Date: Jan 4, 2013 2:04 AM
Author: John Jens
Subject: Re: From Fermat little theorem to Fermat Last Theorem
On Monday, December 31, 2012 1:10:01 PM UTC+2, quasi wrote:

> John Jens wrote

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>

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> >Step 1--> prove a^p + b^p != c^p with a < p ,a,b,c, naturals

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> >Step 2--> extend to rationals , still a < p

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> Step 2 fails.

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> You can scale an integer non-solution down to get a rational

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> non-solution, but that doesn't prove that there are no

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> rational solutions.

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> To prove that there are no rational solutions, it's not

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> acceptable logic to start with an assumed integer solution

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> and scale down to a rational one. Rather, you must start by

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> assuming a rational solution and try for a contradiction.

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> Scaling up fails since when scaling rational a with a < p

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> up to integer A, there is no guarantee that A < p, hence

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> no contradiction.

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>

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> But this has already been explained to you.

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>

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> Bottom line -- your proof is hopelessly flawed.

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>

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> Moreover, your logical skills are so weak that there's

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> no possibility that you can prove _anything_ non-tivial

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> relating to _any_ math problem.

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>

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> Stop wasting your time with mathematical proofs -- your brain

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> isn't wired for that.

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>

>

> quasi

If a^p= c^p- b^p is true for a , b , c ,naturals a < p , is true for a rational , a < p and b , c naturals because c^p- b^p is natural.

We can divide a^p= c^p- b^p with k^p , k rational k > 1 and note (a/k) = q ,

q^p = (c/k)^p - (b/k)^p with q rational q < 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