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Re: From Fermat little theorem to Fermat Last Theorem
Posted:
Jan 4, 2013 2:04 AM
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On Monday, December 31, 2012 1:10:01 PM UTC+2, quasi wrote:
> John Jens wrote
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> >Step 1--> prove a^p + b^p != c^p with a < p ,a,b,c, naturals
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> >Step 2--> extend to rationals , still a < p
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> Step 2 fails.
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> You can scale an integer non-solution down to get a rational
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> non-solution, but that doesn't prove that there are no
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> rational solutions.
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> To prove that there are no rational solutions, it's not
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> acceptable logic to start with an assumed integer solution
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> and scale down to a rational one. Rather, you must start by
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> assuming a rational solution and try for a contradiction.
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> Scaling up fails since when scaling rational a with a < p
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> up to integer A, there is no guarantee that A < p, hence
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> no contradiction.
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> But this has already been explained to you.
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> Bottom line -- your proof is hopelessly flawed.
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> Moreover, your logical skills are so weak that there's
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> no possibility that you can prove _anything_ non-tivial
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> relating to _any_ math problem.
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> Stop wasting your time with mathematical proofs -- your brain
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> isn't wired for that.
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> quasi
If a^p= c^p- b^p is true for a , b , c ,naturals a < p , is true for a rational , a < p and b , c naturals because c^p- b^p is natural.
We can divide a^p= c^p- b^p with k^p , k rational k > 1 and note (a/k) = q ,
q^p = (c/k)^p - (b/k)^p with q rational q < p.
Let?s pick d positive integer , p < d , d?b < c and
assume that d^p+b^p=c^p .
We can find k rational number such d/k < p and we have
(d/k)^p + (b/k)^p = (c/k)^p which is
false of course because d/k < p
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