On Jan 7, 8:18 am, "M_Klemm" <m_f_kl...@t-online.de> wrote: > "John Jens" wrote > > > The reason is to prove FLT . > > Let's split in three steps : > > 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 3--> pick A >= p, assume A^p + b^p = c^p and scaling down to A/k < p > > ,k rational -->contradiction to step 2 > > If a + b ? c>0 because 0<a?b<c implies b ? c < 0 , > > 0 ? a + b ? c < a < p > > then a + b ? c ? 1 and because a + b ? c < a implies a ? 2 and because a < > > p implies p > 2 ... > > .... and using binomial theorem > > Is this intended to be a proof of step 1? > If yes, it is essentially correct, because a^p + b^p = c^p together with a < > p implies > a^p + b^p <= a + b^p < p + b^p < (b+1)^p <= c^p, a contradiction. > > The inequality (b+1)^p <= c^p is however not necessaryly true for rational b > and c with b < c. > > Regards > Michael
I have already told him that regardless of the details of what he is doing that his proof CAN NOT work. I gave the reasons why.
But of course, like all cranks, he just ignores the reviews from experts and just prattles on.