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On the Lighter Side
That CAN'T be true!
Evariste Galois and Fermat's Last Theorem
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Evariste Galois and Fermat's Last Theorem
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Chondrally
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Evariste Galois and Fermat's Last Theorem
«
on:
05/08/2018 10:56:46 »
Fermat's Last Theorem , Unlocking the secret of an ancient mathematical problem. by Amir D. Aczel
probably one of the most evil work's of non-Fiction ever, along with Imvitation to the Mathematics of Fermat-Wiles by Yves Hellegouarch.
The point is that for any n, xt^n+1^n=zt^n can be fit arbitrarily closely for any x=z, as t apporaches infinity
In fact it will fit arbitrarily closely long before t actually gets to infinity.
The other disproof is that Evariste Galais died so young, it could not be right under any circumstances , he did not win the love of a young lady and got no young lady pregnant, which proves the theory, and so Galois theory must be fundamentally false.
there are always complex and/or real solutions to any cubic polynomial or order n polynomial according to the fundamental theorem of algebra and they can be solved numerically in faster time than it takes to solve the cubic or quartic equations that Viete, Cardano, Bombelli and Tartaglia discovered.
There is a book called, the story of the Sqrt(-1)
e^(i*pi/2) = cos(pi/2)+i*sin(pi/2) = i
e^(-i*pi/2) = cos(pi/2) - i*sin(pi/2)=-i
e^(pi/2) = cos(i*pi/2)+i*sin(i*pi/2)=(e^(pi/2)+e^(-pi/2))/2+i*(e^(pi/2)-e^(-pi/2))/2/i = e^(pi/2)
e^(-pi/2) = cos(i*pi/2)-i*sin(i*pi/2)=(e^(pi/2)+e^(-pi/2))/2-i*(e^(pi/2)-e^(-pi/2))/2/i = e^(-pi/2)
e^(i*pi/2)*e^(i*pi/2)= i*i=-1=e(i*pi) consistent
e^(-i*pi/2)*e^(-i*pi/2))=(-i)*(-i)=-1=e^(-i*pi) consistent.
i*i=e^(i*pi/2)*e^(i*pi/2)=e^(i*pi)=-1
so i=-e^(i*pi/2)
i*i=e^(-i*pi/2)*e^(-i*pi/2)=e^(-i*pi)=-1
which disproves one of the Appendices in the story of the sqrt(-1).
«
Last Edit: 05/08/2018 11:37:50 by
Chondrally
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chris
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The Naked Scientist
Re: Evariste Galois and Fermat's Last Theorem
«
Reply #1 on:
05/08/2018 11:28:14 »
Great; but what's the question you are raising about this? Can you please edit the title accordingly, otherwise the moderators will remove the thread.
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