A simple proof of FLT
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From: Obispo de Tolosa on 5 Jun 2010 14:53 And Chip's response was a total non sequitur. You can't just revert to alternative number systems to evade tough issues in the natural one.
From: Jesse F. Hughes on 5 Jun 2010 22:30 Obispo de Tolosa <MathMan345(a)hotmail.com> writes: > And Chip's response was a total non sequitur. You can't just revert > to alternative number systems to evade tough issues in the natural > one. Chip's answer was to the point. I quote it here: ,---- | On May 17, 1:00 pm, Craig Feinstein <cafei...(a)msn.com> wrote: | | > I would like to see a simple proof of FLT. Is there any | > theoretical reason why a simple proof of FLT cannot exist? | | Hi, Craig: | | While no one can rule out the possibility of a "simple" | proof of FLT, it is known that FLT is false in some | number systems that are quite similar to the natural | numbers. So the feeling is that any proof will have a | certain amount of subtlety. | | P. Ribenboim's book, Fermat's last theorem for amateurs, | may interest you. | | regards, chip `---- There is no "reverting" to alternative number systems to evade tough questions in his response. -- Jesse F. Hughes "We need to counter the shockwave of the evildoer by having individual rate cuts accelerated and by thinking about tax rebates." -- George W. Bush, Oct. 4, 2001
From: spudnik on 7 Jun 2010 17:57 the poor man is insisting upon the base of ten, and that is guaranteed to be a barren approach, by all of the work of Fermatttt (Fermat's Little Theorem e.g.; of course, the theorem applies in any base, but ... y'know?). also, the simple idea of assuming that a^n etc. are rational, thence seeking a contradiction, is very well-illustrated in the literature; however, taht is in English and other languages. > Please post your proof, at last! thusNso: maybe AP has a good thought, that speed-trap radars have no need of using such a doppler shift, assuming that it does exist, because mere timing of the radar's return, over some part of a second, is adequate to do the math. and, one always hears of "dppler radar" by weather satellites, whether or not that *inerferometry* is actully required, and the radar-ranging can stil be interpreted in terms of doppler shifts (in the colors of the Weather Channel graphics e.g.). now, why AP does not "beleive" in doppler shifts, apart from the belief in the so-called Hubble *interpretation* of the prevailing redshift-woith-distance effect, seen in the starfield ... you'd have to read his ****, more carefully; and that, ladies & germs, is scatology. --Stop BP's and Waxman's capNtrade arbitrage rip-off! http://wlym.com
From: DRMARJOHN on 6 Jul 2010 11:05 > 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 illustrations, and further defining statements and illustrations will be forthcoming. Martin E. Johnson
From: spudnik on 6 Jul 2010 18:08
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. --BP's Next (or Last) Bailout of Wall St. and the City (of London, the gated community & financial district), Cap and Trade (circa '91, Waxman's Acid Rain bill) --http://wlym.com |