Date: Jan 7, 2013 10:37 AM
Author: John Jens
Subject: Re: From Fermat little theorem to Fermat Last Theorem
On Monday, January 7, 2013 3:18:29 PM UTC+2, M_Klemm 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

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> > 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 ,

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> > 0 ? a + b ? c < a < p

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> > 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

Yes it's only for step 1 when a,b,c naturals.