Friday, October 20, 2006
JSH: Direct demonstration of contradiction
Start with
(5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - sqrt(13))(5 + sqrt(37))/2
where this time I won't go into the derivation but it's found using the same tool that I've posted about before.
Now it is easy to verify that 5 + (sqrt(37) - sqrt(13))/2 is a factor of 2*3*31 and is coprime to (sqrt(37) - sqrt(13))/2, which is a factor of 64*9, with respect to 3 by subtracting one from the other.
Therefore, 5 + (sqrt(37) - sqrt(13))/2 must share the same factors in common with 3 as sqrt(37) + sqrt(13) as that is coprime to sqrt(37) - sqrt(13).
z = 2*31*(sqrt(37) + sqrt(13))/(10 + sqrt(37) - sqrt(13))
should balance those out, and now I find that z is a root of
z^4 + 775z^3 + 1147z^2 - 48050z + 178746 = 0
which is a monic polynomial with integer coefficients showing that z is an algebraic integer.
But now note that
5 + (sqrt(37) - sqrt(13))/2 = 5 - sqrt(13) + (sqrt(37) + sqrt(13))/2
proving that 5 - sqrt(13) and (sqrt(37) + sqrt(13))/2 must have the same factors from 3.
And now letting z = (sqrt(37) - sqrt(13))/(5 - sqrt(13)) I find z this time is a root of
3z^4 + 13z^3 - 51z^2 + 26z + 12 = 0
which is a non-monic polynomial irreducible over rationals—that is, z does not have a rational solution.
So there is an apparent contradiction with the earlier conclusion!!!
And now you can see directly how the coverage problem with the ring of algebraic integers allows the appearance of a direct contradiction.
It is one of the most far reaching yet subtle problems in the history of mathematics and it has taken me YEARS to find an alternate proof that took away functions and variables which were too useful to people disagreeing with me in the past.
So now there are just the numbers and I've said that before but hey, this is hard!!!
I've just figured out yet another way to overturn over a hundred years of some of the most abstruse and complex mathematical ideas—flawed as they are—ever developed.
And I had to do it because you people refused to accept mathematical proof, so I needed something you could SEE directly.
Whew. Feel kind of tired now. Years of effort, so much thought.
(5 + (sqrt(37) - sqrt(13))/2)((sqrt(37) - sqrt(13))/2) = (5 - sqrt(13))(5 + sqrt(37))/2
where this time I won't go into the derivation but it's found using the same tool that I've posted about before.
Now it is easy to verify that 5 + (sqrt(37) - sqrt(13))/2 is a factor of 2*3*31 and is coprime to (sqrt(37) - sqrt(13))/2, which is a factor of 64*9, with respect to 3 by subtracting one from the other.
Therefore, 5 + (sqrt(37) - sqrt(13))/2 must share the same factors in common with 3 as sqrt(37) + sqrt(13) as that is coprime to sqrt(37) - sqrt(13).
z = 2*31*(sqrt(37) + sqrt(13))/(10 + sqrt(37) - sqrt(13))
should balance those out, and now I find that z is a root of
z^4 + 775z^3 + 1147z^2 - 48050z + 178746 = 0
which is a monic polynomial with integer coefficients showing that z is an algebraic integer.
But now note that
5 + (sqrt(37) - sqrt(13))/2 = 5 - sqrt(13) + (sqrt(37) + sqrt(13))/2
proving that 5 - sqrt(13) and (sqrt(37) + sqrt(13))/2 must have the same factors from 3.
And now letting z = (sqrt(37) - sqrt(13))/(5 - sqrt(13)) I find z this time is a root of
3z^4 + 13z^3 - 51z^2 + 26z + 12 = 0
which is a non-monic polynomial irreducible over rationals—that is, z does not have a rational solution.
So there is an apparent contradiction with the earlier conclusion!!!
And now you can see directly how the coverage problem with the ring of algebraic integers allows the appearance of a direct contradiction.
It is one of the most far reaching yet subtle problems in the history of mathematics and it has taken me YEARS to find an alternate proof that took away functions and variables which were too useful to people disagreeing with me in the past.
So now there are just the numbers and I've said that before but hey, this is hard!!!
I've just figured out yet another way to overturn over a hundred years of some of the most abstruse and complex mathematical ideas—flawed as they are—ever developed.
And I had to do it because you people refused to accept mathematical proof, so I needed something you could SEE directly.
Whew. Feel kind of tired now. Years of effort, so much thought.
# posted by James Harris @ 20:45 
