James Harris
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 Posted: Sat Nov 08, 2003 11:26 pm Post subject: Radical math ideas, revisited |
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My first effort at posting a rather basic proof with an astounding
result lead more to replies from confused people than anything else
from what I>ve gathered, so here>s another post that gives more
background information.
The conlusion I have, which is a factorization, is key in showing a
problem with something mathematicians call the ring of algebraic
integers. They define those to be roots of monic polynomials with
integer coefficients, and have had that definition for over a hundred
years.
I just found a neat way to show that the definition leaves out numbers
that *should* be included for completeness.
It>s more than an esoteric issue though as it can be shown that
because certain numbers are left out you can appear to prove two
*different* and contradictory things.
What I>ve done is find a way to factor a polynomial in such a way that
a constant factor, like 49, can be shown to factor out in a certain
way based on a ring property.
That>s a big deal, and it>s how I manage to separate one factorization
out of an infinity of others so that I have the result
(5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) =
300125 x^3 - 18375 x^2 - 360 x + 22
where the a>s are roots of
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).
That factorization stands out from infinity because in rings like the
ring of algebraic integers 7 is not a factor of 22, so I>ve found a
factorization based on a property of only certain rings.
So I use one polynomial to analyze the roots of another polynomial,
and the argument is straightforward,.
And here it is.
1. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, where x is
in the ring of algebraic integers, notice that P(x) has a constant
term that is 1078.
2. It can be shown that
P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3
where you should note that using v = -1 + 49x, gives
P(x) = (v^3+1)(5^3) - 3v(5)(7^2) + 7^3
where the *same* polynomial has been put in a form which allows a
factorization into non-polynomial factors so that I have
P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)
where the a>s are roots of
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).
3. Now let x=0, so
P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)
as the cubic defining the a>s at x=0 is
a^3 - 3a^2, which has roots, 0, 0 and 3, and I>ve picked a_1(0) and
a_2(0) to equal 0, which leaves a_3(0) with a value of 3.
4. Further let a_3(x) = b_3(x) + 3, to keep indices matched. Then I
have
P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7)
P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22).
5. Now P(x) has a factor of 49 as
P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22
which means that
(5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22)
has a factor of 49.
6. However, the constant term of P(x)/49 is 22, which is verified by
again setting x=0, which gives P(0)/49 = 22.
But for two of the factors of P(x), the constant terms is 7, which is
NOT a factor of 22. Therefore, *none* of the constant terms of
P(x)/49 as they multiply to give 22 can have 7 as a factor.
(By saying that 7 is NOT a factor of 22, I>m making a choice as to
where the proof is going. Since I>ve been talking about algebraic
integers, where 7 is NOT a factor of 22, it>s natural to go with a
choice where 7 is NOT a factor of 22.)
Given that the constant terms are independent of x>s value, it must be
the case that dividing P(x) by 49 divides the two constant terms equal
to 7, by 7.
7. But to divide 7 from those constant terms requires dividing
through two of the factors, so
(5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) =
300125 x^3 - 18375 x^2 - 360 x + 22
from reverse use of the distributive property, which gives constant
terms that don>t have 7 as a factor, as required.
Notice that it>s a rather short and direct argument, where if you
accept that 22 does not have 7 as a factor, it>s obvious enough what
the constant terms of the factors must be as you go from 7, 7 and 22,
necessarily to 1, 1, and 22, when you divide P(x) by 49.
James Harris
http://mathforprofit.blogspot.com/ |
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