Fermat's "Last" Theorem Disproved?
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From: cjcountess on 31 Jul 2010 13:22 How much does âFermat's Last Theoremâ, depend on integers being dimensionless if at all? Because if it does, itâs done. And which of these statements from http://mathforum.org/dr.math/faq/faq.fermat.html is actually what Fermat saidâ? this one In the margin of his copy of a book by Diophantus, Pierre de Fermat wrote that it is possible to have a square be the sum of two squares, but that a cube can not be the sum of two cubes, nor a fourth power be a sum of two fourth powers, and so on. Further, he wrote that he had found a truly marvelous proof which the margin was too small to contain. Or this one Fermat's Last Theorem states that xn + yn = zn has no non-zero integer solutions for x, y and z when n > 2. With the first I can even argue as I did at very beginning, that âthe square of a squareâ, is where the theorem breaks down, whether I use the cube as the square of a square or a forth power. And with the second, if the ânon zero positive integersâ, absolutely must be dimensionless, my attempt to disprove the theorem has new life. This is because, even if we use so called âdimensionless pointsâ, to represent the integers, they still form â3 squaresâ as borders of a âPythagorean triangleâ, when laid out 3x3, 4x4, and 5x5, and as such are truly dimensional. Conrad J Countess Again let me humbly ask, "How much does âFermat's Last Theoremâ, depend on integers being dimensionless if at all? BECAUSE IF IT DOES, IT IS DONE.
From: cjcountess on 1 Aug 2010 11:58 On Jul 31, 1:22 pm, cjcountess <cjcount...(a)yahoo.com> wrote: > How much does âFermat's Last  Theoremâ, depend on integers being > dimensionless if at all? > > Because if it does, itâs done. > > And which of these statements fromhttp://mathforum.org/dr.math/faq/faq.fermat.html > is actually what Fermat saidâ? > > this one > > In the margin of his copy of a book by Diophantus, Pierre de Fermat > wrote that it is possible to have a square be the sum of two squares, > but that a cube can not be the sum of two cubes, nor a fourth power be > a sum of two fourth powers, and so on. Further, he wrote that he had > found a truly marvelous proof which the margin was too small to > contain. > > Or this one > Fermat's Last Theorem states that > xn + yn = zn > has no non-zero integer solutions for x, y and z when n > 2. > > With the first I can even argue as I did at very beginning, that âthe > square of a squareâ, is where the theorem breaks down, whether I use > the cube as the square of a square or a forth power. > > And with the second, if the ânon zero positive integersâ, absolutely > must be dimensionless, my attempt to disprove the theorem has new > life. > > This is because, even if we use so called âdimensionless pointsâ, to > represent the integers, they still form â3 squaresâ as borders of a > âPythagorean triangleâ, when laid out 3x3, 4x4, and 5x5, and as such > are truly dimensional. > > Conrad J Countess > > Again let me humbly ask, "How much does âFermat's Last Theoremâ, > depend on integers being dimensionless if at all? > > BECAUSE IF IT DOES, IT IS DONE. Just wanted to restate this: Concerning right triangle with 2 equal side2 = 1, and hypotonuse = sqrt2, the very fact that "sqrt2", is involved, which is not interger, somehow makes it not related to Fermats theorem. But is that the soul reason? If other 2 side are equal "1", this shows also that sense 1 in this case is also not an interger, because (1x1 does not = 1) but instead = "1square unite", as it is geometrical measure of "1 in linear direction x 1 in 90 degree angular direction = 1 square unit". This should be evidence that by extension, all other measurements concerning other triplets, are also geometrical, as nothing separates them but size of number. Conrad J Countess
From: Virgil on 1 Aug 2010 17:05 In article <28020d59-d50f-4360-8c08-8eb085236bf0(a)m17g2000prl.googlegroups.com>, cjcountess <cjcountess(a)yahoo.com> wrote: > Concerning right triangle with 2 equal side2 = 1, and hypotonuse = > sqrt2, the very fact that "sqrt2", is involved, which is not interger, > somehow makes it not related to Fermats theorem. But is that the soul > reason? "interger" should be "integer" "soul" should be "sole" > > If other 2 side are equal "1", this shows also that sense 1 in this > case is also not an interger Just what do you think is the difference between being an "interger" and a not being an "interger"? Is it so much different from being an integer? > because (1x1 does not = 1) but instead = > "1square unite", as it is geometrical measure of "1 in linear > direction x 1 in 90 degree angular direction = 1 square unit". Numbers do not have any implicit unit of measurement associated with them. If a unit of measure is to be associated, it must be explicit or at least implicit in the particular usage, which is NOT the case in FLT. > > This should be evidence that by extension, all other measurements > concerning other triplets, are also geometrical, as nothing separates > them but size of number. Why are they not weights or angles or something other than distances? It may be that in physics or engineering all numbers are assumed to have some associated unit of measurement tied to them, but FLT and the Pythagorean Theorem are in mathematics, and not in either physics or engineering. And in mathematics, numbers are ordinarily assumed to be without any units attached. So why are you assuming otherwise?
From: spudnik on 1 Aug 2010 21:29 the pytahgorean theorem is perfectly dimensional, as he and I both concern ourselves with "circling," instead of "tatragoning." that is, "Einstein's proof" via similarity, which he probably found at the gymnasium in Euclid, is merely diagrammatic as he gave it; the actual construction *is* the lunes proof (Hippocrates', I think, but different than the Oath's .-) > So why are you assuming otherwise? thus: in spite of his slogan about phase-sppace, Minkowski was a fantastic Nd geometer. anyway, it's downright innumerate to worry about it, without actually peeking at l'OEuvre de Fermatttt, but Hipparchus' (or Hippocrates') lunes proof is all that you need for the dimensionality of the 2d pythag. thm., if not the 3d pair of them (quadruplets). the main thing, though, is that Fermat didn't have to prove n=3, since his proof apparently applied to all of the odd primes; only the special case of n=4 does not fall to teh well-known lemma for composite exponents, and this he showed, in one of his rare expositions. thus: too bad, the unit associated with the pound, had to be associated with The newton -- the plagiarist, the spook, the freemason, the corpuscular "theorist" ... --les ducs d'oil! http://tarpley.net/online-books/george-bush-the-unauthorized-biograph... --Light, A History! http://wlym.com/~animations/fermat/index.html
From: cjcountess on 2 Aug 2010 14:17
On Aug 1, 5:05 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <28020d59-d50f-4360-8c08-8eb085236...(a)m17g2000prl.googlegroups.com>, OK: so you want to give me a math lesson, but you'll settel for a spelling leson > "interger" should be "integer" > "soul" should be "sole" > > Just what do you think is the difference between being an "interger" > and a not being an "interger"? Is it so much different from being an > integer? > > Numbers do not have any implicit unit of measurement associated with > them. If a unit of measure is to be associated, it must be explicit or > at least implicit in the particular usage, which is NOT the case in FLT. If the object is to square a value than the implicite unite must be a length or something analogous to a length. > > Why are they not weights or angles or something other than distances? On the quantum level these unites can indeed be unified > It may be that in physics or engineering all numbers are assumed to have > some associated unit of measurement tied to them, but FLT and the > Pythagorean Theorem are in mathematics, and not in either physics or > engineering. And in mathematics, numbers are ordinarily assumed to be > without any units attached. Ordinarily assumed to be, you say, but not nessesarily all the time. So what about the exceptions? > So why are you assuming otherwise? Because I am looking for the exceptions. I think I"ve proved my point Conrad J Countess |