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| Edgar E. Escultura... |
 Posted: Fri Nov 06, 2009 5:03 pm |
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| I have seen this name since 1997 but he has not learned a bit since then. See my response to Justin Benfield; that is the same response to your question. E. E. Escultura |
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| Edgar E. Escultura... |
| Posted: Fri Nov 06, 2009 5:07 pm |
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When one cannot refute or contribute to a comment he can only resort to name-calling that reveals intellectual inadequacy and insecurity because the top is empty.
E. E. Escultura |
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| Edgar E. Escultura... |
| Posted: Fri Nov 06, 2009 5:41 pm |
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| I can only talk about decimals because at least the terminating decimals are well-defined but the real numbers are not; the field axioms that supposedly define them are inconsistent. Thanks, you are one of the very few who make sense on this thread. I think your name rings a bell from the past. E. E. Escultura |
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| fishfry... |
| Posted: Fri Nov 06, 2009 8:50 pm |
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In article
<357014399.28642.1257561544078.JavaMail.root at (no spam) gallium.mathforum.org>,
"Edgar E. Escultura" <escultur36 at (no spam) yahoo.com> wrote:
[quote]1 and 0.99... are distinct objects in the real world called decimals. As
distinct objects they are like apple and orange and to write apple = orange
is certainly nonsense. E. E. Escultura
[/quote]
I agree with you, 1.0000... and .9999 are distinct decimals.
Do you also deny that they represent the same real number? |
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| A... |
| Posted: Sat Nov 07, 2009 5:56 am |
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On Nov 5, 5:21 am, "Edgar E. Escultura" <escultu... at (no spam) yahoo.com> wrote:
[quote]CLARIFICATION OF THE COUNTEREXAMPLES TO FERMAT’S LAST THEOREM
By E. E. Escultura
Although all issues related to the resolution of Fermat’s last theorem have been fully debated worldwide since 1997 and NOTHING had been conceded from my side I have seen at least one post expressing some misunderstanding.. Let me, therefore, make the following clarification:
1) The decimal integers N.99… , N = 0, 1, …, are well-defined nonterminating decimals among the new real numbers [8] and are isomorphic to the ordinary integers, i.e., integral parts of the decimals, under the mapping, d* -> 0, N+1 -> N.99… Therefore, the decimal integers are integers [3]. The kernel of this isomorphism is (d*,1) and its image is (0,0.99…). Therefore, (d*)^n = d* since 0^n = 0 and (0.99…)^n = 0.99… since 1^n = 1 for any integer n > 2.
2) From the definition of d* [8], N+1 – d* = N.99… so that N.99… + d* = N+1. Moreover, If N is an integer, then (0.99…)^n = 0.99… and it follows that ((0.99,..)10)^N = (9.99…)10^N, ((0.99,..)10)^N + d* = 10^N, N = 1, 2, … [8].
3) Then the exact solutions of Fermat’s equation are given by the triple (x,y,z) = ((0.99…)10^T,d*,10^T), T = 1, 2, …, that clearly satisfies Fermat’s equation,
x^n + y^n = z^n, (F)
for n = NT > 2. The counterexamples are exact because the decimal integers and the dark number d* involved in the solution are well-defined and are not approximations.
4) Moreover, for k = 1, 2, …, the triple (kx,ky,kz) also satisfies Fermat’s equation. They are the countably infinite counterexamples to FLT that prove the conjecture false [8]. They are exact solutions, not approximation. One counterexample is, of course, sufficient to disprove a conjecture..
The following references include references used in the consolidated paper [8] plus [2] which applies [8]
References
[1] Benacerraf, P. and Putnam, H. (1985) Philosophy of Mathematics, Cambridge University Press, Cambridge, 52 - 61.
[2] Brania, A., and Sambandham, M., Symbolic Dynamics of the Shift Map in R*, Proc. 5th International
Conference on Dynamic Systems and Applications, 5 (2008), 68–72.
[3] Corporate Mathematical Society of Japan , Kiyosi Itō, Encyclopedic dictionary of mathematics (2nd ed.), MIT Press, Cambridge, MA, 1993
[4] Escultura, E. E. (1997) Exact solutions of Fermat's equation (Definitive resolution of Fermat’s last theorem, 5(2), 227 – 2254.
[5] Escultura, E. E. (2002) The mathematics of the new physics, J. Applied Mathematics and Computations, 130(1), 145 – 169.
[6] Escultura, E. E. (2003) The new mathematics and physics, J. Applied Mathematics and Computation, 138(1), 127 – 149.
[7] Escultura, E. E., The new real number system and discrete computation and calculus, 17 (2009), 59 – 84.
[8] Escultura, E. E., Extending the reach of computation, Applied Mathematics Letters, Applied Mathematics Letters 21(10), 2007, 1074-1081.
[9] Escultura, E. E., The mathematics of the grand unified theory, in press, Nonlinear Analysis, Series A:
Theory, Methods and Applications; online at Science Direct website
[10] Escultura, E. E., The generalized integral as dual of Schwarz distribution, in press, Nonlinear Studies.
[11] Escultura, E. E., Revisiting the hybrid real number system, Nonlinear Analysis, Series C: Hybrid Systems, 3(2) May 2009, 101-107.
[12] Escultura, E. E., Lakshmikantham, V., and Leela, S., The Hybrid Grand Unified Theory, Atlantis (Elsevier Science, Ltd.), 2009, Paris.
[13] Counterexamples to Fermat’s last theorem,http://users.tpg.com.au/pidro/
[14] Kline, M., Mathematics: The Loss of Certainty, Cambridge University Press, 1985.
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]
The following post in this thread was made by E. E. Escultura but I do
not see it on Google Groups, only on mathforum.org:
[quote]I have already explained the contradiction in i many times and one in this thread is my reply to the anonymous
guy. I refer you to it. The counterexample to the trichotomy axiom by Brouwer is Benacerraf, P. and Putnam, H.
(1985) Philosophy of Mathematics, Cambridge University Press, Cambridge, 52 - 61, and my own version is in my
paper, The new real number system and discrete computation and calculus, 17 (2009), 59 ? 84. Both
counterexamles prove that the real numbers are not linearly ordered by "<" (natural ordering) and that an irrational
number is not the limit of a sequence of rational numbers in the standard norm. the Banach-Tarski paradox
stemming from the axiom of choice (a variant of the completeness axiom of the field axioms of the real number
system) is discussed in these papers: Bhaskar, T. G., Kovak, D., Lakshmikantham V. (2006) The Hybrid Set
Theory, Nonlinear Analysis; C-Series, Hybrid Systems and Applications. and, Kline, M. Mathematics: The Loss of
Certainty, Oxford University Press, New York, 1980.
Your question makes sense and deserves a serious response. E. E. Escultura
[/quote]
I am not sure why this is (I don't use mathforum.org--are not all
posts there actually made to the Usenet group?) or whether people with
real newsreaders were able to see this post.
I am also having trouble telling who E. E. Escultura is replying to in
each post, since he does not quote the person he replies to. Mr.
Escultura, it would be nice if you would quote the person you reply to
in your posts, so that when you say "you" it is easier to tell who you
are referring to.
In any case, the paper by Heyting on pgs. 52-61 of the book of
Benacerraf and Putnam is about an intuitionistic foundation for
mathematics, and the difficulties in constructing the real numbers
which he discusses there are only difficulties when using
intuitionistic logic (i.e., one is not allowed to use the argument
"the negation of P is not true implies that P is true" in one's
proofs). There are no contradictions to be found in this paper for
anyone who uses the more standard logic in which ~~P implies P.
The Banach-Tarski paradox is likewise not a contradiction but merely a
counterintuitive fact, one which indicates the need for measure theory
(and the need to distinguish between measurable and non-measurable
sets) in order to have a usable theory of integration.
Why are you claiming that these things represent contradictions in the
construction of the real numbers? They do not, unless you either
insist on using intuitionistic logic, or you insist on ignoring or
misusing measure theory. |
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| fishfry... |
| Posted: Sat Nov 07, 2009 8:38 am |
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In article
<531212911.28663.1257562742387.JavaMail.root at (no spam) gallium.mathforum.org>,
"Edgar E. Escultura" <escultur36 at (no spam) yahoo.com> wrote:
[quote]1/3 is a binary operation that maps the pair of decimals (1,3) to the decimal
0.33...; strictly speaking, 1/3 and 0.33... are not equal; rather, 0.33... is
the image of 1/3. Thus, the equation, 1/3 = 0.33... is sloppy mathematics.
Moreover, one cannot multiply or add nonterminating decimals because either
operation needs the last digit on the right. One can only approximate the
reult. Therefore, the equation, 3(1/3) = 0.99..., is incorrect because the
left side of the equation is only an approximation of the right side.
BTW, this comment is well informed and so I respond in the same manner. E. E.
Escultura
[/quote]
When you teach freshman calculus, do you cover the geometric series?
Doesn't 3/10 + 3/100 + 3/1000 + ... converge to 1/3 without any
reference to the notion of decimals? |
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| Edgar E. Escultura... |
| Posted: Sat Nov 07, 2009 7:38 pm |
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When you teach freshman calculus, do you cover the geometric series?
Doesn't 3/10 + 3/100 + 3/1000 + ... converge to 1/3 without any
reference to the notion of decimals?
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We are just dealing with the basics now. As soon as we have resolved these basic issues we will proceed to calculus and even analysis in general. I have started the rectification of the real number system with the paper, The new real number system and discrete computation and calculus an overview of which is on my website. E. E. Escultura |
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| Edgar E. Escultura... |
| Posted: Sat Nov 07, 2009 10:04 pm |
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First of all 0.999...=1, provided of course we are speaking about the real numbers and not some alternate set of numbers. There are many ways to prove this, if you agree that 0.333....=1/3, then 3*(1/3)=(3/1)*(1/3)=(3*1)/(1*3)=3/3=1/1=1 and 3*(1/3)=3*0.333....=0.999.... which implies 0.999...=1. 0.999... and 1 are distinct notations but a quantity has many possible ways of being written. Another way to prove 0.999...=1 is to use a summation, from the definition of decimal notation 0.999...=(9/(10^1))+((9/(10^2))+(9/(10^3))+... which can be written as a summation of (9/(10^n)) as n goes from 1 to infinity, it isn't too difficult to evaluate this sum.
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1) 1 and 0.99... are distinct mamthematical objects, i.e., distinct decimals, like apple and orange and there is something wrong in the statement apple = orange.
2) Since a decimal is determined or well-defined by its digits nonterminating decimals are ambiguous and ill-defined because we do not know all its digits. Therefore, there is something wrong with the equation 1 = 0.99... since the right side is unknown, ill-defined.
Incidentally, we cannot add or multiply a nonterminating decimal because the operation requires the last digit on the right to carry out it out. we can only approximate. Therefore, the equation is never attained. you might write the equation 1 = the limit of 0.99... in the standard norm but that limit is not the decimal 0.99... it's another concept.
FLT states that x,y,z are positive nonzero integers so the come from the set {1,2,3,4,...}, and n is a positive integer greater than 2. Andrew Wiles successfully proved the theorem in 1994-1995.
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I have posted elsewhere the two errors in Wile's proof:
1) The integers as real numbers are not presently well-defined because the field axioms which supposedly well-define them are inconsistent
2) Complex analysis which he used in the proof is flawed because the imaginary concept i = the root of the equation i^2 + 1 = 0 among the real numbers does not exist. Therefore, i is a vacuous or ill-defined concept. That is why a contradiction can be extracted from it.
E. E. Escultura |
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| Edgar E. Escultura... |
| Posted: Sat Nov 07, 2009 10:13 pm |
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the real joke is that EEE fails to give us a "real one,"
1.0000..., ro compare to a real 0.9999... -- alas.
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1 and 0.99... are distinct decimals, distinct objects like apple and orange and to write apple = orange is simply nonsense. E.E.Escultura |
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| Edgar E. Escultura... |
| Posted: Sat Nov 07, 2009 10:26 pm |
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I don't think that he is smart enough to qualify
as an idiot. Brian Q. Hutchings
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When the top is empty one can only resort to name calling that reveals intellectual inadequacy and racism. I have seen this name a long time ago since 1997 and the guy has not learned a bit; he can only post from the flat of his foot but nothing of substance. He should join the Society of Empty Toppers. E. E. Escultura |
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| Edgar E. Escultura... |
| Posted: Sat Nov 07, 2009 10:31 pm |
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I don't think that he is smart enough to qualify
as an idiot. Pubkeybreaker
When the top is empty one can only resort to name calling that reveals intellectual inadequacy and racism. I have seen this name a long time ago and the guy has not learned a bit; he can only post from the flat of his foot but nothing of substance. He should join the Society of Empty Toppers. E. E. Escultura |
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| Nunemica... |
| Posted: Mon Nov 09, 2009 3:07 am |
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On Nov 5, 11:49 pm, Justin Benfield <Justin.Benfi... at (no spam) gmail.com> wrote:
[quote]First of all 0.999...=1, provided of course we are speaking about the real numbers and not some alternate set of numbers. There are many ways to prove this, if you agree that 0.333....=1/3, then 3*(1/3)=(3/1)*(1/3)=(3*1)/(1*3)=3/3=1/1=1 and 3*(1/3)=3*0.333....=0.999.... which implies 0.999...=1. 0.999...
[/quote]
This is the problem .999 only implies "1" but it is not "1".
If .999 is 1 then 1 must be 1.001
1.001 + .999 = 2
and for 1 to be 1.001 then (zero)0 = 0.001 |
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| Nunemica... |
| Posted: Mon Nov 09, 2009 3:19 am |
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On Nov 9, 5:07 am, Nunemica <tinabarbarar... at (no spam) gmail.com> wrote:
[quote]On Nov 5, 11:49 pm, Justin Benfield <Justin.Benfi... at (no spam) gmail.com> wrote:
First of all 0.999...=1,
[/quote]
Sorry just a slight edit!
[quote]
If .999...= 1 then 1 = 1.001... also
1.001 + .999 = 2[/quote] |
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| Edgar E. Escultura... |
| Posted: Mon Nov 09, 2009 6:20 pm |
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Pubkeybraker:
You're an idiot.- Hide quoted text -
I don't think that he is smart enough to qualify
as an idiot.
Racists need frustrate surgery to remove sourgrapes and placate their inadequacy. E. E. Escultura |
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| Tim Little... |
| Posted: Mon Nov 09, 2009 9:47 pm |
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On 2009-11-09, Nunemica <tinabarbararyan at (no spam) gmail.com> wrote:
[quote]Sorry just a slight edit!
If .999...= 1 then 1 = 1.001... also
[/quote]
What is your definition for the notation "1.001..."? Do you mean that
the pattern "001" in the decimal portion repeats?
- Tim |
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