```Date: Jan 4, 2013 2:04 AM
Author: John Jens
Subject: Re: From Fermat little theorem to Fermat Last Theorem

On Monday, December 31, 2012 1:10:01 PM UTC+2, quasi wrote:> John Jens wrote> > > > >Step 1--> prove a^p + b^p != c^p with a < p ,a,b,c,    naturals> > >Step 2--> extend to rationals , still a < p> > > > Step 2 fails.> > > > You can scale an integer non-solution down to get a rational> > non-solution, but that doesn't prove that there are no > > rational solutions.> > > > To prove that there are no rational solutions, it's not> > acceptable logic to start with an assumed integer solution> > and scale down to a rational one. Rather, you must start by > > assuming a rational solution and try for a contradiction.  > > Scaling up fails since when scaling rational a with a < p> > up to integer A, there is no guarantee that A < p, hence> > no contradiction.> > > > But this has already been explained to you.> > > > Bottom line -- your proof is hopelessly flawed.> > > > Moreover, your logical skills are so weak that there's> > no possibility that you can prove _anything_ non-tivial> > relating to _any_ math problem. > > > > Stop wasting your time with mathematical proofs -- your brain > > isn't wired for that.> > > > quasiIf 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 andassume 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 isfalse of course because d/k < p
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