A simple proof of FLT
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From: DRMARJOHN on 7 Jul 2010 14:18 > > On Jun 5, 10:21 pm, DRMARJOHN <MJOHN...(a)AOL.COM> > > wrote: > > > > > This approach to structuring the problem of FLT > is > > different > > > than the traditional approach.... This is a > > counter-intuitive > > > approach, and thus, I would suggest, has been > > overlooked. > > > > Of course, you've read everything that's ever been > > written about FLT, > > so you know for a fact that in 350 years no one > ever > > thought of > > trying > > what you are suggesting. > > -- > > GM > > If what I am suggesting has been tried before the > following statement would have been made: > > Continuing A^n + B^n = 1. 7-6-10 > > The relationship of A^n to B^n and A to B can be > depicted as two curved surfaces that face each other. > On a graph, the pairs of A^n + B^n =1 are on the > x-axis, with A^n .5 to <1 placed on the axis just > above B^n .5 to < 0, also along the axis. I.e., two > series are placed one above the other on parallel > lines such that .6 + .4 = 1, .7 + .3 = 1, .8 + .2 = > 1, etc. > The y-axis plots the roots for, the bases of, A^n & > B^n . All A's are .793700... and above, but the B's > run from approaching 0 to approaching 1. I.e., all > A's for n=3, 4, 5... and all B's n=3, 4, 5,... . > Every pair of A^n + B^n has an irrational pair of A & > B that gives success. This contrasts to no pair of > rational A & B gives success. > This graph can be considered as containing two types > of curves. There is a curve for all A and B when n=3, > a curve for all A & B when n=4, when n=5, etc.. > There is the vertical curve for .5 for n= 3, 4. > 5...:for .501 and .499 for n = 3, 4, 5... , for .502 > and .498 for n = 3, 4, 5..., for n = 3,4,5... when A > &B are .503, .504, .505 to .999..., .497, .496, > .495, to approaching 0, and all intermediary points. > Now, when the A's & B's are rational, none of these > curves give success. > This description depends on previous statements and > d illustrations, and further defining statements and > illustrations will be forthcoming. > Martin E. Johnson CURVED SURFACE 7-7-10 The example of the numbers that contribute to the two curved surfaces can only be given piecemeal. When A^n=.501,n from 3to9,and B^n=.499, 501 and .499 are matching pairs, i.e., .501+.499=1.00. ---------A------------------ B n=3,-.793753435...------.793647609... 4,----.840938456...------.840854367... 5,----.870585382...------.870515738... 6,----.890928412...------.890869019... 7,----.905749539...------.905697784... 8,----.917026966...------.916981116... 9,----.925895285...------.925854135... at n=3 for .5, A&B= .793700526... at n=9 fro .5, A&B= .925874712...
From: DRMARJOHN on 7 Jul 2010 14:43 > haven't you fallen in to the same game, as before, > wherein this argument ought to work for n=2? > > meaning that it is in no wise a necessary argument, > if even sufficient, because n=2 necessarily hath > rational solutions. > > or some thing! > > Have you answered my question: what do you say to students who ask 'why does the equation for solving the pythagorian equation (at n=2) not apply to n=3?' MJ
From: Arturo Magidin on 7 Jul 2010 19:05 On Jul 7, 5:43 pm, DRMARJOHN <MJOHN...(a)AOL.COM> wrote: > > haven't you fallen in to the same game, as before, > > wherein this argument ought to work for n=2? > > > meaning that it is in no wise a necessary argument, > > if even sufficient, because n=2 necessarily hath > > rational solutions. > > > or some thing! > > Have you answered my question: what do you say to students who ask 'why does the equation for solving the pythagorian equation (at n=2) not apply to n=3?' "What makes you think that an argument that uses the fact that n is 2 throughout in such essential ways would apply to other values of n?" What do you say to students who ask 'why do the psychological theories for humans not apply to termites?" -- Arturo Magidin
From: DRMARJOHN on 8 Jul 2010 05:50 > haven't you fallen in to the same game, as before, > wherein this argument ought to work for n=2? > > meaning that it is in no wise a necessary argument, > if even sufficient, because n=2 necessarily hath > rational solutions. > > or some thing! > > > Now, when the A's & B's are rational, none of these > curves give success. > On 6-14 I responded to your question. Perhaps you are asking me to say why the solutions for the Pythagorean triangle do not apply to FLT. I will say more. First what I said on 6-14: ' "The rules of the Pythagorean triangle establish the x^2+y^2=z^2 that have a solution..All others are non-solutions. Fermat's proof for the 4th exponent of FLT demonstrates that that all solutions for the Pythagorean triangle are non-solutions for exponents above 2. Fermat uses x^2+y^2=z^2, the area of the triangle u=1/2 xy: X^2+Y^2=Z^2, U^2=1/2XY, Z<z, then X(1)^2+Y(1)^2=Z(1)^2, U(1)^2=1/2X(1)Y(1), Z(1)<Z; ETC. These are the steps of the descent which lead to the contradictory situation "you get to 1 and the progression can go no further"(K. Devlin p270-272 Mathematics, the new Golden Age, 1988.) Relevance to how I construct the problem: The area of a Pythagorean triangle is 1/2xy. Two such triangles joined form a rectangle with area 1/2xy+1/2xy=xy, which simplifies to .5+.5=1. Is there a root of .5 that is rational? This is the same question I begin with in A^n+B^n=1, i.e., .5+.5=1. Fermat's proof that x^4+y^4=z^4 has no solutions is based on is based on his demonstration that for a Pythagorean triangle, the area of the triangle can never be an integer, i.e., that 1/2xy can never be a square with integers when x & y are Pythagorean sides.' NOW to use some examples such as 3,4,5 (.6^2 +.8^2=1^2): 36^(1/2)^(1/2)=.36^(1/4) 36^(1/4) x .36^(1/4)=.36^2=.6 36^(1/4) x .6 = .36^(1/4) x .6 36^(1/4) x [.36^(1/4) x .6]= .36^(1/2) x .6 So the expansion of [.36^(1/4)]^4 gives B^2=.6, B^3= B x.6, B^4= .36 the base B is an irrational number Similarly for A=.64^(1/4): A^2=.8, A^3= A x.8, A^4= .64 the base A is an irrational number The same for 5/13 and 12/13. Fermat proved all successes of the exponent 2 are non-successes for exp.4. I would speculate that Fermat squared the square of the Pythagorean sides to get the fourth power he used in his proof, and that he used fractions in his own explorations. The ancients, I have read, used fractions. Number theory also simplifies overwhelming information. 7-8-10 Martin Johnson
From: spudnik on 8 Jul 2010 14:37
I agree, your question is ill-posedness, exemplfied -- and I thank you, for all of we, de newbies! also, the trigona to which you refer as "pythagorean," for any n, are not pythagorean, but they could be called, fermatian, because they are all associated with the "fermat curves," as for n=2, the "right" trigona are inscribed in a semicircle (of course, there are no right or left trigona .-) there is a little book that carries all of that out, with laborious calculation of side-lengths for the fermatian trigona, but it is very sure that the author makes no claim as to Fermat's "last" theorem (et sequentia .-) --my broker says to call your broker about cap&trade, and I'll tell you what happens. http://wlym.com |