Re: Counterexamples to FLT
From: Brian Quincy Hutchings (QncyMI_at_netscape.net)
Date: 12/01/04
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Date: 30 Nov 2004 17:16:50 -0800
congratulation.
I'm using your proof in a short course,
Implausibly Indeniable Proofs of a Really Old Problem. the assumption
will
be that these 4 or 5 or 6 elementary proofs taht I have found,
mostly in the literature, are incorrect,
either in execution or in concept (not "XOR" .-)
that's all of the examples that I'm doing,
thus far.
winners will have their entries submitted
to the annual ... I can't pronounce it, but
it's associated with ... a student of Catalan, or
the U. of Catalan ... have to study his number!
second congratulation.
I'm using the original formatting.
escultur36@hotmail.com ("E. E. Escul;tura") wrote in message news:<200411301800.iAUI0Rf20992@proapp.mathforum.org>...
> I am pleased to summarize the resolution of some issues in
mathematics such as the Ullrich-Israel dilemma: whether 1 = 0.99? and
the status of Wiles? proof of FLT. These are some of the findings: (1)
The number 0.99? is not well-defined in the real number system,
therefore, the equation is nonsense. (2) The real number system is
itself ill-defined because two of its axioms, completeness and the
dichotomy axioms are false. Banach-Tarski and Brouwer constructed
counterexamples to them. Needless to say, without fixing the real
number system FLT remains nonsense and Wiles? ?proof? is nonsense. (3)
The way to fix the real number system is to well-define it by simple
set of axioms without these false axioms. They are as follows: Denote
the new real number system by R*, +, x. Axiom 1: R* contains the basic
integers 0, 1, 2, 3, 4, 4, 5, 6, 7, 8, 9. Axioms 1 and 2: The addition
and multiplication tables of elementary arithmetic. The well-define
addition and multiplication on the inte!
> gers. Then I defined 10 = 1+9. Then a terminating decimal is a
polynomial in decreasing powers of 10 (including negative powers) over
the basic integers. For example, abc.efgh = a10^2 + b10 + c = e/10 +
f/10^2 + g/10^3 + h/10^4. A nonterminating decimal is well-defined
provided every digit is known or computable. Specifically, it has the
form N.abc?s(k)?, where the kth decimal digit s(k) is known or
computable. This number can be written as standard Cauchy sequence:
N.a, N.ab, ?N.abc?s(k), ? Thus most irrationals are ill-defined since
s(k) is unknown in general. At the same time all periodic decimals are
well-defined. The number sqrt(p), p is prime, is well-defined because
its standard Cauchy sequence can be computed. An integer is simply the
integral part of a decimal. This well-defines the integers as a
subspace of the new real number system. Presently, the integers are
not well-defined. This is the main problem of number theory. Brouwer?s
counterexample to the dichotomy a!
> xiom also shows that the irrationals are ill-defined and the real
number system has no natural ordering. The new real number system has
a natural ordering, namely, its lexicographic ordering. Note that
0.99? is well-defined in the new real number system since every digit
is known. In the natural ordering of this space 0.99? < 1. Therefore,
d* = 1 ? 0.99? is well-defined. I call it dark number; it satisfies
the following properties: with two exceptions, for any nonzero number
x, xd* = d*, x + d* = x, d*^N = d* for any integer N, N.99? + d* =
(N+1), d* is the smallest positive new real number and it is unique in
view of the natural ordering of the new real numbers. Here?s a
sampling of the new arithmetic (Applied Mathematics and Computation,
Vol. 138, 2001). Let K be an integer, M.99... and N.99... new
integers. Then (1) K + M.99... = (K+M).99...; (2) K(M.99...) = K(M +
0.99...) = KM + K(0.99...) = KM + (K-1).99...; (3) M.99... + N.99... =
M + N + 0.99... + 0.99 = M + N + 2(0.!
> 99...) = M + N + 1.99...; to verify that 2(0.99...) = 1.99...,
calculate (1.99...)/2 to obtain 0.99... . (4) (M.99...)(N.99...) = (M
+ 0.99...)(N + 0.99...) = MN + M(0.99...) + N(0.99...) + (0.99...)2 =
MN + (M - 1).99... + (N -1).99... + 0.99... = MN + (M + N - 2).99... +
0.99... = MN + (M + N - 1).99... = (MN+M+N -1).99... . These numbers,
including d*, are called new integers because they are isomorphic to
the intgers; therefore, they share the properties of the integers. In
particular, d* behaves like 0 and N.99? behaves like N+1. The
exceptions for x in xd* = d* and x+d* =x are x = 0 and x = N.99?, N =
0, 1, ?, respectively. The countable counterexamples FLT are as
follows: Let x = (0.99?)10^T, y = d*, z = 10^T, T = 1, 2,... Then x^n
+ y^n = (0.99?)10^nT + d* = 10^nT = z^n. Since x, y, z are not equal
to 0, we have here countable counterexamples to FLT. Therefore, the
conjecture is false. Moreover, for any integer k, (kx)^n + (ky)^n =
k^n(x^n + y^n) = k^n(z^n) = (kz)^n.!
> Therefore, the nonzero integers, k(0.99?)10^T, d*, k10^T, k = 1, 2,
?, and k = 1, 2, ?, satisfy Fermat?s equation and are also
counterexamples to Fermat?s FLT. We have here countable unbounded
counterexamples to Fermat?s last theorem. My original resolution of
FLT (Nonlinear Studies, Vol. 5) uses the notion of Cauchy
representation and modular equality in the sense of the new
nonstandard analysis. For details, visit my websites:
http://www.users.bigpond.com/pidro/home.htm,
http://home.iprimus.com.au/pidro/. See also Nonlinear Studies, Vol. 5,
1998, Applied Mathematics and Computation, Vols. 130 and 138 as well
as the following threads in the archives of MathForge.net:
Constructivist principles ?; Simpler Proof of FLT.
--Advice, 0.05; free, if wrong!
http://tarpley.net/bush5.htm
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