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 June 15th, 2013, 01:45 PM #11 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Fermat's last theorem ( half A4 evidence ) Often when people who do know abstract mathematics look at a formal mathematics proof, it looks like gibberish. Unfortunately, I suspect that some of those people feel that if they write gibberish, they are writing a mathematical proof!
 June 16th, 2013, 03:29 AM #12 Newbie   Joined: Jun 2013 Posts: 11 Thanks: 0 Re: Fermat's last theorem ( half A4 evidence ) high school level picture below : ( If You will not understand we can call You "IDIOT" ) http://2.bp.blogspot.com/-1EUKTmDQULw/U ... MG2347.JPG
June 16th, 2013, 03:52 AM   #13
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Re: Fermat's last theorem ( half A4 evidence )

Quote:
 Originally Posted by tesla2 high school level picture below : ( If You will not understand we can call You "IDIOT" ) http://2.bp.blogspot.com/-1EUKTmDQULw/U ... MG2347.JPG
Before learning something of mathematics, it is necessary to learn much of good manners.

 June 16th, 2013, 05:15 AM #14 Math Team   Joined: Apr 2012 Posts: 1,579 Thanks: 22 Re: Fermat's last theorem ( half A4 evidence ) You seem to be suggesting that if A + B - C = 0, then B - C "cancels out" and therefore A = 0. But that is obviously false. If A + B - C = 0, then A + B = C. Indeed, that's what you correctly derived A + B - C = 0 from. So if you are going to "cancel out" anything, you "cancel out" the entire left hand side of the equation and get 0 = 0, which is trivial, but true, not false. Just consider ANY simple example. 5 + 7 = 12 is a true statement and therefore 5 + 7 -12 = 0 is a true statement and it does NOT entail that 5 = 0. So it looks like your attempt to prove FLT rests on some wildly faulty high school algebra. Your claim has nothing to do with the special conditions on the numbers that FLT considers, but a sweeping claim that ANY equation of the form A + B = C implies that A = 0, which is simply absurd. If you were right, then ALL basic addition would be impossible unless one of the numbers added is 0. So 0 + 5 = 5 would be okay according to your claim. But since 1?0, 1 + 5 = 6 would be false, but it's true/
 June 20th, 2013, 03:08 AM #15 Newbie   Joined: Jun 2013 Posts: 11 Thanks: 0 Re: Fermat's last theorem ( half A4 evidence ) below link = 1 000 000 USD prize similar equation and problem >>> http://www.businessinsider.com/beale-co ... _247967430 For small help I can offer 10 % of award details later . MAROSZ EVIDENCE BEAL PRIZE CONJECTURE http://4.bp.blogspot.com/-KvpwdW3l4Ss/U ... MG2362.JPG Line III we have two functions ( right ) and (left ) We can make two graphs and we can have three situation : I zero common points ==> not exist A that can solve A^x +B^x =C^z success !!! to get Beal Prize - parallel situation or (right ) graph is more wide 1- I showed two the same parallel graph situation ( d=e and t=n and g=n ) 2- (right equation - graph ) is more wide (d>e and t>g>n ) II only one common point success !!! to get Beal Prize I showed that we can have situation where exist only one solution ( one golden A ) that can solve equation A^x +B^x =C^z .Two graphs can cross only in one point III all points are common ==> we have theorem that fit to all A always exist "+1" ! we will never be able IDEAL cover two different graphs THE END OF EVIDENCE Fermat's last theorem problem http://1.bp.blogspot.com/-Iokdrpsk70M/U ... MG2360.jpg Post's Author : http://www.tesla4.blogspot.com/
June 20th, 2013, 03:31 AM   #16
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Re: Fermat's last theorem ( half A4 evidence )

Quote:
 Originally Posted by tesla2 below link = 1 000 000 USD prize similar equation and problem >>> http://www.businessinsider.com/beale-co ... _247967430 For small help I can offer 10 % of award details later . MAROSZ EVIDENCE BEAL PRIZE CONJECTURE http://4.bp.blogspot.com/-KvpwdW3l4Ss/U ... MG2362.JPG Line III we have two functions ( right ) and (left ) We can make two graphs and we can have three situation : I zero common points ==> not exist A that can solve A^x +B^x =C^z success !!! to get Beal Prize - parallel situation or (right ) graph is more wide 1- I showed two the same parallel graph situation ( d=e and t=n and g=n ) 2- (right equation - graph ) is more wide (d>e and t>g>n ) II only one common point success !!! to get Beal Prize I showed that we can have situation where exist only one solution ( one golden A ) that can solve equation A^x +B^x =C^z .Two graphs can cross only in one point III all points are common ==> we have theorem that fit to all A always exist "+1" ! we will never be able IDEAL cover two different graphs THE END OF EVIDENCE Fermat's last theorem problem http://1.bp.blogspot.com/-Iokdrpsk70M/U ... MG2360.jpg Post's Author : http://www.tesla4.blogspot.com/
Since you have discovered everything, I will leave my studies

June 20th, 2013, 04:59 PM   #17
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Re: Fermat's last theorem ( half A4 evidence )

Quote:
 Originally Posted by mente oscura Since you have discovered everything, I will leave my studies
:P

Tesla's offer of 10% to anyone who "helps" him win the Beal prize is some of the best unintentional humor I've seen in a while.

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