"Fermat's Last Theorem" Disproved?
in [Math]
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From: David C. Ullrich on 15 Jul 2010 08:36 On Wed, 14 Jul 2010 13:13:34 -0600, Virgil <Virgil(a)home.esc> wrote: >In article <epnr36ltg0v8ev0bh6qulo1c76213h2i85(a)4ax.com>, > David C. Ullrich <ullrich(a)math.okstate.edu> wrote: > >> 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... > >Would not the finding of 3 such integers be a disproof by counterexample >of FLT? Good question. Leads to another question: Are you familiar with the expression "whoosh..."?
From: Frederick Williams on 15 Jul 2010 09:20 cjcountess wrote: > > Has anyone taken two cubes, created from two equal and right angular > lines of triangle, and one made from cube of hypotenuse, and measured > their volumes, to see if the �c� cube equals the �a� cube + the �b� > cube? > > Has this ever been done that anyone knows of ? > > That would be the proof. a, b and c need to be positive integers to disprove FLT. > So yes although (a^2)^2 = a^4 in normal mathematics from this > perspective (a^2)^2= a^3. At the risk of sounding impolite, _that_ is utter bollocks. The case n = 3 of FLT was (almost) dealt with by Euler in the sixteenth century, Legendre filled a gap. A proof may be found in Hardy & Wright section 13.4. They also give a reference to Dickson's History for Legendre's contribution. Wikipedia has copious references. -- I can't go on, I'll go on.
From: Thomas Nordhaus on 15 Jul 2010 09:38 Frederick Williams schrieb: > The case n = 3 of FLT was (almost) dealt with by Euler in the sixteenth > century Make that the eighteenths century. -- Thomas Nordhaus
From: Chip Eastham on 15 Jul 2010 09:45 On Jul 14, 12:59 pm, Mark Murray <w.h.o...(a)example.com> wrote: > On 4/07/2010 15:37, cjcountess wrote: > > > 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 > > 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. Apart from the OP's failure to construct solutions of a^3 + b^3 = c^3, there is also Conrad's failure to address the requirement that a,b,c be nonzero integers. Certainly nonzero real a,b,c exist that solve the equation, even if Conrad's geometry does not correspond to a solution. --c
From: Maarten Bergvelt on 15 Jul 2010 09:47
On 2010-07-15, Thomas Nordhaus <thnord2002(a)yahoo.de> wrote: > Frederick Williams schrieb: > >> The case n = 3 of FLT was (almost) dealt with by Euler in the sixteenth >> century > > Make that the eighteenths century. > No, Euler was such a genius, he proved it before his mother was born. -- Maarten Bergvelt |