
Re: From Fermat little theorem to Fermat Last Theorem
Posted:
Jan 4, 2013 2:04 AM


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 nonsolution down to get a rational > > nonsolution, 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_ nontivial > > relating to _any_ math problem. > > > > Stop wasting your time with mathematical proofs  your brain > > isn't wired for that. > > > > 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

