"Fermat's Last Theorem" Disproved?
in [Math]
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From: Virgil on 14 Jul 2010 15:13 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?
From: curious george on 14 Jul 2010 16:05 "cjcountess" <cjcountess(a)yahoo.com> wrote in message news:f63cb198-95e1-4ad2-8fd9-1e76a2aba43b(a)q12g2000yqj.googlegroups.com... �FERMAT�S LAST THEOREM�, disproved? >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 >Conrad J Countess can you clarify what you mean by " since a square of each square is a cube " I am sure you do not mean (a^2)^2= a^3.
From: cjcountess on 14 Jul 2010 17:18 On Jul 14, 4:05 pm, "curious george" <bu...(a)bunch.net> wrote: > "cjcountess" <cjcount...(a)yahoo.com> wrote in message > > news:f63cb198-95e1-4ad2-8fd9-1e76a2aba43b(a)q12g2000yqj.googlegroups.com... > FERMATS LAST THEOREM, disproved? > > >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 > >Conrad J Countess > > can you clarify what you mean by " since a square of each square is a cube " > I am sure you do not mean (a^2)^2= a^3. .. When I say square the square I am looking at it geometrically this way. Let me be clear. Just as you take a 1 dimensional line and square it to make a 2 dimensional surface, it is as layering line upon line of equal measure, until they stack up to equal to length of original line. Now if we lay that 2 dimensional surface accumulation of equal stacked lines, flat and extend the entire surface into the 3rd dimension also to that same height, it is as extending each of the lines used to create the flat 2 d surface, into the 2d in the 90 degree angular direction to the flat surface the same amount, and the accumulative effect should be that of a cube. The volume of the cube made from a, + the volume of the cube made from b, = the volume of the cube made from c. 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. So yes although (a^2)^2 = a^4 in normal mathematics from this perspective (a^2)^2= a^3. It is analogous to saying that just as (1x1=1) in linear math, (1 unit length x 1 unite length in 90 degree angular direction = 1 square inch) in geometry and (1 velocity vector in linear direction x 1 equal and 90 degree angular velocity vector creates v^2 and a balance of centrifugal/centripetal forces for circular motion in more dynamic forms of mathematics. I don't know of any other way of saying it except as the square of the square when looked at this way even thought it conflicts with conventional math. But with appropriate explanation it should be clear that the cube can be seen as the square of a square. Conrad J Countess
From: Virgil on 14 Jul 2010 18:17 In article <73543717-90cd-4a70-ac7e-5fa82c5eeee3(a)j13g2000yqj.googlegroups.com>, cjcountess <cjcountess(a)yahoo.com> wrote: > On Jul 14, 4:05�pm, "curious george" <bu...(a)bunch.net> wrote: > > "cjcountess" <cjcount...(a)yahoo.com> wrote in message > > > > news:f63cb198-95e1-4ad2-8fd9-1e76a2aba43b(a)q12g2000yqj.googlegroups.com... > > �FERMAT�S �LAST �THEOREM�, disproved? > > > > >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 > > >Conrad J Countess > > > > can you clarify what you mean by " since a square of each square is a cube " > > I am sure you do not mean (a^2)^2= a^3. > > . > When I say square the square I am looking at it geometrically this > way. > > Let me be clear. Just as you take a 1 dimensional line and square it > to make a 2 dimensional surface, it is as layering line upon line of > equal measure, until they stack up to equal to length of original > line. > > Now if we lay that 2 dimensional surface �accumulation of equal > stacked lines�, flat and extend the entire surface into the 3rd > dimension also to that same height, it is as extending each of the > lines used to create the flat 2 d surface, into the 2d in the 90 > degree angular direction to the flat surface the same amount, and the > accumulative effect should be that of a cube. > > The volume of the cube made from �a�, + the volume of the cube made > from �b�, = the volume of the cube made from �c�. > > > 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. > > So yes although (a^2)^2 = a^4 in normal mathematics from this > perspective (a^2)^2= a^3. Nonsense! To get from a line to a square (or, more generally, a rectangle) one moves the line in a direction perpendicular to the line, which increases the dimension by 1. Similarly, moving a square (or rectangle) in a direction perpendicular to it sweeps out a cube (rectangular parallelopiped or box shape), but there is no "square of a square" in that process anywhere. And your non-math certainly does not prove or disprove anything about anything mathematical.
From: cjcountess on 14 Jul 2010 19:22
On Jul 14, 6:17 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <73543717-90cd-4a70-ac7e-5fa82c5ee...(a)j13g2000yqj.googlegroups.com>, > > > > cjcountess<cjcount...(a)yahoo.com> wrote: > > On Jul 14, 4:05 pm, "curious george" <bu...(a)bunch.net> wrote: > > > "cjcountess" <cjcount...(a)yahoo.com> wrote in message > > > >news:f63cb198-95e1-4ad2-8fd9-1e76a2aba43b(a)q12g2000yqj.googlegroups.com.... > > > ³FERMAT²S LAST THEOREM², disproved? > > > > >If the Pythagorean theorem still holds, (the square of ³a^2² + the > > > >square of ³b^2², = the square of c^2), and sense the square of each > > > >square is a cube, a^3 + b^3 = c^3 > > > >Conrad J Countess > > > > can you clarify what you mean by " since a square of each square is a cube " > > > I am sure you do not mean (a^2)^2= a^3. > > > . > > When I say square the square I am looking at it geometrically this > > way. > > > Let me be clear. Just as you take a 1 dimensional line and square it > > to make a 2 dimensional surface, it is as layering line upon line of > > equal measure, until they stack up to equal to length of original > > line. > > > Now if we lay that 2 dimensional surface ³accumulation of equal > > stacked lines², flat and extend the entire surface into the 3rd > > dimension also to that same height, it is as extending each of the > > lines used to create the flat 2 d surface, into the 2d in the 90 > > degree angular direction to the flat surface the same amount, and the > > accumulative effect should be that of a cube. > > > The volume of the cube made from ³a², + the volume of the cube made > > from ³b², = the volume of the cube made from ³c². > > > 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 ³c² cube equals the ³a² cube + the ³b² > > cube? > > > Has this ever been done that anyone knows of ? > > > That would be the proof. > > > So yes although (a^2)^2 = a^4 in normal mathematics from this > > perspective (a^2)^2= a^3. > > Nonsense! > > To get from a line to a square (or, more generally, a rectangle) one > moves the line in a direction perpendicular to the line, which increases > the dimension by 1. > > Similarly, moving a square (or rectangle) in a direction perpendicular > to it sweeps out a cube (rectangular parallelopiped or box shape), Precisely but > there is no "square of a square" in that process anywhere. Wrong When we square a 1 D line, we extend that 1 D line at right angle into 2 dimensions, and when we cube a line we further extend that same 2 D square into 3 dimensions in right angular direction, in effect repeating the same process we did with the line, to the square. Thus in fact we are squaring the square, just as we squared the line. Its the same process, call it what you want. > > And your non-math certainly does not prove or disprove anything about > anything mathematical. But I proved the theorem untrue with my non conventional math. Comment on the measurement of volume, will the volume of "a^3" + the volume of "b^3" = the volume of "c^3"? Conrad J Countess |