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| MoeBlee |
 Posted: Fri Aug 25, 2006 10:51 pm |
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Tony Orlow wrote:
[quote:79f1c358ca]
Logic determines truth. Induction is more than just a form of proof.
[/quote:79f1c358ca]
Logic determines VALIDITY. And induciton is indeed more than just a
form of proof. But your concept of induction is uninformed,
superficial, and confused.
MoeBlee |
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| MoeBlee |
| Posted: Fri Aug 25, 2006 10:51 pm |
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MoeBlee wrote:
[quote:893cee42b1]With the axiom of choice (which is one method),
every set has a cardinality. That cardinality is a cardinal number. And
that cardinal number is an aleph indexed by an ordinal.
[/quote:893cee42b1]
CORRECTION: That should be, in ZFC, every set has a cardinality. That
cardinality is either a natural number or an aleph (and, of course,
every aleph is indexed by an ordinal). |
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| Dik T. Winter |
| Posted: Fri Aug 25, 2006 10:51 pm |
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In article <44ef3b88@news2.lightlink.com> Tony Orlow <tony@lightlink.com> writes:
[quote:9e9a90f41c]Dik T. Winter wrote:
....
There is no such specific natural number. It is when we have them
all, but as there is no largest number, this can not be achieved by
taking them one by one.
The set of all naturals numbers consists of only natural numbers. There
is NO natural number where the count becomes infinite. So there is no
point in the set, even if you COULD get to the "last" one, where any
infinite set has been achieved.
And there is no point in the set where you have the complete set. Yes.
Indeed. So what?
So, if there is no point in the set which can even remotely be
considered infinitely far from the beginning, what makes it actually
infinite?
[/quote:9e9a90f41c]
What the set of all finite numbers makes infinite? The axiom of infinitys
states that the set of all finite (natural) numbers does exist. From that
is is easy to prove that that set is not finite, hence infinite.
[quote:9e9a90f41c]If no element of the set can be an infinite number of steps
from the start, you may not be able to find an end.
[/quote:9e9a90f41c]
And indeed, when you go step by step you will not get at the end.
[quote:9e9a90f41c]But does that mean
it's "greater than" every finite, or only "greater than or equal"?
[/quote:9e9a90f41c]
What is the difference? Assuming you mean "aleph-0" when you write "it",
it is easily proven that:
aleph-0 is greater than or equal to each natural
gives the theorem:
aleph-0 is greater than each natural.
Because: suppose aleph-0 >= each n in N. Now suppose in addition that it
is equal to some particular n. Well, n + 1 is in N, and so aleph-0
should also be larger or equal to n + 1. Hence it can not be equal to n.
So it is not equal to any n at all. And so aleph-0 > each n in N.
[quote:9e9a90f41c]Indeed. The set of all natural numbers is just sufficient.
No, it is far too great. If you have a countably infinite number of bit
positions, then you have an uncountably infinite set of strings. Where
bit positions are indexed by the naturals, the naturals are the power
set of the number of bit positions,
[/quote:9e9a90f41c]
Wrong. This is plain nonsense. Suppose there are three bit positions.
The set of naturals {1, 2, 3} is sufficent to index them. In what way
is {1, 2, 3} the power set of the number of bit positions (3)?
[quote:9e9a90f41c]No one is obligated to accept the theory at all. Whether it is proven
to be "correct" or not, as I have no idea what "correct" in this context
means. Is Euclidean geometry "correct"? Is hyperbolic geometry "correct"?
Is elliptic geometry "correct"?
Ah, now you bring up a prime example. Euclid set down laws for flat 2D
geometry, and questioning those axioms led to new shapes for space.
Accrdingly, the axioms of set theory might work together to describe a
system, but it is not impossible that entirely other systems might arise
from different starting assumptions.
[/quote:9e9a90f41c]
And indeed, I never did state the opposite. But if you want to get at a
new system, provide axioms, definitions, and whatever. Tell us what
axioms to retain and what axioms to reject. And if there are axioms to
be rejected, come up with alternatives.
[quote:9e9a90f41c]But what is the case is that if you accept the axioms, you also have
to accept what follows from the axioms.
Yes, I understand that, and much to the consternation of some, I don't.
[/quote:9e9a90f41c]
Yes, much consternation, I can understand that. So apparently you are
accepting the axioms, but not what follows from the axioms. What kind
of logic are you using?
[quote:9e9a90f41c]Rusin had the gall to tell me that if I don't accept that there are an
infinite number of finite naturals, then I will join JSH and others on
his "kill list". I don't claim that my conclusions are derived purely
from ZFC or NBG, but that there are more fundamental concerns which
contradict both, and that some other prioritization of principles needs
to happen. Proper subsets are smaller. The addition of a single element
needs to be reflected in the size of the set. Infinite values are larger
than finite values. Things like that.
[/quote:9e9a90f41c]
Well, you are stating such. So provide a framework. Either within the
accepted axioms, or without it. But if you want to remain within the
accepted axioms, you should also accept what follows from these axioms.
And if you want to go outside, provide your set of axioms.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |
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| MoeBlee |
| Posted: Fri Aug 25, 2006 10:51 pm |
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Tony Orlow wrote:
[quote:c35804bcb8]If it is a theorem that every set has a cardinality, then it would be
inconsistent if it were also a theorem that there exists a set without
a cardinality. But the fact that we don't have a procedure to always
announce what SPECIFIC cardinality a set has does not contradict any
theorem of set theory.
However, if we have a set which we can prove does NOT have any
cardinality within the system, then there's a hole, yes?
[/quote:c35804bcb8]
If, using only first order logic with identity, you can prove from ZFC
that there exists a set that does not have a cardinality, then you will
have proven that ZFC (and just ZF for that matter, since ZFC is
relatively consistent with ZF) is inconsistent.
[quote:c35804bcb8]So, is "every set has a cardinality, either finite or an indexed aleph"
a theorem or not?
[/quote:c35804bcb8]
It is a theorem of ZFC. And every aleph is indexed, so you don't need
to mention that for this purpose.
And by the way, that reminds me that I made a mistake in not qualifying
that only INFINITE sets have an aleph as their cardinality. Finite sets
have a natural number as their cardinality and infinite sets have an
aleph as their cardinality.
And if I mentioned ZC before, it should have been ZFC, since (as far as
I know, we also need the axiom schema of replacement to have all the
ordinals we need).
So, yes, it is a theorem of ZFC that every finite set has a natural
number as its cardinality and every infinite set has an aleph as its
cardinality.
[quote:c35804bcb8]You suggest that the axiom of choice makes it so.
[/quote:c35804bcb8]
And I should have mentioned that the axiom schema of replacement also
is need (as far as I know) in addition to the Z axioms that are not
derivable from the axiom schema of replacement.
[quote:c35804bcb8]I
say that the set of bit positions required by the binary naturals does
not have such a cardinality.
[/quote:c35804bcb8]
First, define, from the language of ZFC, every term you use in that
formulation and every term you use to prove that formulation. Second,
show an argument using only first order logic with identity and that
proceeds only from axioms and already established theorems of ZFC. Then
we'll talk about it.
MoeBlee |
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| Dik T. Winter |
| Posted: Fri Aug 25, 2006 10:51 pm |
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In article <44ef36ae$1@news2.lightlink.com> Tony Orlow <tony@lightlink.com> writes:
[quote:d966f4a9f7]Dik T. Winter wrote:
....
So you can compute all solutions of a polynomial equation even of
higher than fourth degree in finite time? I doubt that.
I did not state that. I said that the numbers were computable, where
I use the mathematical sense of computable.
Such that one can specify which finite number of iterations will get one
within a specific finite range of accuracy, gven a specific method of
approximation? It's a limit concept, really, yes?
[/quote:d966f4a9f7]
Computer scientist you are? How wrong you are. The very first definition
is that a number is computable if there is a Turing machine so you can give
it a specific integer n and it will calculate all digits from the first
to the n-th. There is no limit concept involved at all. All algebraic
numbers are computable, as are a host of non-algebraic numbers (e, pi).
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |
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| Dik T. Winter |
| Posted: Fri Aug 25, 2006 10:51 pm |
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In article <virgil-B3927D.12230325082006@news.usenetmonster.com> Virgil <virgil@comcast.net> writes:
[quote:3d7d1de164]In article <1156501289.435365.119480@75g2000cwc.googlegroups.com>,
mueckenh@rz.fh-augsburg.de wrote:
....
The problem is not indexing but "indexing without covering".
That is easy to prove impossible for any finite natural number in unary
representation. And, alas, there are only finite natural numbers.
I have no idea what "indexing without covering" means, and until it has
a clear definition, I will continue to state that indexing all of them
in any way indexes all of them.
[/quote:3d7d1de164]
What I have been able to find was that with indexing of digit WM means
that every digit position can be indexed by a natural number. With
covering WM means that all positions to a certain one can be indexed
by a natural number. What WM is asserting is that when a number can
be totally indexed, it also can be totally covered. Quantifier dislexia
disguised.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |
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| Dik T. Winter |
| Posted: Fri Aug 25, 2006 10:51 pm |
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In article <44ef3da9@news2.lightlink.com> Tony Orlow <tony@lightlink.com> writes:
[quote:3d85bb5297]Dik T. Winter wrote:
.....
Why do you need an axiom for that? Why is it
not derivable logically?
Because without the axiom of infinity the set of naturals need not exist,
and indeed, you can build a completely logical system with the negation
of the axiom of infinity and with all other axioms remaining. It is
similar to the parallel axiom in geometry.
But without an axiom of infinity, it is demonstrable that, given the
axiom of internal infinity (continuity), x<z -> x<y<z, that any finite
interval includes an infinite number of points. Start with the line, and
identify points. There's infinity.
[/quote:3d85bb5297]
Your axiom uses things that are not defined. What is the *meaning* of
"x<z"?
[quote:3d85bb5297]You both do not have an linearisation of the reals. You both are doing
other things.
Describe those "other things". How are they "other"?
[/quote:3d85bb5297]
You are expanding the reals in some undefined matter.
[quote:3d85bb5297]The second proof of the uncountability of the reals is not about reals?
That's news. What was it about, then, in your opinion?
Sorry, I already explained that in previous articles to which you have
replied. But if you do not read the articles you reply to there is
something seriously wrong in the discussion.
It is about digital representation, which is the same as power set, even
in other number bases, which means more than two states of inclusion per
member.
[/quote:3d85bb5297]
Darn. Try to read. Cantor's proof is not about reals, it is neither
about digital representations. It is about none of the things you are
mentioning. But nevertheless you maintain that it is news that it is
not about the reals, while you read what I wrote?
[quote:3d85bb5297]Larger than any finite. The set of naturals is as large as, but no
larger than, every natural.
That is not a definition, because it makes no sense. "The set of naturals
is as large as every natural"?
It is no larger than all naturals
[/quote:3d85bb5297]
That is something completely different again.
[quote:3d85bb5297]From that: "The set of naturals is as
large as 1", "The set of naturals is as large as 2". What is the meaning
of these statements?
That is when you substitute "every", meaning "each", for "all". Careful.
[/quote:3d85bb5297]
Yes, you should be careful in what you mean, and not use a word that has
multiple meanings so that you can be misunderstood. So I will refrase:
[quote:3d85bb5297]Larger than any finite. The set of naturals is as large as, but no
larger than, all naturals.
Is that what you intended? In that case you just stated a tautology.[/quote:3d85bb5297]
[quote:3d85bb5297]Then I don't know what proof you are talking about. When people say
"Cantor's second", they are generally referring to his second proof of
the uncountablility of the reals based on the diagonal argument, as
opposed to the first, based on an unreachable intermediate value.
But they are wrong. The proof was *not* about the uncountability of the
reals. The diagonal proof Cantor provided was not about that. It was
a proof about the things I outlined just above.
It was about power set and digital representation, which are identical.
It was about symbolic sets.
[/quote:3d85bb5297]
You finally did read it? If so, you really should improve your German.
[quote:3d85bb5297]I thought it was clear that I was using a notion of infinite, like WM,
from a quantitative standpoint, rather than set-theoretic.
Without definition.
Greater than any finite. Simple enough?
[/quote:3d85bb5297]
So the cardinality of the naturuals is infinite?
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |
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| Virgil |
| Posted: Sat Aug 26, 2006 2:25 am |
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In article <44ef9c12@news2.lightlink.com>,
Tony Orlow <tony@lightlink.com> wrote:
[quote:44ab7c9b8b]MoeBlee wrote:
Tony Orlow wrote:
No, it all rests on the notions of identity and equality. As Leibniz
pointed out, when the properties of two objects are all exactly the
same, then they are the same object. So, when we say two numbers are
equal, that means all properties of the two are equal.
Ha! The fallacy of reversing implication right there! An example of
just about the most basic fallacy.
When you say "a=b", you say "A a A b P(a)=P(b)". The two are equivlent
statements, and therefore imply each other.
[/quote:44ab7c9b8b]
When one says of sets A and B that A = B, that is equivalent to
For all x (x \in A <==> x \in B). At least in ZF, ZFC and NBG.
[quote:44ab7c9b8b]
No, the indiscernibility of identicals does NOT imply the identity of
indiscernibles. You need both implications; you can't derive one from
the other. And, in first order logic, one direction can be posited only
in the semantics not in the axioms.
You prove two quantities equal by showing there is no difference, do you
not?
[/quote:44ab7c9b8b]
When one says of sets A and B that A = B, that is equivalent to
"For all x (x \in A <==> x \in B). At least in ZF, ZFC and NBG."
http://en.wikipedia.org/wiki/Axiom_of_extension |
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| Virgil |
| Posted: Sat Aug 26, 2006 2:30 am |
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In article <44ef9c5a$1@news2.lightlink.com>,
Tony Orlow <tony@lightlink.com> wrote:
[quote:22513997e5]MoeBlee wrote:
Tony Orlow wrote:
Yes, and the universe is consistent by definition, so math should be
consistently overall as well.
Unless the universe is a set of sentences, the notion of consistency
does not even apply.
The universe is governed by the properties of the elements within it,
which properties are statements true about those elements.
[/quote:22513997e5]
But as those "properties" noumenon and we can only observe phenomema in
the "real world". And they can never meet. |
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| Virgil |
| Posted: Sat Aug 26, 2006 2:40 am |
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In article <44ef9f9c@news2.lightlink.com>,
Tony Orlow <tony@lightlink.com> wrote:
[quote:3e2ebdbf25]MoeBlee wrote:
Tony Orlow wrote:
Given a set x, can we always determine card(x)?
I'm guessing, but I'm pretty sure that there is not an algorithm that
will tell you the ordinal index of the aleph that is the cardinality of
any given set.
Good guess, except I have no idea what your driving at.
[/quote:3e2ebdbf25]
He is driving at the fact that for infinite ordinals there are infinite
sets of different ones having the same cardinality. |
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| Virgil |
| Posted: Sat Aug 26, 2006 2:49 am |
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In article <44efa44f@news2.lightlink.com>,
Tony Orlow <tony@lightlink.com> wrote:
[quote:522e0d7778]Virgil wrote:
In article <44eef7f0@news2.lightlink.com>,
Tony Orlow <tony@lightlink.com> wrote:
MoeBlee wrote:
Tony Orlow wrote:
MoeBlee wrote:
Please say what sentence and its negation you believe are both theorems
of set theory.
I'm not sure how this would be proven in set theory (I don't think it
is),
So you don't know that set theory is inconsistent (by 'inconsistent' I
mean the usual definition).
but it appears to be a belief, anyway, that all sets can be
classified through cardinality.
With suitable axioms, we can define a cardinality operation 'card' so
that we get a theorem: card(x)=card(y) <-> x equinumerous with y.
Given a set x, can we always determine card(x)?
Depends. In ZF m not necessarily, but in ZFC or NBG, at least
theoretically yes.
No, but you are obligated to define a cardinality for this set which is
consistent, if you claim the theory is consistent. You can't.
For each index value there is a natural whose binary string requires
that index value.
Thus anything less than N is too small.
Thus N is required, with cardinality Card(N).
And yet, N is infinitely too large.
[/quote:522e0d7778]
It may be to large for TO, but it is just the right size for
mathematicians.
TO has this nasty habit of swallowing camels and straining at gnats, at
least with respect to mathematical issues. |
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| Virgil |
| Posted: Sat Aug 26, 2006 3:02 am |
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In article <44efa5dc@news2.lightlink.com>,
Tony Orlow <tony@lightlink.com> wrote:
[quote:599fdeac3a]Virgil wrote:
In article <44eefa93@news2.lightlink.com>,
Tony Orlow <tony@lightlink.com> wrote:
On the contrary, that set of bit positions has cardinality equal to that
of N. The fact that the same set of bit positions is capable of more
does not mean that there need be any smaller set which is sufficient.
That depends on how many more it produces.
[/quote:599fdeac3a]
Does To claim to be able to do it with any smaller index set?
If not then the index set specified as being required is the set that is
required.
[quote:599fdeac3a]
Then TO should be able to give the precise number of binary bits
required to represent every natural up through , say, ten, but not to
represent any larger naturals.
If he can not then his whole argument fails.
No, one needs to be able to specify to the nearest bit.
[/quote:599fdeac3a]
So TO's argument fails, as expected.
[quote:599fdeac3a]
Unless TO can give the number of bits to count up to ten but no farther,
the issue of unused strings is a straw man.
To cover 10 but not 20: 4.
[/quote:599fdeac3a]
How about to cover 10 but not 11?
[quote:599fdeac3a]Unless TO can do every natural with a finite set of bits, an infinite
set is required.
Not that the same thing happens for the rationals and decimal
representations. Infinitely many decimal places are required.
But, which infinity?
[/quote:599fdeac3a]
The smallest one is big enough. Anything smaller is too small and
anything larger is larger than needed.
[quote:599fdeac3a]Everybody but TO accepts Card(N) as necessary and (more than) sufficient.
Excessively sufficient, with no smaller sufficient alternative.
[/quote:599fdeac3a]
Precisely.
[quote:599fdeac3a]
If TO claims to have a workable system, and wishes those claims to be
treated with anything but distain,then he must present that system in
its entirety with some better evidence than he has so far produced that
it actually does work.
I know.
[/quote:599fdeac3a]
Then until you have such a system you know what you will deservedly get.
[quote:599fdeac3a]So, TO, what is YOUR definition of TRUTH?
Truth is the value put on a statement, from 0 to 1. Truth is consistency
with reality. There are a few ways to talk about truth.
If truth is merely a value one puts on a statement, everyone's "truths"
will be different, which is not very useful.
In that sense, many truth values are false.
[/quote:599fdeac3a]
Not according to those who hold them to be truths. Without some
independent standard, there can be no universal truth at all, only
personal truths.
[quote:599fdeac3a]Any logical tautology derived only from a gIven set of axioms is TRUE in
that axiom system.
This is form of relative truth, and mathematics can be certain of no
other.
Logic determines truth.
[/quote:599fdeac3a]
Logic determines only relative truth ( this is true if that is true). |
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| Virgil |
| Posted: Sat Aug 26, 2006 3:11 am |
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In article <44efa6c5@news2.lightlink.com>,
Tony Orlow <tony@lightlink.com> wrote:
[quote:bf41b21251]Either every finite natural has a finite bit string and no infinite set
of positions is required
No inifinite set is requires fora any natural but for any finite number[/quote:bf41b21251]
of bits, there is a natural that requires more than that many bits, so
there cannot be any finite bound on the set of bit needed.
In standard set theories, an unbounded set of naturals is infinite by
any standard definition. When TO produces a system in which infinite
sets are finite, only then will his counterclaims be given any
consideration.
[quote:bf41b21251]Yes, given the current understanding, any unbounded set is infinite.
[/quote:bf41b21251]
And will remain so until TO's mythical system appears, and possibly even
then.
[quote:bf41b21251]Find any dictionary anywhere that says otherwise, that allows TO's
self-contradictory unbounded but finite.
Dictionaries reflect the widest and most accepted usage of words.
They also, if they are any good, include most, if not all, special
meanings. So if it cannot be found in any of them, even the OED, it is
because it ain't so.
Oh.[/quote:bf41b21251] |
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| Virgil |
| Posted: Sat Aug 26, 2006 3:19 am |
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In article <44efab93@news2.lightlink.com>,
Tony Orlow <tony@lightlink.com> wrote:
[quote:737c97bf15]If it is a theorem that every set has a cardinality, then it would be
inconsistent if it were also a theorem that there exists a set without
a cardinality. But the fact that we don't have a procedure to always
announce what SPECIFIC cardinality a set has does not contradict any
theorem of set theory.
However, if we have a set which we can prove does NOT have any
cardinality within the system, then there's a hole, yes?
[/quote:737c97bf15]
If there were a hole, TO is too stupid to find it.
And the index set argument doesn't fly.
[quote:737c97bf15]So, is "every set has a cardinality, either finite or an indexed aleph"
a theorem or not?
[/quote:737c97bf15]
In what system?
[quote:737c97bf15]You suggest that the axiom of choice makes it so. I
say that the set of bit positions required by the binary naturals does
not have such a cardinality.
[/quote:737c97bf15]
As that set is N, TO is wrong, at least in ZF, ZFC and BG.
TO is trying to pull off an old swindle of his that falsely declares
endless sets finite. The set of bit postitions is endless because for
every bit position, say the nth, there is a natural, 2^(n+1) that
requires a larger one. Thus there is no required bit position that can
be the last one last or the largest one possible, and the set of bit
positions is the same as the set of naturals. |
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| Guest |
| Posted: Sat Aug 26, 2006 6:34 am |
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MoeBlee schrieb:
[quote:2f1f0e345f]mueckenh@rz.fh-augsburg.de wrote:
In real mathematics (as opposed to matheology)
we have the valid foundation: What cannot be addressed, that does not
exist.
Do you have axioms for this "real mathematics"?
[/quote:2f1f0e345f]
Abolish Axioms. Acquire And Ask An Abacus.
What it says and what you can derive from that by pure logic: That is
mathematics.
Regards, WM |
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