On Saturday, December 29, 2012 1:00:44 PM UTC+2, John Jens wrote: > If a > p > > ... > > So we can?t find naturals 0<a?b<c with a<p with p odd > > prime to satisfy a^p+b^p=c^p. > > > > We can extend this to a , b , c rational numbers , > > 0<a?b<c and a<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 ,which implies that > > > > we can?t find positive integers to satisfy > > a^p+b^p=c^p with p > 2.
I won't go back to warn you about possible mistakes in your "proof", rather I will only tell you that if you have any hope to be taken seriously by anyone seriously related to mathematics you better type down your stuff with LaTeX, either in your blog or in a PDF file, otherwise many mathematicians, like me say, won't even take the time to try to read all that in ASCII.