will love
it!!!
Regards, WM
WM
> On 25 May, 20:51, William Hughes <wpihug...@hotmail.com> wrote:
>
> > On May 25, 2:04 pm, LudovicoVan <ju...@diegidio.name> wrote:
>
> > > To attack it from an other side: given that in the finite (over the
> > > height of the tree) the number of paths is lesser than the number of
> > > nodes, how can we have that the opposite relation holds in the
> > > infinite?
>
> > Simple, in the infinite tree there are no nodes that cannot
> > be found in some finite tree. However. in the infinite tree there
> > are many (uncountably many) paths that are not found in any
> > finite tree.
>
> along a parent-child trajectory? If so, what are you talking about
> William?
>
> > trees, are countable. It is the infinite paths that are uncountable,
>
> You are have restated the thesis: your thesis, for which I still see
> no support.
Small wonder. There is no support. Consider my Proof A. Of course I
use infinite paths and append each one to a node. This establishes the
fact hat there are not more paths than nodes.
Regards, WM
herbzet
lwa...@lausd.net wrote:
> So "bully" is the only
> five-letter word that I'll initiate in a thread.
Obviously, that makes you a bully.
A very conflicted bully.
--
hz
WM
> On May 26, 7:56 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > On 26 Mai, 06:38, calvin <cri...@windstream.net> wrote:
>
> > > The complete binary tree contains all possible infinite sequences
> > > of 0 and 1, both repeating and non-repeating, all assumed to
> > > follow the binary point.
>
> > If such sequences exist at all, then that statement is certainly true.
>
> We know that many such sequences exist. The repeating
> sequences, representing all of the rational numbers
> betwen zero and one certainly exist.
>
> And we know that certain non-repeating, but well defined,
> sequences exist, such as .01001000100001000001..., so
> we know that sequences for certain irrational numbers
> betwen zero and one also exist.
>
> What remains are the sequences that are neither repeating
> nor well defined (to the extent of our knowledge about
> them).
>
> Of these, some are computable (in theory) to any position,
> for example the part of pi that exceeds 3. So I think you
> have to admit that those sequences exist also.
Disregarding physical constraints, as we will do here, I fully agree.
These numbers exist. The set of all of them is countable.
>
> Finally, what remains are those sequences that are neither
> repeating, nor well defined, nor computable (yet).
There is no yet required, even in parentheses. The set of all finite
alphabets is countable. The set of all words is countable. The set of
all languages is countable. This proves that the set of all numbers
that can be computed or individually identified is countable too ---
in spite of the fact that the contrary bis believed by some Fools Of
Matheology.
This set (including all existing alphabets, dictionaries, provable
theorems, and threads of sci.logic) is completely written here:
0
1
00
01
10
11
...
(better: would be completely written here, if it was possible to
complete an infinite set.) Of course there is no diagonalization
possible.
> I take
> it that those are the sequences whose existence you doubt.
> Am I right?
You are right.
>
>
>
> > > Thus it contains all real numbers
> > > between zero and one, and that set of real numbers is
> > > uncountable.
>
> > The paths of the binary tree are not more abundant than the countable
> > set of nodes. This theorem stands firm as you may see from the
> > helpless attempts to contradict it collected in the original post.
>
> That is not a theorem; it is an assertion, which does not
> stand firm because it has not been proven. Nor will it
> ever be proven, because it is not true.
It is easy to specify attributes of non existing things and then to
believe in them. Nevertheless, in this case, this claim can be
disproved based on the following fact:
If Cantor's proof is correct, then an infinite path is the limit of
its finite initial segments.
But not all paths are the limits of their finite initial segments.
Of course you need not believe that this is a contradiction. You need
not even believe that 1 > 2 is a contradiction or SUM(N) = 0 or 0 >
oo.
But I do so, and I encourage all mathematicians and non-mathematicians
to join this position.
Regards, WM
WM
> In article
> <95d34fcc-0210-45e2-836f-9e2194d20...@g19g2000vbi.googlegroups.com>,
>
>
>
>
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 26 Mai, 12:51, David C. Ullrich <dullr...@sprynet.com> wrote:
> > > On Mon, 25 May 2009 06:00:18 -0700 (PDT), LudovicoVan
>
> > > <ju...@diegidio.name> wrote:
> > > >On 25 May, 12:57, David C. Ullrich <dullr...@sprynet.com> wrote:
> > > >> On Sun, 24 May 2009 22:28:27 -0700 (PDT), LudovicoVan
>
> > > >> <ju...@diegidio.name> wrote:
> > > >> >[...]
>
> > > >> >Extending to the infinite case
>
> > > >> Oops. You can't just "extend" an inequality valid for natural
> > > >> numbers to the infinite case (at least not without justification).
>
> > > >There is no need to justify that inequality in my argument.
>
> > > Very convenient.
>
> > Is tere any need, in your opinon, to justify the followig inequality?
>
> > SUM[n --> oo] 2 - 1 - 1 = SUM[n --> oo] 0 =< SUM[n --> oo] 2^-n <
> > 2^aleph_0
>
> As that inequality is irrelevant to any property of a complete infinite
> binary tree, at least outside of WM's dingbat world, there is no need to
> justify that particular irrelevancy.
You need not emphasize the fact that you have not yet understood.
>
>
>
> > > And probably "the" error - assuming that we can
> > > simply extend this or that to the infinite case is
> > > precisely the error in many "proofs" that various
> > > uncountable sets are countable.
>
> > Among these errors one of the first class, seems to me, to be Cantor's
> > theroem and its basic assumption, namely that sqrt(3) = (1,7, 1,73,
> > 1,732, ...).
>
> For the reals of Cantor's first uncountability theorem, a real is either
> an equivalence class of Cauchy sequences or a Dedekind cut, and not
> merely a sequence.
For Cantor and his theorem a real is the limit of a sequence, called
Fundamentalreihe.The diagonal of an infinite list is certainly the
limit of a sequence.
Regards, WM
Regards, WM
Herbert Newman
> I admit that I was wrong to call you "crazy."
This may indeed be the case. But imho it's not wrong (but correct) to call
YOU crazy and/or a crank. :-)
Herb
Aatu Koskensilta
> This may indeed be the case. But imho it's not wrong (but correct) to
> call YOU crazy and/or a crank. :-)
What's the point of this smiley? All this shouting at people, calling
them assholes and what not, apparently makes you very happy, but perhaps
you could keep such things to yourself in the future, smiling contently
in private.
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon mann nicht sprechen kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
WM
>
>
>
>
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 24 Mai, 21:01, Marshall <marshall.spi...@gmail.com> wrote:
> > > On May 24, 10:57 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > Yes; that is what I should have said. Sometimes this excessive
> > > information requirement results in a very large number, but in
> > > other cases it could be between 0 and 1.
>
> > So it is.
> > Pi is defined by many finite formulas.
> > But if a finite formula is lacking, as unavoidably must be the case
> > for most reals, then it is impossible to define or to use them as
> > individuals.
> > The decimal expansion of pi or even that of 1/33 = 0.030303... does
> > not exist _in the real world_, because not all natural numbers
> > required as indices (counting the places) are available, be it in
> > written, thought or what ever form.
>
> WM seems to believe that without a decimal expansion, a number cannot
> exist,
Virigil seems to be incapable to understand written text. I wrote
above:
"if a finite formula is lacking".
> whereas it is only the decimal expansion (a special form of name
> for the number) which does not exist.
You are wrong. I confess that 0.030303... = 1/33 = one divided by 33 =
3/99 does exist. What does not exist, in MatheRealism, is the complete
decimal expansion. (Because there is a first place that cannot be
decided whether it is occupied by 0 or 3, because its index natural
number cannot be expressed in the real world --- nor can the digits of
1/33 up to that place).
> > There are two reasons: First set theorists have conquered mathematics
> > claming that only mathematics based upon set theory (and only in
> > formalized form) is real mathematics.
>
> Category theorists claim all mathematics is based on categories.
> Logicians claim all mathematics is based solely on logic.
> And so on.
But the truth is, that mathematics is based on natural numbers,
geometrical notions and logical rules (I mean the correct logic, of
course).
>
> > Second, there are some scholars who try to prove theorems about finite
> > numbers by means of infinite cardinals.
>
> Name one.
Goodstein, H. Friedman.
>
> > Naysayer Hippasos von Metapont showed that the pythagorean theorem
> > "everything is number" (or ratio) was wrong (without introducing
> > irrational numbers, he left that to Eudoxos).
>
> The Pythagoreans took "everything is number" as an axiom, not as a
> theorem provable only derived from other axioms. And finding that axiom
> systems have problems occurs several times.
Yes, most recently I proved the axiom of infinity is wrong.
>
> > Naysayer Nicole d'Oresme denied established knowledge, then the Bible,
> > and proposed the heliocentric system, contradicting the Bible.
> > (Fortunately for him, nobody cared at that time.)
>
> Denying physical assumptions is irrelevant.
Denying assumptions is commendable in every domain, if the assumptions
are wrong.
>
> > The other way round: Christian Doppler stated his theory by 1850.
>
> Physics again, irrelevant again.
Not at all. Dopplers formulas are mathematics describing waves that
are described by mathematics. 1852 Petzval, one of the leading
Austrian mathematicians at that time, wrote: Dopplers theory cannot be
said to be without value because it has a negative value in that it
detracts so many intelligent minds from serious reasearch. It is
superficial and lacks profoundness. Petzval wanted to describe the
effect by a set of coupled differential equations.
Dopplers priciple is about as lucid as the binary tree.
Regards, WM
Herbert Newman
> Herbert Newman <nomail@invalid> writes:
>>
>> This may indeed be the case. But imho it's not wrong (but correct) to
>> call YOU crazy and/or a crank. :-)
>>
> What's the point of this smiley?
It indicates that I was happy when writing that sentence. :-)
(You know, simple truths make me happy.)
Herb
P.S. I guess it's time to killfile you again. :-)
WM
>
>
>
>
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 25 Mai, 08:57, William Elliot <ma...@rdrop.remove.com> wrote:
> > > On Sun, 24 May 2009, WM wrote:
> > > > Theorem: The complete infinite binary tree has only countably many
> > > > infinite paths.
>
> > > > 0.
> > > > / \
> > > > 0 1
> > > > / \ / \
> > > > 0 10 1
> > > > ...
>
> > > You are trying to prove the impossible as the conjectured theorem is false.
>
> > If so, then we have to tolerate the follwing situation:
> > The real number sqrt(3) is the limit of a rational sequence.
>
> That is only one of the properties of sqrt(3).
>
> > Every real number is the imit of a rational sequence.
> > Because not all real numbers are the limits of rational sequences.
>
> Name one which is not!
>
all paths is countable. The set of sequences of finite paths is
countable, because all finite paths are countable and the sequences
generated by them cannot surpass the set of paths continued in the
union of the sequence of all finite trees.
>
>
> > I do not tolerate that situation.
>
> Since it is a phony "situation" of your own making, why not?
It is of Cantor's making.
He recognized that the complete infinite binary tree is not the limit
of the complete infinite zsequence of finite binary trees.
Regards, WM
Aatu Koskensilta
> P.S. I guess it's time to killfile you again. :-)
Apparently the thought brings you great happiness, for some reason. Go
for it, then.
WM
> > I agree. There is another argument. For every appended path we inrease
> > the number of nodes by aleph_0, the number of paths by 1. This should
> > suddenly switch when finishing the infinite?
>
> To "finish" the job of constructing a complete infinite binary tree in
> this way, by appending new paths one at a time, one must a priori assume
> that the set of paths being appended is countable,
That is proved by the set of nodes being countable.
> so your argument is
> circular and therefore invalid.
No. The argument could show that the construction is insufficient to
cover the complete infinite binary tree. Then we would have another
proof of uncountability of the reals. Alas, I can show that there is
no node of any path missing. Hence, there is no path missing.
>
> > Ad if it cannot be finished, then the whole discussion about
> > cardinalities is in vain.
>
> Why? WM misreads the evidence again!
>
> Being unable to finish by adding one path at a time is evidence that the
> set of paths cannot be counted out one at a time like the naturals can
> be counted.
That is complete rubbish. The naturals cannot be counted out in that
manner either. But being able to finish by adding one path in half
time of the previous one, is evidence that the set of paths is
countable. (Even doing it in 99 % of time of the previous one would
show this.)
Regards, WM
WM
> But instead of continuing with
> the discussion of Potential Infinity, which I believe could be an
> interesting theory, WM has gone back to the Binary Tree.
There I show that only potential infinity is free of contradictions.
>
> Since I don't accept that the inconsistency of ZFC, if it's ever
> proved, will be proved using binary trees,
Why? Any rational reason?
Regards, WM
Martin Musatov
You simply make no sense. And I say this respectfully as an academic
challenge. The entire sect of "math culture" marvels at its
limitations like a snake charmer marvels at a snake. For surely, "We
cannot know...", or "It has been shown you cannot count...",
seriously, really, why? I have no good answers to why sets are deemed
"uncountable", why the binary representation of the word "i-n-f-i-n-i-
t-e" has to carry the same weight as the word, when literally all it
means to a computer is the processing of eight letters, which varies
by processor speed. Seriously, mathematics is full of idiots who
belittle newcomers, who insult anyone who disagrees, and then winks at
each other like they are in some sort of cool club only those who
worship idiots like Godel and insist on impossibilities as cornerstone
behaviors. Seriously, and again, I say this with all due respect and
exclusion to the exceptions (God bless them), the behavior of the
mathematics community makes me sick and you should all grow up, be
civil, and explain things with reasonable answers instead of hiding
bejind terms of vague proportion and recititng status quoe theories
like Parrots on the shoulders of complexity pirates raking the good
soil with crude discourse and ill-behavior. Mathematics, grow up,
start making plain sense, and stop defending relics just because you
appreciate the plunder.
Affirmed in every way, Martin Musatov
Martin Musatov
You simply make no sense. And I say this respectfully as an academic
challenge. The entire sect of "math culture" marvels at its
limitations like a snake charmer marvels at a snake. For surely, "We
cannot know...", or "It has been shown you cannot count...",
seriously, really, why? I have no good answers to why sets are deemed
"uncountable", why the binary representation of the word "i-n-f-i-n-i-
t-e" has to carry the same weight as the word, when literally all it
means to a computer is the processing of eight letters, which varies
by processor speed. Seriously, mathematics is full of idiots who
belittle newcomers, who insult anyone who disagrees, and then winks at
each other like they are in some sort of cool club only those who
worship idiots like Godel and insist on impossibilities as cornerstone
behaviors. Seriously, and again, I say this with all due respect and
exclusion to the exceptions (God bless them), the behavior of the
mathematics community makes me sick and you should all grow up, be
civil, and explain things with reasonable answers instead of hiding
bejind terms of vague proportion and recititng status quoe theories
like Parrots on the shoulders of complexity pirates raking the good
soil with crude discourse and ill-behavior. Mathematics, grow up,
start making plain sense, and stop defending relics just because you
appreciate the plunder.
Affirmed in every way, Martin Musatov
P.S. What the "f" is "potential infinity". Finite or infinite is like
being pregnant, get it? You are or you are not, but there is no
middle. You annoy the patience out of even me sometimes.
Martin Musatov
You are totally covering your a!s. While I of course appreciate the
brief and civil conversation always appreciated, do you think I really
who was the first to use the word "bully" on Sci.Math or whether their
name was galathaea? No, I do not. It is about the mathematics and I
am tired from those who try to confuse it and distract from it by
juggling their little laughs, wink, and insult variables to dillute
truth in a sorry attempt to save what little knowledge they have,
though they cling to it and pride themselves upon it like gold
sheathe. If I post a proof of FLT or P=NP people do not respond with
legitimate critiques, instead they insult or steer the conversation to
some unrelated personal conjecture unrelated to the mathematics
problem at hand. Why is this behavior so consistent? Could it be the
puppet master atop the polynomial "math" hierarchy might be
benefitting from this juvenile "war"? Sure, it could and I suggest it
is. Mathematics is power, mathematics controls information, influences
innovation, makes wealth, keeps secrets, rules economies, and governs
how we look at basic human development including biology and medicine!
Now I did not like Math in high school and college. I took Math for
Liberal Arts and was a Finance turned Screenwriting major, but
something drew me back into the realm of logic and reason on a very
basic level and I was shocked to see the condition it was in. I am not
a parent or an old "crank" (I detest that term and insist anyone who
uses it has no place in mathematics), I am 30 years old, and work as a
web developer and freelance writer. Who is pulling the strings on the
Hebrew old world Jewish variable laden hierarchy is what I would like
to know. I am not a racist but when I post a proof P=NP on a math
board and suddenly my name is being dragged across a Nazi hate thread
calling me "Fartin Martin" Satan-vomit and accusing me of being
aligned with such ignorance, and suddenly my proof is posted on an
anti-Jewish thread on an Israeli website talking about Hamas bombing
Israel, I want to know: why?
MoeBlee
> Lately, there has been a war taking place in many different
> threads here on sci.math.
Lately, as usual for at least the last ten years or so.
> There are two sides to this debate,
> namely the "bullies" and the "cranks." The so-called "cranks"
> are those who reject some part of standard theory, while the
> "bullies" are those who call the "cranks" out for it.
Would you PLEASE stop lying?!
We don't call the cranks out merely for rejecting ZFC, but rather for
their illogical, uninformed, vague, circular, dogmatic, stubborn, etc.
MANNER of arguing about the subject.
This has been pointed out to you over and over and over, yet you
persist in the canard that we "bully" cranks merely for some
intellectual stance they have in opposing ZFC.
> Whose side am I on in this debate? I've been referred to as a
> "meta-crank," because I've expressed disagreement with the
> categorization of certain posters as "cranks."
NO! You did it AGAIN! You're called a crank about the subject of
cranks because your ARGUMENTS on the subject are illogical and
unresponsive, not just because you disagree with the categorization.
MoeBlee
Virgil
<df036122-6a08-48c4...@f19g2000yqo.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
Since such "matheology" only exists in WM's tiny mind, no matter.
The difficulty is that WM wants somehow to sneak in countable
infiniteness so as to "construct" his version of an infinite binary
tree, but does not realize that allowing arbitrary countably infinitely
long paths automatically and necessarily allows and even necessitates
uncountably many of them.
AS long as WM sticks to strictly finite paths, his trees may behave as
he wants them to, but once he allows infinite paths, things are no
longer under his control, but take on a life of their own.
--
Virgil
Virgil
<ec0198fe-8be4-4754...@s12g2000yqi.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 25 Mai, 21:51, William Hughes <wpihug...@hotmail.com> wrote:
> > On May 25, 2:04 pm, LudovicoVan <ju...@diegidio.name> wrote:
> >
> > > To attack it from an other side: given that in the finite (over the
> > > height of the tree) the number of paths is lesser than the number of
> > > nodes, how can we have that the opposite relation holds in the
> > > infinite?
> >
> > Simple, in the infinite tree there are no nodes that cannot
> > be found in some finite tree. �However. in the infinite tree there
> > are many (uncountably many) paths that are not found in any
> > finite tree. �Or put another way, the finite paths, the paths found in
> > finite
> > trees, are countable. �It is the infinite paths that are uncountable,
>
> In consequence you are arguing that the cardinal number of the path
> (infinite binary sequence) 1/pi
What makes WM think that any path has a cardinal number that is
different from the cardinal of any other path?
Having a cardinal number is a property of a set, If one regards paths as
sets f odes or sets of edges, or almost any sort of set, then in a
complete tree, whether finite or infinite, all paths have the same
cardinal.
So it os only WM, with his deliberate ambiguity, that can refer to the
"cardinal" of a real number like 1/pi as if it were different from the
cardinal of any other real number.
> i.e., |{1/pi}|, is larger than the
> cardinal number of the set of its finite initial segments (because
> they all can be found in finite trees). So we have
>
> |{1/pi}| = 1 > aleph_0
>
> Again a fine result! And so counterintuitive!! Logicians will love
> it!!!
Logicians will first see that it only occurs in WM's anti-mathematics,
but not in any real mathematics. Then they may be amused by it.
--
Virgil
Virgil
<42641228-0c3b-4e4e...@j18g2000yql.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
That would only be the case if no more than one path need be associated
with any node, but, in fact, there is NO surjection from nodes to paths.
As Cantor proved.
If WM claims otherwise, let him display such a surjection.
Cantor displays sufficient reason do show none exists.
>
> Regards, WM
--
Virgil
Martin Musatov
Critique the mathematics:
P=NP marty....@gmail.com - April 1st, 2009 - (view article)
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http://knowledgerush.com/kr/jsp/db/board.jsp?id=39473
A Rippling Proof of P = NPA Rippling Proof of P = NP. DR. E. A. M
group. May 31, 2005. P = NP. We proceed by induction on the polynomial
P. Base case trivial (by symbolic evaluation) ...http://
dream.inf.ed.ac.uk/events/liber-amicorum-2005/rebundy/dream3.pdf - -
Cached - Similar pages
http://dream.inf.ed.ac.uk/events/liber-amicorum-2005/rebundy/dream3.pdf
Stepcase:
ARipplingProofofP=NP
DR
E
AMgroup
May31,2005
P=NP
WeproceedbyinductiononthepolynomialP
Basecasetrivial(bysymbolicevaluation)
ax
P
"
=
N
ax
P
"
Ripplinginthestepcaseisblocked.ProceedbycasesplitonN
Zerobranch
Weakfertilise
ax(NP)=0
(ax
P)
"
Speculatelemmaassociativityoftimes-whichinstantiatesNto0
Nowtreatsucessorbranch
axnnnn...
axnnnn...
axnnnn...
P
"
P
"
P
"
=
=
=0(
axnnnn...nn.n...
P=P
whichfollowsbyreflexivityofequality¤
axnnn...
P
"
P
"
(Suc0)axnnn...
)
P
"
8XPM9-7F9HD-4JJQP-TP64Y-RPFFV
762HW-QD98X-TQVXJ-8RKRQ-RJC9V
http://mathforum.org/kb/thread.jspa?threadID=1929733&tstart=0
2*2=4 and 3*3=9 then 4*9=36+1=37 is prime.
4*4=16 and 9*9=81 then 16*81=1296+1=1297 is prime.
16*16=256 and 81*81=6561 then 256*6561=1679616+1=1679617 is prime.
256*256=65536 and 6561*6561=43046721 then
65536*43046721=2821109907456+1=2821109907457 is prime.
Musatov's theorem for Constant Prime: (2^2)*(3^2)+1.
Successive squaring as above.
I Just Proved [P=NP] and I get to announce it on
Usenet.
Source: http://coding.derkeiler.com/Archive/General/comp.theory/200904/msg00122.html
From: Martin Michael Musatov <marty.musatov@xxxxxxxxx>·
Date: Sun, 26 Apr 2009 07:35:05 0700 (PDT)·
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Date: 1 Apr 2004 10:30:59 0800
Making use of a new type of modeltheoretic tool the Boolean Sieve
we have been able to construct a Ptime algorithm for SAT, thus
providing a resolution to one of the most famous, longstanding open
problems of Theoretical Computer Science. A detailed, but accessible
and informal, general overview of the Boolean Sieve method (more
information can be found here by carrying out a Google groups search
under "Boolean Sieve" and "Mathematician's Algorithm"). However, a
brief description will be provided below of the method, some
applications outside the specific context of SAT, as well as an
overview of how it was applied to SAT. Opportunity providing, an
abstract or possibly even an online copy of the submitted paper (just
accepted for publication) will be made available at the above Web
site.
What is a Boolean Sieve? Basically, it is a construct that is
generated from a set of models, for an axiomfree theory ("free
theory"), that are defined to filter out the possible logical
relations between a set of statements which could be rendered in that
theory. A possible application may be to seek out significant
axiomatizations that may be applied to the set of operations and
predicates in the underlying free theory. The term "filter" is more
than appropriate given the nature of the formal machinery behind the
method.
For instance, consider Group Theory. An interesting (but not well
known) fact is that groups can be defined by their inverse operation
I Just Proved [P=NP] and I get to announce it on Usenet.
I Just Proved [P=NP] and I get to announce it on Usenet.1
(division), just as well by multiplication. The underlying free theory
is an algebraic sort with the following set of operations:
() |> 1 (identity)
(a, b) |> a/b (quotient)
So, it then becomes natural to ask: what are the logical relations
between the possible statements that could be made over the underlying
free theory. Such a situation is precisely the kind of circumstance
where one would use the Boolean Sieve method.
What one does is write down a bunch of statements (ideally, including
a set of statements that we already know from prior considerations
would completely characterize a group), and then select a bunch of
models for the free theory (which in the case at hand may or may not
actually be groups). Each model should have the property that each
statement has a truth value whose evaluation in that model can be done
"efficiently".
The result is a set of raw data from which a profile can be assembled.
The method of integrating all the basic facts is the Boolean Sieve,
itself. The result of applying the Sieve is an efficient
characterization, as a set of Horn clauses, of the Boolean lattice
generated by the statements. >From there (for instance) one could read
off the significant relations and possible axiomatizations, e.g.,
(a/c)/(b/c) = a/b; a/a = 1; a/1 = a
or for Abelien groups:
a(bc) = c(ba); a(ab) = b; 11 = 1.
More generally, a Boolean Sieve will allow us to filter out the
possible relations between a set of statements. The Sieve is called
Complete for that set, if all possible relations are constructed by
the Sieve. What we've actually done is resolve a generalization of SAT
(i.e., determine the validity of a Horn clause involving Boolean
formulas over Nvariables) by defining a process (that is N^3 in
complexity) that generates a complete Boolean Sieve that is N^3 in
size.
Why N^3? Well, this is where it gets interesting: the method for
generating the complete Boolean Sieve is essentially a disguised
version of the Earley parsing algorithm for contextfree grammars! The
significance and nature of this link remains a total mystery to us.
Currently, we are investigating extensions of the Boolean Sieve which
will provide a basis for modeltheoretic theorem proving methods or
"Semantic Theorem Proving". As any expert mathematician will be able
to relate, such an appropach has a far more direct bearing on the way
mathematicians actually approach problems. They will take a stock set
of examples, run a set of possible statements through the examples
(oftentimes subconsciously) and "magically" arrive at a set of
conjectures. We conjecture that the latent method behind this process
is none other than the Boolean Sieve, itself. We even speculate that
"mathematical intuition", itself, may be nothing more than the by
product of this subconscious process. Thus, for instance, one could
develop a more honed "intuition" by having a larger stock of ready
made examples "under the belt", so to say.
Needless to say, these developments will go far beyond the specifics
of the P = NP problem, as most anyone would have been able to
I Just Proved [P=NP] and I get to announce it on Usenet.
I Just Proved [P=NP] and I get to announce it on Usenet.2
Defending truth firmly, no-nonsense truth well-told, Martin Michael
Musatov
WM
> AS long as WM sticks to strictly finite paths, his trees may behave as
> he wants them to, but once he allows infinite paths, things are no
> longer under his control, but take on a life of their own.
The only condition to stick to is that paths do not exist without
nodes. Paths simply are sequences of nodes. This becomes clear by the
fact that every node contributes one bit to at least one of the 2^N
sequences. Cantor claims that the number of these sequences is
uncountable. He claims further that the limit of the sequence of
finite trees is not the infinite tree. (By the way: What is the
difference between the infinite tree and the union of all finite
trees? Nothing.) But he claims that pi is the limit of the finite
initial segments. These two claims suggest the question: How many
paths are limits in the tree and why not less or more. This number,
according to Cantor, must be X with 1 =< X < 2^N. How to determine it?
Regards, WM
Martin Musatov
Martin Musatov wrote:
> lamb...@inter.net wrote:
> > On May 26, 12:25 pm, Virgil <virg...@nowhere.com> wrote:
> > > In article
> > > <f5b171b3-3270-4594-8894-2ef663df0...@3g2000yqk.googlegroups.com>,
> > > WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > On 25 Mai, 10:58, William Elliot <ma...@rdrop.remove.com> wrote:
> > > > > Then start proving they're *countable* by using the usual axiomatic basis.+
> > > > I do just this. I show they are countable by not using the
> > > > axiomatic basis. Are you an alias of Musatov? No I am Musatov.
> > > WM deludes himself if he thinks that he has established any of his
> > > claims to the satisfaction of anyone but himself. Or perhaps MMM+
> >
> > With the old thread having passed the Google limit of a thousand
> > posts WM has started a new thread. But instead of continuing with
> > the discussion of Potential Infinity, which I believe could be an
> > interesting theory, WM has gone back to the Binary Tree. The
> > theory that a binary tree could be countable could also be an
> > interesting theory, but it's already been pointed out that WM
> > doesn't believe in actually infinite sets, including the infinite
> > binary tree, and this is actually intended as an attempt to prove
> > that ZFC is inconsistent.
> >
> > Since I don't accept that the inconsistency of ZFC, if it's ever
> > proved, will be proved using binary trees, I have no further
> > comment on binary trees.
> >
> > But I've noticed that a new so-called "crank" has appeared in
> > this thread, namely Martin Musatov. Most of Musatov's previous
> > posts attempt to prove either FLT or P=NP, and thus his posts are
> > similar in tone to JSH's. I defend neither JSH nor Musatov. I am similar in tone to none.
> >
> > And so it seems interesting that WM would accuse one of the
> > standard theorists as being an alias of Musatov, which is
> > followed by:
> >
> > > WM is more like MMM than anyone else posting in this thread, he's just
> > > like MMM but half upside down.
> >
> > So this Musatov is obviously someone so "cranky" that not even
> > the other "cranks" want to be associated with him, and WM
> > associates Musatov with his other opponents. Musatov is likely
> > an example of what Dudley, the authority on "cranks," would
> > consider to be "crazy," which is beyond being a "crank."
>
> Why do you call me that? I am merely testing the limits of a theory. --
> MMM I will answer that for them. They do that because it allows avoiding confronting mathematics of the exchnage. Note:
I follow no one and defy categorization by your own sets.
*Musatov*:the binary set presents a factual divisor between binary and
the naturals buy segregating decimal values in algorithmic base. The
path of the binary tree is countable because it can be contained and
manipulated an infinite number of ways like the algorithm. For a clear
analogy the binary tree is like a stick shift tranmsmission.
Mathematics has not yet accepted the idea of "overdrive". Signed in
truth, Martin Musatov
Virgil
<c10321a7-b0bc-466a...@o18g2000yqi.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> > Of these, some are computable (in theory) to any position,
> > for example the part of pi that exceeds 3. So I think you
> > have to admit that those sequences exist also.
>
> Disregarding physical constraints, as we will do here, I fully agree.
> These numbers exist. The set of all of them is countable.
> >
> > Finally, what remains are those sequences that are neither
> > repeating, nor well defined, nor computable (yet).
>
> There is no yet required, even in parentheses. The set of all finite
> alphabets is countable. The set of all words is countable. The set of
> all languages is countable. This proves that the set of all numbers
> that can be computed or individually identified is countable too ---
> in spite of the fact that the contrary bis believed by some Fools Of
> Matheology.
Those wh object to WM's anti-mathematics do not claim that there
uncountably many "names" for numbers, but do not see any conclusive
reason for requiring every number to have a name, and a number of good
reason why not.
WM seems obsessed by names and seems to think nothing tht does not have
one can exist. Others merely note that unnamed things are difficult to
discuss.
>
> This set (including all existing alphabets, dictionaries, provable
> theorems, and threads of sci.logic) is completely written here:
> 0
> 1
> 00
> 01
> 10
> 11
> ...
> (better: would be completely written here, if it was possible to
> complete an infinite set.) Of course there is no diagonalization
> possible.
Since it does not contain even 2, it is incomplete.
> > > The paths of the binary tree are not more abundant �than the countable
> > > set of nodes.
This only holds for finite trees. Once one gets to infinite trees, WM
gets lost.
> > > This theorem stands firm as you may see from the
> > > helpless attempts to contradict it collected in the original post.
> >
> > That is not a theorem; it is an assertion, which does not
> > stand firm because it has not been proven. �Nor will it
> > ever be proven, because it is not true.
>
> It is easy to specify attributes of non existing things and then to
> believe in them. Nevertheless, in this case, this claim can be
> disproved based on the following fact:
>
> If Cantor's proof is correct, then an infinite path is the limit of
> its finite initial segments.
That is something that may be true in WM's MathUnrealism, but is not
necessarily the case anywhere else, at least without a better definition
of "limit" that WM' ambiguities produce.
If the initial segments are to be considered merely as sets (of nodes or
edges, totally ordered by subset inclusion), then the any infinite path
is the union of a totally ordered by inclusin) sequence of infinitely
many finite paths.
> But not all paths are the limits of their finite initial segments.
If not all are such unins, then which ones are not the unions, in the
sense described above, of their finite initial segments, WM?
>
> Of course you need not believe that this is a contradiction. You need
> not even believe that 1 > 2 is a contradiction or SUM(N) = 0 or 0 >
> oo.
I believe that all of them hold in WM's world, since in such an
inconsistent world every thing is both true and false.
--
Virgil
WM
> You simply make no sense. And I say this respectfully as an academic
> challenge.
Then let us change the topic. Here is a challenge for you:
XXIII
____ = II
VII
There you see the numbers 23, 7, and 2, realized by matchsticks, in a
false equation. 23/7 is not 2. You are allowed to change the position
of one and only one matchstick to reduce the error as much as
possible.
Regards, WM
Virgil
<ee311630-a57f-499e...@z19g2000vbz.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 26 Mai, 18:14, A Nony Mouse <vir...@nowhere.com> wrote:
> > In article
> > <95d34fcc-0210-45e2-836f-9e2194d20...@g19g2000vbi.googlegroups.com>,
> >
> >
> >
> >
> >
> > �WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 26 Mai, 12:51, David C. Ullrich <dullr...@sprynet.com> wrote:
> > > > On Mon, 25 May 2009 06:00:18 -0700 (PDT), LudovicoVan
> >
> > > > <ju...@diegidio.name> wrote:
> > > > >On 25 May, 12:57, David C. Ullrich <dullr...@sprynet.com> wrote:
> > > > >> On Sun, 24 May 2009 22:28:27 -0700 (PDT), LudovicoVan
> >
> > > > >> <ju...@diegidio.name> wrote:
> > > > >> >[...]
> >
> > > > >> >Extending to the infinite case
> >
> > > > >> Oops. You can't just "extend" an inequality valid for natural
> > > > >> numbers to the infinite case (at least not without justification).
> >
> > > > >There is no need to justify that inequality in my argument.
> >
> > > > Very convenient.
> >
> > > Is tere any need, in your opinon, to justify the followig inequality?
> >
> > > SUM[n --> oo] 2 - 1 - 1 = SUM[n --> oo] 0 =< SUM[n --> oo] 2^-n �<
> > > 2^aleph_0
> >
> > As that inequality is irrelevant to any property of a complete infinite
> > binary tree, at least outside of WM's dingbat world, there is no need to
> > justify that particular irrelevancy.
>
> You need not emphasize the fact that you have not yet understood.
I am proud of the fact that I do not understand how the above nonsense
is allegedly relevant to complete infinite binary trees outside of WM's
world of MathUnrealism.
And what holds inside that world is of no interest to anyone except WM
himself.
> >
> >
> >
> > > > And probably "the" error - assuming that we can
> > > > simply extend this or that to the infinite case is
> > > > precisely the error in many "proofs" that various
> > > > uncountable sets are countable.
> >
> > > Among these errors one of the first class, seems to me, to be Cantor's
> > > theroem and its basic assumption, namely that sqrt(3) = �(1,7, 1,73,
> > > 1,732, ...).
> >
> > For the reals of Cantor's first uncountability theorem, a real is either
> > an equivalence class of Cauchy sequences or a Dedekind cut, and not
> > merely a sequence.
>
> For Cantor and his theorem a real is the limit of a sequence, called
> Fundamentalreihe.The diagonal of an infinite list is certainly the
> limit of a sequence.
Irrelevancies again.
WM is mixing up theorems. Cantor has one theorem proving that the set of
all reals cannot be counted by the naturals and another that the set of
all binary sequences cannot be counted by the naturals.
His so-called "diagonal-proof" theorem does not even mention reals.
And in any case, WM has no valid criticism of either proof.
--
Virgil
Virgil
<42dbed67-1129-4114...@u8g2000yqn.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 26 Mai, 21:04, Virgil <virg...@nowhere.com> wrote:
> > In article
> > <f8af62ae-bec2-47a8-baf6-1d886c6ff...@g1g2000yqh.googlegroups.com>,
> >
> >
> >
> >
> >
> > WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 24 Mai, 21:01, Marshall <marshall.spi...@gmail.com> wrote:
> > > > On May 24, 10:57 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > > Yes; that is what I should have said. Sometimes this excessive
> > > > information requirement results in a very large number, but in
> > > > other cases it could be between 0 and 1.
> >
> > > So it is.
> > > Pi is defined by many finite formulas.
> > > But if a finite formula is lacking, as unavoidably must be the case
> > > for most reals, then it is impossible to define or to use them as
> > > individuals.
> > > The decimal expansion of pi or even that of 1/33 = 0.030303... does
> > > not exist _in the real world_, because not all natural numbers
> > > required as indices (counting the places) are available, be it in
> > > written, thought or what ever form.
> >
> > WM seems to believe that without a decimal expansion, a number cannot
> > exist,
>
> Virigil seems to be incapable to understand written text. I wrote
> above:
> "if a finite formula is lacking".
There are all sorts of numbers, like pi and e and sqrt(p) for p not a
square of a natural, for which no finite formula in WM's sense exists,
which are in common and everyday use in mathematics.
>
> > whereas it is only the decimal expansion (a special form of name
> > for the number) which does not exist.
>
> You are wrong. I confess that 0.030303... = 1/33 = one divided by 33 =
> 3/99 does exist. What does not exist, in MatheRealism, is the complete
> decimal expansion. (Because there is a first place that cannot be
> decided whether it is occupied by 0 or 3, because its index natural
> number cannot be expressed in the real world --- nor can the digits of
> 1/33 up to that place).
What is constrained by having to conform to WM's "real world" is not
mathematics. It is engineering.
>
> > > There are two reasons: First set theorists have conquered mathematics
> > > claming that only mathematics based upon set theory (and only in
> > > formalized form) is real mathematics.
> >
> > Category theorists claim all mathematics is based on categories.
> > Logicians claim all mathematics is based solely on logic.
> > And so on.
>
> But the truth is, that mathematics is based on natural numbers,
> geometrical notions and logical rules (I mean the correct logic, of
> course).
Wm has no access to either truth or correct logic.
According to the sort of "geometrical notions" WM would accept, only
Euclidean geometry would be allowed.
> >
> > > Second, there are some scholars who try to prove theorems about finite
> > > numbers by means of infinite cardinals.
> >
> > Name one.
>
> Goodstein, H. Friedman.
> >
> > > Naysayer Hippasos von Metapont showed that the pythagorean theorem
> > > "everything is number" (or ratio) was wrong (without introducing
> > > irrational numbers, he left that to Eudoxos).
> >
> > The Pythagoreans took "everything is number" as an axiom, not as a
> > theorem provable only derived from other axioms. And finding that axiom
> > systems have problems occurs several times.
>
> Yes, most recently I proved the axiom of infinity is wrong.
Nowhere but in your own mind. And such "localized" proofs are not proofs.
> >
> > > Naysayer Nicole d'Oresme denied established knowledge, then the Bible,
> > > and proposed the heliocentric system, contradicting the Bible.
> > > (Fortunately for him, nobody cared at that time.)
> >
> > Denying physical assumptions is irrelevant.
>
> Denying assumptions is commendable in every domain, if the assumptions
> are wrong.
WM claims that some are wrong, but his logic is so flawed and his terms
so ambiguously used as to prevent acceptance by those with less flawed
logic or accustomed to less ambiguity of terminology.
> >
> > > The other way round: Christian Doppler stated his theory by 1850.
> >
> > Physics again, irrelevant again.
>
> Not at all. Dopplers formulas are mathematics describing waves that
> are described by mathematics.
Are those "waves" alleged to be purely mathematical or are they at least
partly physical? If at all physical, it is physics.
--
Virgil
WM
> WM is mixing up theorems. Cantor has one theorem proving that the set of
> all reals cannot be counted by the naturals and another that the set of
> all binary sequences cannot be counted by the naturals.
>
> His so-called "diagonal-proof" theorem does not even mention reals.
But if applied to numbers, then the diagonal number is the limit of
its finite initial segments.
>
> And in any case, WM has no valid criticism of either proof.
In case of the binary tree, the paths (= real numbers of the unit
interval) are not the limits of their finite initial segmentsm because
they all, covering and containing the complete set of nodes of the
infinite binary tree, belong to a countable set.
Therefore, the number of paths that are limits, must surpass 1 but
must not reach 2^1aleph_0. How can that number be determined???
Regards, WM
Virgil
<c25207c9-b655-4502...@k2g2000yql.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 26 Mai, 21:12, Virgil <virg...@nowhere.com> wrote:
> > In article
> > <b21cf8a7-1f96-42ef-b512-12edc537c...@h2g2000yqg.googlegroups.com>,
> > �WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > If so, then we have to tolerate the follwing situation:
> > > The real number sqrt(3) is the limit of a rational sequence.
> >
> > That is only one of the properties of sqrt(3).
> >
> > > Every real number is the imit of a rational sequence.
> > > Not every real number is the limit of a rational sequence.
> > > Because not all real numbers are the limits of rational sequences.
> >
> > Name one which is not!
> >
> If all paths are limits of sequences of finite paths, then the set of
> all paths is countable. The set of sequences of finite paths is
> countable, because all finite paths are countable and the sequences
> generated by them cannot surpass the set of paths continued in the
> union of the sequence of all finite trees.
WM fails to support his claim that
WM: Not every real number is the limit of a rational sequence.
WM: Because not all real numbers are the limits of rational sequences.
but tries to hide his failure by going off on a tangent.
I repeat, if not every real is the limit of a rational seqeunce, as WM
claims, NAME ONE THAT IS NOT, WM!!!
> >
> >
> > > I do not tolerate that situation.
> >
> > Since it is a phony "situation" of your own making, why not?
>
> It is of Cantor's making.
> He recognized that the complete infinite binary tree is not the limit
> of the complete infinite zsequence of finite binary trees.
Can WM cite Cantor ever saying or writing that the complete infinite
binary tree was, or was not, related to finite binary trees in any way?
Absent such a citation, I will conclude that WM is flat out lying.
--
Virgil
Virgil
<3232acad-e917-42fd...@z7g2000vbh.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 26 Mai, 21:18, Virgil <virg...@nowhere.com> wrote:
>
> > > I agree. There is another argument. For every appended path we inrease
> > > the number of nodes by aleph_0, the number of paths by 1. This should
> > > suddenly switch when finishing the infinite?
> >
> > To "finish" the job of constructing a complete infinite binary tree in
> > this way, by appending new paths one at a time, one must a priori assume
> > that the set of paths being appended is countable,
>
> That is proved by the set of nodes being countable.
I have seen no such proof, and do not believe one can exist outside of
WM's personal world of MathUnrealism.
>
>
> > so your argument is
> > circular and therefore invalid.
>
> No. The argument could show that the construction is insufficient to
> cover the complete infinite binary tree.
> Then we would have another
> proof of uncountability of the reals. Alas, I can show that there is
> no node of any path missing.
Then do so.
> >
> > > Ad if it cannot be finished, then the whole discussion about
> > > cardinalities �is in vain.
> >
> > Why? WM misreads the evidence again!
> >
> > Being unable to finish by adding one path at a time is evidence that the
> > set of paths cannot be counted out one at a time like the naturals can
> > be counted.
>
> That is complete rubbish. The naturals cannot be counted out in that
> manner either.
Do you mean that when one pairs off 1 with 1 and 2 with 2 and 3 with 3
and so on, that there are some naturals left unpaired by this process?
Which ones?
But being able to finish by adding one path in half
> time of the previous one, is evidence that the set of paths is
> countable. (Even doing it in 99 % of time of the previous one would
> show this.)
You can not shown that this process will actually add all paths in
finite time unless you have already proved that the set of paths is
countable, which you have not done.
Every one of WM's attempts to prove countability of the set of all paths
of a complete binary tree, or the set of all reals, or the set of all
infinite binary sequences assumes its conclusion at some point, so is
circular or invalid.
It is perfectly possible to develop a consistent mathematics in which
all sets are finite, and I understand it has been done a number of times
by a number of people, but it is not possible without assuming anything,
the way WM wants to do it.
--
Virgil
Virgil
<27882e3c-7090-47f4...@z7g2000vbh.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 27 Mai, 00:24, lwal...@lausd.net wrote:
> > But instead of continuing with
> > the discussion of Potential Infinity, which I believe could be an
> > interesting theory, WM has gone back to the Binary Tree.
>
> There I show that only potential infinity is free of contradictions.
Shown to the satisfaction of no one at all except WM himself.
> >
> > Since I don't accept that the inconsistency of ZFC, if it's ever
> > proved, will be proved using binary trees,
>
> Why? Any rational reason?
Even if ZFC eventually turns out to be provably inconsistent, it has not
been and never will be proven by WM.
--
Virgil
Virgil
<842052cc-4b93-40b2...@n19g2000vba.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 27 Mai, 21:36, Virgil <virg...@nowhere.com> wrote:
>
> > AS long as WM sticks to strictly finite paths, his trees may behave as
> > he wants them to, but once he allows infinite paths, things are no
> > longer under his control, but take on a life of their own.
>
> The only condition to stick to is that paths do not exist without
> nodes.
There are a lot of other conditions to stick to if one wants one's trees
to behave like the trees of http://en.wikipedia.org/wiki/Binary_tree.
> Paths simply are sequences of nodes. This becomes clear by the
> fact that every node contributes one bit to at least one of the 2^N
> sequences.
Actually to at least 2^N of those 2^N sequences ( and not to another 2^N
of them).
> Cantor claims that the number of these sequences is
> uncountable.
He claims that any attempt to "count" them (surject the set of naturals
onto the set of all those seqeunces) omits at least one sequence, and,
furthermore, he proves it.
> He claims further that the limit of the sequence of
> finite trees is not the infinite tree.
I am not aware of Cantor anywhere claiming that.
Can WM give a citation for his claim?
> (By the way: What is the
> difference between the infinite tree and the union of all finite
> trees? Nothing.)
Unless one defines those finite trees as sets so that the set of all of
them is well-ordered by inclusion, the union of them need not exist at
all, much less be a tree.
> But he claims that pi is the limit of the finite
> initial segments.
I am not aware of Cantor anywhere claiming that.
Can WM give a citation for for his claim?
> These two claims suggest the question: How many
> paths are limits in the tree and why not less or more. This number,
> according to Cantor, must be X with 1 =< X < 2^N. How to determine it?
I am not aware of Cantor anywhere claiming that.
Can WM give a citation for for his claim?
What Cantor does claim is that any given list of binary sequences one
can construct a binary sequence not in the given list.
And he proves it.
--
Virgil
Virgil
<edf1a683-a281-472d...@y7g2000yqa.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 27 Mai, 22:15, Virgil <virg...@nowhere.com> wrote:
>
> > WM is mixing up theorems. Cantor has one theorem proving that the set of
> > all reals cannot be counted by the naturals and another that the set of
> > all binary sequences cannot be counted by the naturals.
> >
> > His so-called "diagonal-proof" theorem does not even mention reals.
>
> But if applied to numbers, then the diagonal number is the limit of
> its finite initial segments.
The first proof does not have any "diagonal" numbers in it at all, and
the "diagonal" proof does not have any non-natural numbers in it.
> >
> > And in any case, WM has no valid criticism of either proof.
>
> In case of the binary tree, the paths (= real numbers of the unit
> interval) are not the limits of their finite initial segmentsm because
> they all, covering and containing the complete set of nodes of the
> infinite binary tree, belong to a countable set.
Often claimed by WM but never proven by him, and false in any system
allowing a complete infinite binary tree to exist at all.
>
> Therefore, the number of paths that are limits, must surpass 1 but
> must not reach 2^1aleph_0. How can that number be determined???
It reaches 2^aleph_0 in any system which allows a a complete infinite
binary tree to exist at all. All of WM's pseudo-trees are incomplete.
--
Virgil
Michael Press
"Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> lwa...@lausd.net writes:
>
> > So this Musatov is obviously someone so "cranky" that not even
> > the other "cranks" want to be associated with him, and WM
> > associates Musatov with his other opponents. Musatov is likely
> > an example of what Dudley, the authority on "cranks," would
> > consider to be "crazy," which is beyond being a "crank."
>
> I doubt it. Martin seems to be a troll, just fooling around to see
> how much attention he can generate. One hint: his first P=NP post was
> a recycled years-old April Fools joke by a different poster. That's
> not the behavior of a crank. That's the behavior of a practical
> joker.
>
> (His second joke, involving a link to an unreadable text which he
> pretended contained a coherent argument, was much more original and
> hence funnier. He had me fooled until I realized he had earlier
> recycled that April Fool's post.)
I spotted the looney two paragraphs in.
Did not follow the link. Ignored all its
posts since.
--
Michael Press
Martin Musatov
>Does a V count as two match sticks? Bevause it has the geometry of two matchsticks but if I move only 1 I then have \ I.I see simply this:
IIIIIII/IIII=I+III
I do not count like the Roman's nor do I pretend to simulate my
algorithm in their Numerals.
But so you can see the arrangement:
2: 3: 4: 5: 6: 7: 8: 9: 10: 11: 12: 13: 14: 15: 16: 17: 18: 19: 20:
21: 22: 23: 24: 25: 26: 27:...
In which case:
21/7=3 via
17:
23
William Hughes
> On 25 Mai, 18:49, William Hughes <wpihug...@hotmail.com> wrote:
>
> > (the difference being that there are no "infinite nodes"
> > but there are infinite paths)
>
> Oh yeah. There are paths stretching even into that domain where no
> nodes exist.
Absolute Piffle. You know this is not what was meant, but you
persist in this nonsense.
A finite path is a subset of nodes, all of which are at finite
levels, that has a last node.
An infinite path is a subset of nodes, all of which are at finite
levels, that does not have a last node.
Please acknowledge. Outside of Wolkenmuekenheim, the infinite tree
contains
infinite paths.
- William Hughes
WM
> An infinite path is a subset of nodes, all of which are at finite
> levels, that does not have a last node.
>
> Please acknowledge. the infinite tree contains
> infinite paths.
Of course. And every path contains an infinite number of nodes such
that for *every* pair of given paths, A and B, A contains an
infinitude of nodes that do not belong to B, and vice versa.
Therefore there cannot be less nodes than paths.
If you are in doubt that the tree can be constructed from infinite
paths, each of which contains aleph_0 nodes that do not belong to any
other path, then simply look at the construction given here
http://www.hs-augsburg.de/~mueckenh/GU/GU12.PPT#365,30,Folie 30
and map all the nodes that belong to the path p_0 = 0.000... onto this
path p_0. All the nodes that belong to the next constructed path p_1 =
0.111... except those that belong to p_0 (here only the first one) map
on p_1. All the nodes that belong to p_2 = 0.0111..., except those
that belong to already constructed paths, map on p_2, and so on. So
you have aleph_0 nodes mapped on every path that are not mapped on any
other path.
Your claim that there were more paths than nodes is falsified.
Regards, WM
WM
> WM wrote:
> > On 27 Mai, 19:57, Martin Musatov <marty.musa...@gmail.com> wrote:
>
> > > You simply make no sense. And I say this respectfully as an academic
> > > challenge.
>
> > Then let us change the topic. Here is a challenge for you:
>
> > XXIII
> > ____ = II
> > VII
>
> > There you see the numbers 23, 7, and 2, realized by matchsticks, in a
> > false equation. 23/7 is not 2. You are allowed to change the position
> > of one and only one matchstick to reduce the error as much as
> > possible.
> > Regards, WM
> >Does a V count as two match sticks?
Yes, so does the X.
Because it has the geometry of two matchsticks but if I move only 1 I
then have \ I.I see simply this:
>
> IIIIIII/IIII=I+III
> I do not count like the Roman's nor do I pretend to simulate my
> algorithm in their Numerals.
> But so you can see the arrangement:
> 2: 3: 4: 5: 6: 7: 8: 9: 10: 11: 12: 13: 14: 15: 16: 17: 18: 19: 20:
> 21: 22: 23: 24: 25: 26: 27:...
> In which case:
> 21/7=3 via
> 17:
> 23
Here two simple tricks moving one matchstick only:
XXII
____ = III (small error)
VII
and
XXIII
____ = II (small error)
XII
But the best one, resulting in an error of less than 1 % is
XXII _
____ = II (one matchstick used to write pi, very small error)
VII
Regards, WM
William Hughes
> On 28 Mai, 13:33, William Hughes <wpihug...@hotmail.com> wrote:
>
> > An infinite path is a subset of nodes, all of which are at finite
> > levels, that does not have a last node.
>
> > Please acknowledge. the infinite tree contains
> > infinite paths.
>
> Of course.
Interersting, And yet when the statement is made that
the infinite tree contains infinite paths you reply with
the sarcastic comment
WM: Oh yeah. There are paths stretching even
WM: into that domain where no nodes exist.
So we have in Wolkenmuekenheim logic
The infinite binary tree does not contain
infinite paths
The infinite binary tree does contain
infinite paths
- William Hughes
>
William Hughes
Ah a Wolkenmuekenheim meaning of "best"
XXII
---- =/= II (one matchstick used to change equals to not-equals, no
error)
VII
- William Hughes
Virgil
<ab9e9609-e519-4854...@r34g2000vba.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 28 Mai, 13:33, William Hughes <wpihug...@hotmail.com> wrote:
>
> > � � �An infinite path is a subset of nodes, all of which are at finite
> > � � �levels, that does not have a last node.
> >
> > Please acknowledge. � the infinite tree contains
> > infinite paths.
>
> Of course. And every path contains an infinite number of nodes such
> that for *every* pair of given paths, A and B, A contains an
> infinitude of nodes that do not belong to B, and vice versa.
> Therefore there cannot be less nodes than paths.
As usual, WM's argument is a non-sequitur.
>
> If you are in doubt that the tree can be constructed from infinite
> paths
That it contains infinite paths is clear, but if WM's "construction" is
limited to appending paths sequentially, it is doomed to failure, as no
sequence of paths can contain all paths. As Cantor proved.
>
> Your claim that there were more paths than nodes is falsified.
Perhaps so in in WM's world of MAthUnrealism, but nowhere else.
--
Virgil
WM
> On May 28, 9:59 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > On 28 Mai, 13:33, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > An infinite path is a subset of nodes, all of which are at finite
> > > levels, that does not have a last node.
>
> > > Please acknowledge. the infinite tree contains
> > > infinite paths.
>
> > Of course.
>
> Interersting, And yet when the statement is made that
> the infinite tree contains infinite paths you reply with
> the sarcastic comment
>
> WM: Oh yeah. There are paths stretching even
> WM: into that domain where no nodes exist.
Only when you try to imply that there are not infinitely many nodes or
that there should be more infinite paths than the infinitely many
finite paths from which the infinite paths are made (by unions which,
in the infinite binary tree, are linear).
Regards, WM
WM
much as possible. (By the way, I am not the author of this puzzle.)
Regards, WM
WM
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 28 Mai, 13:33, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > An infinite path is a subset of nodes, all of which are at finite
> > > levels, that does not have a last node.
>
> > > Please acknowledge. the infinite tree contains
> > > infinite paths.
>
> > Of course. And every path contains an infinite number of nodes such
> > that for *every* pair of given paths, A and B, A contains an
> > infinitude of nodes that do not belong to B, and vice versa.
> > Therefore there cannot be less nodes than paths.
>
> As usual, WM's argument is a non-sequitur.
>
>
>
> > If you are in doubt that the tree can be constructed from infinite
> > paths
>
> That it contains infinite paths is clear, but if WM's "construction" is
> limited to appending paths sequentially, it is doomed to failure, as no
> sequence of paths can contain all paths. As Cantor proved.
>
Alas it can contain all nodes as I proved, and that is the whole tree!
Regards, WM
Virgil
<cf20410a-ce3a-4200...@h2g2000yqg.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
If there can be infinitely many nodes then there can be infinitely long
paths.
Given any attempt to count those paths, Cantor has a method for finding
one which was omitted in that counting.
In fact, an easy modification of Cantor's argument shows that there are
always AT LEAST AS MANY UNCOUNTED as have been counted, but as WM was
quite incapable of understanding it the last time it was presented, I do
not include it here.
--
Virgil
Virgil
<e28f9992-32a7-43fc...@e21g2000yqb.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 28 Mai, 20:19, Virgil <virg...@nowhere.com> wrote:
> > In article
> > <ab9e9609-e519-4854-affb-e1c84f9aa...@r34g2000vba.googlegroups.com>,
> >
> > �ｿｽWM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 28 Mai, 13:33, William Hughes <wpihug...@hotmail.com> wrote:
> >
> > > > �ｿｽ �ｿｽ �ｿｽAn infinite path is a subset of nodes, all of which are at finite
> > > > �ｿｽ �ｿｽ �ｿｽlevels, that does not have a last node.
> >
> > > > Please acknowledge. �ｿｽ the infinite tree contains
> > > > infinite paths.
> >
> > > Of course. And every path contains an infinite number of nodes such
> > > that for *every* pair of given paths, A and B, A contains an
> > > infinitude of nodes that do not belong to B, and vice versa.
> > > Therefore there cannot be less nodes than paths.
> >
> > As usual, WM's argument is a non-sequitur.
> >
> >
> >
> > > If you are in doubt that the tree can be constructed from infinite
> > > paths
> >
> > That it contains infinite paths is clear, but if WM's "construction" is
> > limited to appending �ｿｽpaths sequentially, it is doomed to failure, as no
> > sequence of paths can contain all paths. As Cantor proved.
> >
>
> Alas it can contain all nodes as I proved, and that is the whole tree!
One can easily have all nodes without having all paths, or "the whole
tree"!
The set of only those paths having at most a finite number of right
branchings clearly includes all nodes, as a node has at most a finite
number of branchings either way, but not all paths, as most of them
have infinitely many branchings both ways.
So WM is wrong again, and one CAN have all nodes without having all
paths.
--
Virgil