> 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.