"Fermat's Last Theorem" Disproved?
in [Math]
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From: cjcountess on 14 Jul 2010 10:37 FERMATS LAST THEOREM, disproved? Take a square that is cut in half diagonally, to give us a 90 degree triangle, of two equal right angular sides, and one longer hypotenuse, or slanted side. Label each of the right angular straight sides, a and b, and the longer slanted hypotenuse, c. Next, applying the Pythagorean Theorem, square each of the 3 sides, turning each of the one dimensional lines to 2 dimensional squares, and we find that, a^2 + b^2 = c^2 Now to take it to a higher level, in order to test Fermat's Last Theorem, which states that this ( a^n + b^n = c^n), formula, only works for numbers of which the exponent is = to 2. Lay each of the 2 D sides on their sides and square them, also upward, at 90 degrees, which instead of extending the whole 1D line upward a length equal to its self equaling a 1 inch square, will extend the 2D square upward a length equal also to itself.. If the Pythagorean theorem still holds, (the square of a^2 + the square of b^2, = the square of c^2), and sense the square of each square is a cube, a^3 + b^3 = c^3 "FERMAT'S LAST THEOREM" which was recently said to be proven, is not. Conrad J Countess
From: Frederick Williams on 14 Jul 2010 10:57 cjcountess wrote: > > �FERMAT�S LAST THEOREM�, disproved? > > Take a square that is cut in half diagonally, to give us a 90 degree > triangle, of two equal right angular sides, and one longer hypotenuse, > or slanted side. > > Label each of the right angular straight sides, �a� and �b�, and the > longer slanted hypotenuse, �c�. > > Next, applying the Pythagorean Theorem, square each of the 3 sides, > turning each of the one dimensional lines to 2 dimensional squares, > and we find that, a^2 + b^2 = c^2 > > Now to take it to a higher level, in order to test �Fermat's Last > Theorem�, which states that this ( a^n + b^n = c^n), formula, only > works for numbers of which the exponent is = to �2�. > > Lay each of the �2 D� sides on their sides and square them, also > upward, at 90 degrees, which instead of extending the whole �1D� line > upward a length equal to its self equaling a 1 inch square, will > extend the 2D square upward a length equal also to itself.. > > If the Pythagorean theorem still holds, (the square of �a^2� + the > square of �b^2�, = the square of c^2), and sense the square of each > square is a cube, a^3 + b^3 = c^3 > > "FERMAT'S LAST THEOREM" which was recently said to be proven, is not. > > Conrad J Countess So now find three integers greater than nought that satisfy a^3 + b^3 = c^3. -- I can't go on, I'll go on.
From: David C. Ullrich on 14 Jul 2010 12:01 On Wed, 14 Jul 2010 15:57:08 +0100, Frederick Williams <frederick.williams2(a)tesco.net> wrote: >cjcountess wrote: >> >> �FERMAT�S LAST THEOREM�, disproved? >> >> Take a square that is cut in half diagonally, to give us a 90 degree >> triangle, of two equal right angular sides, and one longer hypotenuse, >> or slanted side. >> >> Label each of the right angular straight sides, �a� and �b�, and the >> longer slanted hypotenuse, �c�. >> >> Next, applying the Pythagorean Theorem, square each of the 3 sides, >> turning each of the one dimensional lines to 2 dimensional squares, >> and we find that, a^2 + b^2 = c^2 >> >> Now to take it to a higher level, in order to test �Fermat's Last >> Theorem�, which states that this ( a^n + b^n = c^n), formula, only >> works for numbers of which the exponent is = to �2�. >> >> Lay each of the �2 D� sides on their sides and square them, also >> upward, at 90 degrees, which instead of extending the whole �1D� line >> upward a length equal to its self equaling a 1 inch square, will >> extend the 2D square upward a length equal also to itself.. >> >> If the Pythagorean theorem still holds, (the square of �a^2� + the >> square of �b^2�, = the square of c^2), and sense the square of each >> square is a cube, a^3 + b^3 = c^3 >> >> "FERMAT'S LAST THEOREM" which was recently said to be proven, is not. >> >> Conrad J Countess > >So now find three integers greater than nought that satisfy a^3 + b^3 = >c^3. Why? I don't see what that has to do with Fermat's Last Theorem...
From: Mark Murray on 14 Jul 2010 12:59 On 4/07/2010 15:37, cjcountess wrote: > �FERMAT�S LAST THEOREM�, disproved? > > > Take a square that is cut in half diagonally, to give us a 90 degree > triangle, of two equal right angular sides, and one longer hypotenuse, > or slanted side. > > Label each of the right angular straight sides, �a� and �b�, and the > longer slanted hypotenuse, �c�. > > Next, applying the Pythagorean Theorem, square each of the 3 sides, > turning each of the one dimensional lines to 2 dimensional squares, > and we find that, a^2 + b^2 = c^2 Or, if you want to be a bit more accurate, and keeping the sides separate a^2 + a^2 = 2 a^2 2 sides of length a, and a hypotenuse of length sqrt(2 a^2), where sqrt(x) means "the positive square root of x". Thus, you have 2 squares of area a^2, that do indeed equal the area of the square on the hypotenuse - 2 a^2. > Now to take it to a higher level, in order to test �Fermat's Last > Theorem�, which states that this ( a^n + b^n = c^n), formula, only > works for numbers of which the exponent is = to �2�. > > Lay each of the �2 D� sides on their sides and square them, also > upward, at 90 degrees, which instead of extending the whole �1D� line > upward a length equal to its self equaling a 1 inch square, will > extend the 2D square upward a length equal also to itself.. For each side of length a, we have a cube of volume a^3. For the hypotenuse, we have a cube of volume sqrt(2 a^2)^3. 2 a^3 is not equal to sqrt(2 a^2)^3. > If the Pythagorean theorem still holds, (the square of �a^2� + the > square of �b^2�, = the square of c^2), and sense the square of each > square is a cube, a^3 + b^3 = c^3 The square of a square is NOT a cube. It is a power of FOUR. Apart from that, the above statement is somewhere betwen wrong and meaningless (hard to tell which due to unclear writing). > "FERMAT'S LAST THEOREM" which was recently said to be proven, is not. Not even close. M -- Mark "No Nickname" Murray Notable nebbish, extreme generalist.
From: spudnik on 14 Jul 2010 14:39
nice demonstration of "doubling the hexahedron," en passant! > For the hypotenuse, we have a cube of volume sqrt(2 a^2)^3. thus&so: <deletives impleted> just don't leave a time-tunnel in the vicinity of your grandfather, if he is still alive, because he might configure what you "were about" to do, and hi to the future to prevent you, or the past to give a condom to your dad. "Granpa, it was going to be an accident ... I mean...." "But, Dad, we're Catholic!" > Scientific concensus today isn't your great grandaddy's scientific thus&so: grammar is just a part of the three Rs, the minimum you have to know, to be a literate slave -- and what some so-called Republicans call, "the basics," to impart learning-disorders amongst the rabble's youth, viz Murray and What's-his-name. thus&so: first of all, bloodletting has some current back-up ... or, at least, leeches are pretty useful in surgery. secondly, someone "above" made some statement about graphs (that is, quantification) in the harder sciences (although it seems that the soft ones use tons of statistical algorithms), and I'd like to cite the NYTimes weatherpage as a source of subliminal justification for the algorithms of the GCMers. the more qualitative aspect of that page, is the daliy vignettes on various things about weather -- n'est, mesoclimate. my random reading of this shows that cold records are at least as common as hot records, whereby goes my primary (nonquant) take on the phrase, global warming. just say, the climate, she a-changin', and rest easy! > errors as blood letting "scientists" is ridiculous. --Rep. Waxman's "new" cap&trade, same as his circa '91?... Is the House Banking Bill, before Senate, cap&trade?... les ducs d'oil! http://tarpley.net |