September 13th, 2015, 05:44 PM  #21  
Senior Member Joined: Sep 2010 Posts: 221 Thanks: 20  Quote:
 
September 13th, 2015, 07:25 PM  #22  
Senior Member Joined: Dec 2007 Posts: 687 Thanks: 47  Quote:
These sort of elementary logical flaws mixing necessary and sufficient conditions, and confusion between equality and equivalence, thrive in all socalled FLT Proofs I see in this forum, I wonder whether they come from the same source...  
September 13th, 2015, 10:03 PM  #23 
Senior Member Joined: Apr 2015 From: Barto PA Posts: 170 Thanks: 18 
Well, that's one actual critique so far. Here's another: "Fermat's Last Theorem (main case: n is prime): For integers A, B, C and prime n>2 the equal $\displaystyle A^n+B^n=C^n$ does not exist." That's fine except n = 2 is also prime and I can not see where the argument accounts for this case. [See my next note, below.] "The essence of the contradiction: If Fermat’s equality $\displaystyle A^n+B^n=C^n$ exists.." I interpret this to mean that if there are any integer solutions to A^n + B^n = C^n for n > 2 then... "... then A+BC=0 and $\displaystyle A^n+B^n<C^n$." A^n + B ^n < C^n is true for any n > 1 if A + B = C; so how exactly does the argument not contradict the case n = 2? Further in the paper: "So, let us assume that for a prime number n>2, relatively prime A, B, C, and A'[or B']≠0 1°) $\displaystyle A^n=(CB)P$,..." This seems to assume the number A is always the product of two different factors. If that is the assumption then no  there is no reason why A can not be prime. That's the best I can do because it's the most I can almost (!) understand. 
September 13th, 2015, 11:23 PM  #24 
Senior Member Joined: Dec 2012 Posts: 1,010 Thanks: 24 
I felt many time in similar errors. The most difficoult problem in FLT is to arrive at a conclusion that cannot live space to round cicles...(or classic stupid errors). I was finally arrived ad the concernig that the only way is to use a more simple modular algebra (more than the one udes by Wiles) and prove the point that is the non vanishing mixed product from n>2 that cause the proof of flt. I invent the complicatemodulus algebra, but I turn around for several years since I'm not able to clearly prove what I've in mind and what seems to me "logic". Logic is not a proof... a reduction as absurdum is a proof, or a counterexample. ...And is what I hope I found this morning starting from: $\displaystyle (C^n/B^n  A^n/B^n) = 1$ using my step sum in the proper manner Going to the limit and proving that in the case n>2 the mixed product seems doesen't vanish living us another equation that contracdict the (1). Of course for n=2 the absence of the mixed product let the solution be possible. I work on and in the next week I let you know... Thanks Ciao Stefano 
September 14th, 2015, 12:58 AM  #25 
Senior Member Joined: Dec 2012 Posts: 1,010 Thanks: 24  Pls turn in: "the solution can be found JUST going to the limit, so variable in R (not in Q, not in N) when the mixed product terms will vanish..." n=2 works since there are just 2 terms so no "mixed" terms (equal to flat derivate). Thanks Ciao Stefano 
September 14th, 2015, 01:23 AM  #26  
Senior Member Joined: Dec 2007 Posts: 687 Thanks: 47  Quote:
You will easily see that when $P\Rightarrow Q$ is true, it also can be the case that $P$ is false, so many people commit fallacies such as: since the $implication$ is true, so is the antecedent $P$, what is false. Or: since the $implication$ is true, so is the antecedent $P$ and the consequent $Q$, what is also false. Or: since the $implication$ is true and the consequent $Q$ is also true, so is the antecedent $P$, FALSE. Now lets see a silly example: $P=$"somebody is alive" and $Q=$"somebody was a baby". Now, it is sufficient to say that somebody is alive to conclude that somebody was a baby in the past. It is necessary to have been a baby in order to be alive (we still don't engineer complete adults in genetic labs). Then $P\Rightarrow Q$ is a true implication. But suppose that you know someone deceased. $P\Rightarrow Q$ and $Q$ is also true of that person. It does NOT follow that the person is alive. Not, put $P=''A+BC=0''$ and $Q=''A^n+B^nC^n=0''$, then assuming that $A^n+B^nC^n=0$ is true, then $P\Rightarrow Q$ is emptly true (it may be the case that $A+BC\not=0$ but that wont matter). But if we say that $Q\Rightarrow P$, we are saying that $P=A+BC=0$ is a necessary condition for $Q=A^n+B^nC^n=0$ to be true, the same way that "somebody was a baby" is necessary to infer that "somebody is alive", what is plainly false. Last edited by almahed; September 14th, 2015 at 01:25 AM.  
September 15th, 2015, 04:06 AM  #27  
Senior Member Joined: Oct 2011 Posts: 133 Thanks: 1  magazines Quote:
= I sent the proof to Polish mathematical journal.  
September 15th, 2015, 04:19 AM  #28 
Senior Member Joined: Oct 2011 Posts: 133 Thanks: 1  contradiction You're right, but I'm also right: A + BC = 0. The contradiction in this is.

September 15th, 2015, 04:38 AM  #29  
Senior Member Joined: Oct 2011 Posts: 133 Thanks: 1  n=2 Quote:
2) Yes.  
September 15th, 2015, 06:35 AM  #30 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 933 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  

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