On Sunday, December 30, 2012 9:51:08 PM UTC+2, quasi wrote: > John Jens wrote: > > > > >I pick a < p ,proved that do not exist a , b , c > > >rational numbers with 0<a=<b<c and a<p to satisfy > > >a^p+b^p=c^p > > > > No, you never proved the above claim. > > > > You only thought you did. > > > > First you tried to show that the equation > > > > a^p + b^p = c^p > > > > has no solutions in integers a,b,c,p subject to the > > conditions 0 < a <= b < c, p > 2, a < p. > > > > For the sake of argument, let's allow that claim. > > > > Then you attempted to extend to positive rationals a,b,c. To do > > that, you scale a,b,c down, dividing each by a positive integer > > large enough so the new value of a is less than p. Then you > > claim a contradiction since you already showed that a < p is > > impossible. But you showed that for positive integer values of > > a, not for positive rational values of a, so (barring circular > > reasoning) you don't have your claimed contradiction. > > > > quasi
I intentionally choose a < p ,there's no crime to choose an a natural, a < p. If for a < p ,a,b,c, naturals , we can multiply the inequality a^p + b^p != c^p with any rational number, it will remain a inequality.