( three mass , lever and velocity ) 30 of June 2013
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Fermat's last theorem
( three mass , lever and velocity ) 30 of June 2013
EN >
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Fermat's last theorem
This is a joke, right? Surely you know that the prize for a proof is no longer offered? "Fermat's last theorem" has already be proven.
And I find it difficult to believe that anyone thinks "Fermat's last theorem" has any connection to "cars" or "Archimedes' lever".
Has to be a joke!
JOKE ?
I already work for company ( industry ) we had problem
there was one electric car and this car must keep constant velocity ( real problem not joke ? )
during car trip carbon go to inside "crate trucks" ( carbon use gravitation forces )
We have a car and we have body that change mass ( Engineer must fight with real problem in real life )
it was my inspiration
Archimedes = Joke ? not three masses = joke ? not
in classical mechanic exist problem that can not be solve by nomberse
( beam statically determinable ) everyone who graduate technical universyty know this problem
if we have this problem we use the "drawing methode" we can not use nombers
I see many job for people who like mathematic - here is not a joke
problem is very hard Time is very important for mathematic right now You not understand this that time can be "joker card "
Fermat was Genius -- Relativity can destroy mathematic --- Fermat saw problem before many of us Joke Yes He made joke !!! but problem is
that simple three mass will be gate to many new secret )
MAROSZ ( today 16:00 polish local time ) ---- > TIME is VERY IMPORTANT for MATHEMATIC ! and all equations !
here I want to show Ytube
(other inventor from Poland - Mr Lucjan Łągiewka - city - Kowary - Poland "
he designed new susspension type for F1 bolides ( Mercedes Mclaren has got his bumper inside susspension )
One Idiot from Cambridge try copied his Idea but he was and is Idiot and he did't and don't know what realy did Mr Lagiewka
Mr Łagiewka ----> for me three mass = secret !!!
Marosz Joke ? Ytube about picture EN
Who is Marosz joke maker ?
I'm 32 years old Engineer and Inventor -----> my design vision city can fly and use wire --------> > SIMPLE PATENTS
new generation GPS system ( time space dimmensions ) are absolute ( without absolute time mathematic not exist ) --------> Montions of the Solar System
mathematica ------> Fermat's last theorem
Why I made this TXT ?
This what I wrote below can be naive but : I live in the ghetto (small Polish town 25% unemployment) and I see no future for my ART I hope that someone from Universities will recognize that I'm not IDIOT and my Ideas have power
- please sent me mail (polski.tesla2(...)gmail.com ) offer me help
I'm ready start build copy Mr Tesla ( Earth Quake tool ) I need place ( labolatory ) and people who have open mind ( mathemtic physics ) We can start very long trip in the Universe without fuel on board
we can easy stop " stop and wait " Earth and Sun will go .
I can learn You how to use Sun but ( kinetic energy not hot !!!) I already designed car that use AIR similar to this car:
but 50x cheaper and 4x better distance
My version of engine can propel big ship too ...
Look, truth be told I'm getting tired of dealing with you. This is a Help Forum. We help people with Math difficulties...We do not serve as a platform for getting your world saving ideas out into the public domain.
Knock it off or you'll be banned. Again.
-Dan
I'm probably going to regret this, having been sucked into some of Tesla's previous posts, and even though I know he doesn't listen to reason here goes:
Tesla2: in your scenario you have mass of car 3 = mass of car 1 + mass of car 2: m_3 = m_1+ m_2. So the two sides will balance if L_3 = L_1 = L_2. At least I think that's what you're getting at. Let's call the moment arm = L. So you have:
Torque of cars 1 and 2 = m_1L_1 + m_2L_2 = (m_1+m_2)L
Torque of car 3 = (m_3)L_3 = m_3L
Set these equal:
(m_1+m_2)L = (m_3)L, and since m_3 = (m_1 + m_2) you get:
(m_1+m_2)L = (m_1+m_2)L
This is self evident, so no major breakthrough there.
Now let's let L be a function of t, such as L(t) = At^n, where A is a constant and n = some integer. Balancing torques gives:
m_1 A t^n + m_2 A t^n = m3 A t^n
You seem to think that this has something to do with Ferma's Last Teoremt, but it doesn't. Fermat is of the form X^2 + Y^2 = Z^2 where X, Y and Z are integers, but here you have something that is of the form X t^n + Y t^n = Z t^n, where X, Y and Z are constants: X = (m_1)A, Y=(m_2)A, and Z = (m_3)A. Time t is whatever value you want (not necessarily an integer). What does this have to do with Fermat's Last Theorem? Answer: nothing.
I was looking at the first page of the OP's blog (cause I'm a glutton for punishment) and found this gem:
2^(4/3) = 16^(1/3) = 4^(1/2) = 2.
Which of course is wrong. But by relying on this he "shows" that since 2^3 + 2^3 = (2^4/3))^3, and since he thinks 2^(4/3) is an integer, he has a violation of Fermat's Last Theorem. But of course (a) 2^(4/3) is not equal to 2 (it's value is about 2.52), and (b) his "proof" would also say that 2^3 + 2^3 = 2^3, which even he should have noticed makes it clearly wrong.
Mr Tesla2 - you don't need to worry about how sad Andrew Wiles would be to receive your email.
Then later he claims a violation of the Beal conjcture, which states that A^x + B^y= C^z has no solutions for A,B,C,x,y,z positive integers and x,y,z > 2 unless A, B, and C have a common prime factor. He does this from the fact that 1^3 + 2^5 = 33 = (33^(1/3))^3 . Apparently Mr. Tesla believes that the cube root of 33 is an integer. Yeesh.
Sorry somethink is ok here ?
2^(4/3) = 2^[4*(1/3)] II Law of exponents ( below link book )
http://image.mathcaptain.com/cms/ima...-exponents.JPG
2^4*(1/3) = 16^(1/3 ) ( Law of exponents no II , m , n have got the same rights so we can exchange position )
please take calculator put 16 and (2x click ) ^ ,^ =2 ?
2^(4/3) = 16^(1/3) = 4^(1/2) = 2 ?
all is in book
Fermat was joke maker ! one post above
we see that people have huge problem with exponent ( even experts )
so many years we have problem ? ( at first 2^(4/3) look strange but it is 2 )
similar 11^(4/3) =11 .... 231^(4/3) =231 ..... 55^(4/3) =55
Hello Tesla. I see you have changed your ID - are you trying to disassociate yourself from your previous posts?
How do you get ?? This is just plain wrong. Note that 16 = 4^2, so you can do this: . This does not equal 2. You can easily check this: if 16^(1/3) = 2 then 2^3 = 16, but as I sure you know 2^3 = 8, not 16.
None of the experts here seem to have any problems with how to work with exponents. I strongly suggest that in the future if you think you have found a way to outsmart Fermat, or Beal, or Andrew Wyle, or Einstein, or Galileo (as in other threads) you CAREFULLY check your work before posting, because every time you have posted something in the past 9 months either here or on PhysicsHelpForum.com your posts have been riddled with schoolboy errors.
This is just so wrong. If , then try raising each side to the third power. On the left hand side you have : , and on the right hand side you have . So what you are trying to tell us is that . Divide both sided by and you get . So if you believe that , then you also believe that 11 = 1. The only time that n^(4/3) = n is if n=1, or n=0.