"Fermat's Last Theorem" Disproved?
in [Math]
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From: Mike Terry on 14 Jul 2010 19:55 "cjcountess" <cjcountess(a)yahoo.com> wrote in message news:f63cb198-95e1-4ad2-8fd9-1e76a2aba43b(a)q12g2000yqj.googlegroups.com... > �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, Well it doesn't. Pythagorean is proved only when squares are involved, not cubes. So that's the end of your disproof. Mike. > (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: Rick Decker on 14 Jul 2010 21:19 On 7/14/10 7:22 PM, cjcountess wrote: <snip> >> >>> 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 ? > Presumably (since I just did, for example). > > But I proved the theorem untrue with my non conventional math. > Non-conventional, perhaps; wrong, certainly. > Comment on the measurement of volume, will the volume of "a^3" + the > volume of "b^3" = the volume of "c^3"? Certainly not. In fact, a^3 + b^3 will be *less* than c^3 using your construction, for any (non-degenerate) right triangle with shorter sides of length a, b, and hypotenuse c. > Regards, Rick
From: Virgil on 14 Jul 2010 23:26 In article <af16ef18-8ac1-4c53-a81b-ac0d42ccbff5(a)i28g2000yqa.googlegroups.com>, cjcountess <cjcountess(a)yahoo.com> wrote: > 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... > > > > �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), > > 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. That would, at most, only makes it untrue in non-conventional mathematics, , but has no effect on its truth in conventional mathematics. Where the current proof of FLT remains in full force. > > Comment on the measurement of volume, will the volume of "a^3" + the > volume of "b^3" = the volume of "c^3"? Not if a, b and c are all to be integral. If you were ever able to produce integers a, b and c such that a^3 + b^3 = c^3, you would only then have a case, but you can't and you don't.
From: Tim Little on 15 Jul 2010 00:31 On 2010-07-14, cjcountess <cjcountess(a)yahoo.com> 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? Yes, they have. It doesn't. - Tim
From: bert on 15 Jul 2010 03:35
On 15 July, 00:22, cjcountess <cjcount...(a)yahoo.com> wrote: > 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"? A few points for you, which previous answerers may have, I think, not made completely clear: (1) Mathematicians separate "numbers" into several categories: integer, rational, irrational, algebraic, transcendental. You don't, and you should have. (2) Fermat's Last Theorem applies to INTEGERS (and by a simple mathematical argument, also to rationals). The numbers in your counter-example are algebraic, so it doesn't apply to FLT. (3) Just for your interest, the Pythagorean example with the diagonal of a square is one which has no solution in integers or rationals, only in algebraic numbers. This was proved by Euclid about 500 BC, and G.H. Hardy in his autobiographical "A Mathematician's Apology" quotes the proof as an example of what mathematics is about and how it's really done. (4) Similarly, what Andrew Wiles has proved is that while Fermat's equation has solutions in algebraic numbers (which you have also shown), it has none in integers or in rationals. -- |