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| Edgar E. Escultura... |
 Posted: Thu Nov 05, 2009 12:24 am |
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Two Fatal Defects of Andrew Wiles’ Proof of FLT
By E. E. Escultura
1) The field axioms of the real number system are inconsistent; Felix Brouwer and this blogger provided counterexamples to the trichotomy axiom and Banach-Tarski to the completeness axiom, a variant of the axiom of choice. Therefore, the real number system is ill-defined and FLT being formulated in it is also ill-defined. What it took to resolve this conjecture was to first free the real number system from contradiction by reconstructing it as the new real number system on three simple consistent axioms and reformulating FLT in it. With this rectification of the real number system, FLT is well-defined and resolved by counterexamples proving that it is false. (Main reference: Escultura, E. E., The new real new real number system and discrete computation and calculus, Neural, Parallel and Scientific Computations, 17 (2009), 59 – 84).
2) The other fatal defect is that the complex number system that Wiles used in the proof being based on the vacuous concept i is also inconsistent. The element i is the vacuous concept: the root of the equation x^2 + 1 = 0 which does not exist and is denoted by the symbol i = sqrt(-1) from which follows that,
i = sqrt(1/-1) = sqrt 1/sqrt(-1) = 1/i = i/i^2 = -i or
1 = -1 (division of both sides by i),
2 = 0, 1 = 0, i = 0, and, for any real number x, x = 0,
and the entire real and complex number systems collapse. The remedy is in the appendix to the paper, The generalized integral as dual to Schwarz distribution, in press, Nonlinear Studies.
Another example of a vacuous concept is the greatest integer. Let N be the greatest integer. By the trichotomy axiom one and only one of the following axioms holds: N < 1, N = 1, N > 1. The first inequality is clearly false. If N > 1, then N^2 > N, contradicting the choice of N. therefore N = 1. This is the original statement of the Perron paradox and it is blamed on the vacuous concept N. In general, any vacuous concept yields a contradiction.
E. E. Escultura
Research Professor
V. Lakshmikantham Institute for Advanced Studies
GVP College of Engineering, JNT University
Madurawada, Vishakhapatnam, AP, India
http://users.tpg.com.au/pidro/ |
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| Dirk Van de moortel... |
| Posted: Thu Nov 05, 2009 6:07 am |
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Edgar E. Escultura <escultur36 at (no spam) yahoo.com> wrote in message
1117368396.14753.1257416680869.JavaMail.root at (no spam) gallium.mathforum.org
[quote]Two Fatal Defects of Andrew Wiles’ Proof of FLT
By E. E. Escultura
1) The field axioms of the real number system are inconsistent; Felix Brouwer and this blogger provided counterexamples to the
trichotomy axiom and Banach-Tarski to the completeness axiom, a variant of the axiom of choice. Therefore, the real number system
is ill-defined and FLT being formulated in it is also ill-defined. What it took to resolve this conjecture was to first free the
real number system from contradiction by reconstructing it as the new real number system on three simple consistent axioms and
reformulating FLT in it. With this rectification of the real number system, FLT is well-defined and resolved by counterexamples
proving that it is false. (Main reference: Escultura, E. E., The new real new real number system and discrete computation and
calculus, Neural, Parallel and Scientific Computations, 17 (2009), 59 – 84).
2) The other fatal defect is that the complex number system that Wiles used in the proof being based on the vacuous concept i is
also inconsistent. The element i is the vacuous concept: the root of the equation x^2 + 1 = 0 which does not exist and is denoted
by the symbol i = sqrt(-1) from which follows that,
i = sqrt(1/-1) = sqrt 1/sqrt(-1) = 1/i = i/i^2 = -i or
1 = -1 (division of both sides by i),
2 = 0, 1 = 0, i = 0, and, for any real number x, x = 0,
and the entire real and complex number systems collapse. The remedy is in the appendix to the paper, The generalized integral as
dual to Schwarz distribution, in press, Nonlinear Studies.
Another example of a vacuous concept is the greatest integer. Let N be the greatest integer. By the trichotomy axiom one and only
one of the following axioms holds: N < 1, N = 1, N > 1. The first inequality is clearly false. If N > 1, then N^2 > N,
contradicting the choice of N. therefore N = 1. This is the original statement of the Perron paradox and it is blamed on the
vacuous concept N. In general, any vacuous concept yields a contradiction.
E. E. Escultura
Research Professor
V. Lakshmikantham Institute for Advanced Studies
GVP College of Engineering, JNT University
Madurawada, Vishakhapatnam, AP, India
http://users.tpg.com.au/pidro/
[/quote]
On Usenet he can write whatever he wants about his current
title, but it looks like the institute forced him to include the
Emeritus qualifier on his website :-)
Yes, people like this are allowed on the street.
Dirk Vdm |
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| MMM... |
| Posted: Thu Nov 05, 2009 9:44 am |
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Double idiot squared, plus infinity, plus one, squared, tied in a sack, and
thrown over the back of a donkey. Times two. |
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| Ken Quirici... |
| Posted: Thu Nov 05, 2009 1:01 pm |
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On Nov 5, 5:24 am, "Edgar E. Escultura" <escultu... at (no spam) yahoo.com> wrote:
[quote]Two Fatal Defects of Andrew Wiles’ Proof of FLT
By E. E. Escultura
1) The field axioms of the real number system are inconsistent; Felix Brouwer and this blogger provided counterexamples to the trichotomy axiom and Banach-Tarski to the completeness axiom, a variant of the axiom of choice. Therefore, the real number system is ill-defined and FLT being formulated in it is also ill-defined. What it took to resolve this conjecture was to first free the real number system from contradiction by reconstructing it as the new real number system on three simple consistent axioms and reformulating FLT in it. With this rectification of the real number system, FLT is well-defined and resolved by counterexamples proving that it is false. (Main reference: Escultura, E. E., The new real new real number system and discrete computation and calculus, Neural, Parallel and Scientific Computations, 17 (2009), 59 – 84).
2) The other fatal defect is that the complex number system that Wiles used in the proof being based on the vacuous concept i is also inconsistent. The element i is the vacuous concept: the root of the equation x^2 + 1 = 0 which does not exist and is denoted by the symbol i = sqrt(-1) from which follows that,
i = sqrt(1/-1) = sqrt 1/sqrt(-1) = 1/i = i/i^2 = -i or
[/quote]
Actually, could somebody explain why this is NOT
valid:
i = sqrt(1/-1) = sqrt 1/sqrt(-1) = 1/i,
therefore
i^2 = 1,
therefore -1 = 1?
I tried a few examples like
sqrt(9/4) = sqrt(9)/sqrt(4) = 3/2 = 1.5
OR
sqrt(9/4) = sqrt(2.25) = 1.5.
So what is my molasses mind missing?
[quote]
1 = -1 (division of both sides by i),
2 = 0, 1 = 0, i = 0, and, for any real number x, x = 0,
and the entire real and complex number systems collapse. The remedy is in the appendix to the paper, The generalized integral as dual to Schwarz distribution, in press, Nonlinear Studies.
Another example of a vacuous concept is the greatest integer. Let N be the greatest integer. By the trichotomy axiom one and only one of the following axioms holds: N < 1, N = 1, N > 1. The first inequality is clearly false. If N > 1, then N^2 > N, contradicting the choice of N. therefore N = 1. This is the original statement of the Perron paradox and it is blamed on the vacuous concept N. In general, any vacuous concept yields a contradiction.
E. E. Escultura
Research Professor
V. Lakshmikantham Institute for Advanced Studies
GVP College of Engineering, JNT University
Madurawada, Vishakhapatnam, AP, Indiahttp://users.tpg.com.au/pidro/[/quote] |
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| CapCity... |
| Posted: Thu Nov 05, 2009 4:16 pm |
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"MMM" <wako at (no spam) wako.net> wrote in message news:hcuobh$m8f$1 at (no spam) aioe.org...
[quote]Double idiot squared, plus infinity, plus one, squared, tied in a sack,
and thrown over the back of a donkey. Times two.
[/quote]
= EEE = 3E
Right?
[quote]
[/quote] |
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| A... |
| Posted: Thu Nov 05, 2009 5:54 pm |
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On Nov 5, 7:32 pm, Tim Little <t... at (no spam) little-possums.net> wrote:
[quote]On 2009-11-05, Ken Quirici <kquir... at (no spam) yahoo.com> wrote:
Actually, could somebody explain why this is NOT
valid:
i = sqrt(1/-1) = sqrt 1/sqrt(-1) = 1/i,
The second equality is unjustified. It requires a property of sqrt
that holds in the reals, but cannot be extended to the whole complex
field.
When defining sqrt as a function, you have to choose which roots to
take for the nonzero values. With reals the most useful choice is the
set of all positive roots, as they are closed under multiplication.
However in the complex field there is no possible choice that is
closed under multiplication. At best you can say that
sqrt(x) sqrt(y) = +/- sqrt(x y).
So the original "paradox" becomes:
i = sqrt(1/-1) = +/- sqrt(1) / sqrt(-i) = +/- 1/i,
which is quite true.
- Tim
[/quote]
More generally, when one defines i as the square root of -1, one
chooses ONE of the two square roots of -1 in the complex numbers to be
i; the other is necessarily -i. The "paradox" above, and some portion
of Escultura's misunderstandings, is the result of failing to
distinguish between the two square roots of -1 in the complex numbers. |
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| Tim Little... |
| Posted: Thu Nov 05, 2009 7:32 pm |
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On 2009-11-05, Ken Quirici <kquirici at (no spam) yahoo.com> wrote:
[quote]Actually, could somebody explain why this is NOT
valid:
i = sqrt(1/-1) = sqrt 1/sqrt(-1) = 1/i,
[/quote]
The second equality is unjustified. It requires a property of sqrt
that holds in the reals, but cannot be extended to the whole complex
field.
When defining sqrt as a function, you have to choose which roots to
take for the nonzero values. With reals the most useful choice is the
set of all positive roots, as they are closed under multiplication.
However in the complex field there is no possible choice that is
closed under multiplication. At best you can say that
sqrt(x) sqrt(y) = +/- sqrt(x y).
So the original "paradox" becomes:
i = sqrt(1/-1) = +/- sqrt(1) / sqrt(-i) = +/- 1/i,
which is quite true.
- Tim |
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| Edgar E. Escultura... |
| Posted: Thu Nov 05, 2009 10:06 pm |
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Just as those who cannot do mathematics philosophise about it or write about its history those who cannot rebut a comment talk about something else.
E. E. Escultura |
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| Edgar E. Escultura... |
| Posted: Thu Nov 05, 2009 10:11 pm |
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| When one cannot rebut a comment he resorts to name calling that only reveal intellectual inadequacy and emptiness at the top. E. E. Escultura |
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| Edgar E. Escultura... |
| Posted: Thu Nov 05, 2009 10:19 pm |
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| Actually, traditional mathematicians choose one of the two square roots of -1 called principal value to avoid the contradiction. But this does not negate the fact that one can derive a contradiction from i. The root of this problem is the vacuous nature of i being the root of x^2 + 1 which does not exist. E. E. Escultura |
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| Edgar E. Escultura... |
| Posted: Thu Nov 05, 2009 10:29 pm |
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| This is the only serious comment of this thread so far and it's deeply appreciated. The source of the contradiction, however, is that the concept i = the root of the equation x^2 + 1 = 0 among the real numbers does not exist. i.e., i is ill-defined. E. E. Escultura. |
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| Ken Quirici... |
| Posted: Fri Nov 06, 2009 5:32 am |
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On Nov 5, 10:54 pm, A <anonymous.rubbert... at (no spam) yahoo.com> wrote:
[quote]On Nov 5, 7:32 pm, Tim Little <t... at (no spam) little-possums.net> wrote:
On 2009-11-05, Ken Quirici <kquir... at (no spam) yahoo.com> wrote:
Actually, could somebody explain why this is NOT
valid:
i = sqrt(1/-1) = sqrt 1/sqrt(-1) = 1/i,
The second equality is unjustified. It requires a property of sqrt
that holds in the reals, but cannot be extended to the whole complex
field.
When defining sqrt as a function, you have to choose which roots to
take for the nonzero values. With reals the most useful choice is the
set of all positive roots, as they are closed under multiplication.
However in the complex field there is no possible choice that is
closed under multiplication. At best you can say that
sqrt(x) sqrt(y) = +/- sqrt(x y).
So the original "paradox" becomes:
i = sqrt(1/-1) = +/- sqrt(1) / sqrt(-i) = +/- 1/i,
which is quite true.
- Tim
More generally, when one defines i as the square root of -1, one
chooses ONE of the two square roots of -1 in the complex numbers to be
i; the other is necessarily -i. The "paradox" above, and some portion
of Escultura's misunderstandings, is the result of failing to
distinguish between the two square roots of -1 in the complex numbers.
[/quote]
Well the above two replies make eminent good sense!
It's so easy when someone explains it to you. |
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| A... |
| Posted: Fri Nov 06, 2009 5:53 am |
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On Nov 6, 3:19 am, "Edgar E. Escultura" <escultu... at (no spam) yahoo.com> wrote:
[quote]Actually, traditional mathematicians choose one of the two square roots of -1 called principal value to avoid the contradiction. But this does not negate the fact that one can derive a contradiction from i. The root of this problem is the vacuous nature of i being the root of x^2 + 1 which does not exist. E. E. Escultura
[/quote]
What contradiction do you claim that one can derive from the existence
(in the complex numbers) of a square root of -1? There was an attempt
you made already in this thread but it was only based on your failure
to distinguish between i and -i. You also claim that Brouwer, Banach,
and Tarski constructed counterexamples to the generally recognized
properties of the real numbers. What are these counterexamples? |
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| fernando revilla... |
| Posted: Fri Nov 06, 2009 8:55 am |
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E. E. Escultura wrote:
[quote]2) The other fatal defect is that the complex number
system that Wiles used in the proof being based on
the vacuous concept i is also inconsistent. The
element i is the vacuous concept: the root of the
equation x^2 + 1 = 0 which does not exist
[/quote]
But it's well known that R^2 together with adequate operations
sum and product, provide to R^2 a field structure which contains
a sub field isomorphic to the standard (R, +, *). So there are objects,
for instance i:=(0,1), that satisfies i^2+1=0.
So, it seems you are denying the most pure essence about algebraic
concepts.
Regards. |
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| Brian... |
| Posted: Fri Nov 06, 2009 8:57 am |
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On Nov 6, 1:29 am, "Edgar E. Escultura" <escultu... at (no spam) yahoo.com> wrote:
[quote]This is the only serious comment of this thread so far and it's deeply appreciated. The source of the contradiction, however, is that the concept i = the root of the equation x^2 + 1 = 0 among the real numbers does not exist. i.e., i is ill-defined. E. E. Escultura.
[/quote]
Correct me if I'm wrong, you can define C to be R[x]/<x^2+1>, the
quotient ring of the set of polynomials with real coefficients modded
out by the ideal generated by x^2+1. The isomorphism sends x to i.
There is nothing ill-defined about that.
You can also define C by the set of ordered pairs of real numbers with
specific definitions for addition and multiplication for ordered pairs
of real numbers. Again, nothing ill-defined about that.
Moreover, you seem to be suggesting that since a polynomial equation
has no real roots implies that C is ill-defined. Maybe you can explain
why I'm wrong but that's like saying that since 3x+1=-1 has no roots
among the integers that Q is ill-defined, and like saying that since
x^2 = 2 has no rational roots that the set of irrationals are ill-
defined. There is a progression of number sets that allow for more
polynomial equations to be "solved" (ie, roots found), starting with
N, then Z, Q, R, and finally C. That C is algebraically closed
implies that polynomials will not yield further number sets. |
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