Rupert
Posts:
3,810
Registered:
12/6/04


Re: JSH: End of an error
Posted:
Dec 8, 2006 10:44 PM


Rupert wrote: > jstevh@msn.com wrote: > > The widespread usage of flawed ideas in number theory has put me in the > > position of not just proving the errors and getting publication of > > proof in a peer reviewed math journal, but necesitated the finding of a > > second proof as well due to extraordinary resistance to such a shocking > > result. > > > > I have finally found the long sought after second proof. > > > > Start with > > > > P(x) = 175x^2  15x + 2 > > > > and do your analysis on the factorization > > > > P(x) = (f(x) + 2)*(g(x) + 1) > > > > where f(0) = g(0) = 0, where the functions are otherwise to be > > determined. > > > > So I have > > > > 175x^2  15x + 2 = (f(x) + 2)*(g(x) + 1) > > > > and next I do some simple algebra starting with multiplying both sides > > by 7: > > > > 7*(175x^2  15x + 2) = 7*(f(x) + 2)*(g(x) + 1) > > > > now I reorder in a special way on the left side and pick one way to > > multiply by 7 on the right: > > > > (49x^2  14x)5^2 + (7x1)(7)(5) + 7^2 = (f(x) + 2)*(7g(x) + 7) > > > > That work is done so that you have 5 and 7 on the left like a > > polynomial with variables, except of course 5 and 7 are not variable, > > but the operations are valid to this point. > > > > But now I need to balance the right side so let > > > > f(x) = 5a_1(x) + 5 > > > > and > > > > 7g(x) =5a_2(x) > > > > which gives > > > > (49x^2  14x)5^2 + (7x1)(7)(5) + 49 = (5a_1(x) + 7)*(5a_2(x) + 7) > > > > allowing me to find a solution of the a's where they are roots of > > > > a^2  (7x1)a + (49x^2  14x) = 0. > > > > The problem though is in getting to this pointusing some VERY simple > > algebraI multiplied g(x) times 7 to get 5a_2(x), so it makes sense > > that a_2(x) has 7 as a factor. > > > > Only if you have some reason to think that g(x) is an algebraic > integer. You have no reason to think that whatsoever. > > > BUT if you plug in some numbers like let x=1, you have > > > > a^2  6a + 35 = 0 > > > > which solves as > > > > a = (6 +/ sqrt(104))/2 > > > > and the result is that one of the roots has 7 as a factor. > > > > But provably neither do in the ring of algebraic integers. > > > > In arguing against this result posters have claimed that neither root > > has 7 itself as a factor, but instead each root shares factors in > > common with 7, but from the start with > > > > 7g(x) =5a_2(x) > > > > if a_2(x) does NOT have 7 itself as a factor then g(x) must divide off > > some factors in common with 7. > > > > But going back to > > > > 175x^2  15x + 2 = (f(x) + 2)*(g(x) + 1) > > > > note that 175x^2  15x + 2 is always even when x is an integer. > > > > But now let > > > > f(x) = 2*g'(x) and g(x) = f'(x)/2 > > > > and making the substitutions and simplifying gives > > > > 175x^2  15x + 2 = (g'(x) + 1)*(f'(x) + 2) > > > > and the same approach as before gives the same solution for the a's, > > No, it does not. > > We have > > f(x)=5a_1(x)+5 > g(x)=5a_2(x)/7 > > where the a's are roots of > > a^2(7x1)a+(49x^214x)=0. > > Now, if we let g'(x)=f(x)/2 and f'(x)=2g(x), we have > > f'(x)=2(5a_2(x)/7)=5b_1(x)+5 where b_1(x)=2(a_2(x)/7)1 > g'(x)=(5a_1(x)+5)/2=5b_2(x)/7 where b_2(x)=7(a_1(x)+1)/2. > > We have a_1(x)+a_2(x)=7x1, a_1(x)a_2(x)=49x^214x. > > By working out b_1(x)+b_2(x) and b_1(x)b_2(x), we can get an equation > of which the b's will be roots. It will not be the same equation. >
My apologies, this is not the way to get the minimum polynomial of the b's. The two b's will have different minimum polynomials.
b_1(x) will be a root of
((7/2)(b+1))^2(7x1)((7/2)(b+1))+(49x^214x)=0
whereas
b_2(x) will be a root of
((2/7)b1)^2(7x1)((2/7)b1)+(49x^214x)=0.
As I say, I assume this is what you are trying to do, namely you are letting g'(x)=f(x)/2 and f'(x)=2g(x), and then you are defining b's so that
f'(x)=5b_1(x)+5 7g'(x)=5b_2(x).
If that is what you are doing, then the b's will not be roots of the same equation as the a's.
If you are doing something else, you'd better explain what. > > where now > > > > 7g'(x) = 7*f(x)/2 = 5a_2(x) > > > > and f(x) would have to divide out factors from 7, so there would be a > > contradiction. > > > > Therefore the claim that g(x) can simply divide off factors from 7, so > > that 7*g(x) does not have 7 as a factor is refuted. > > > > Note this claim is an objection raised to protect the current role of > > the ring of algebraic integers as provably 7*g(x) does NOT have 7 as a > > factor in that ring, with nonzero integer x as noted above. > > > > So what is the resolution to this weirdness? > > > > Well, consider evens and notice that 2 is not a factor of 6 with only > > evens because 3 is not even. > > > > So it's not very complicated since the problem with the ring of > > algebraic integers is about exclusion. > > > > Some numbers are excluded because they are not root of monic > > polynomials with integer coefficients, so you can get weird stuff, and > > an analogy is with evens where odds are excluded, so 2 is not a factor > > of 6 with only evens. > > > > Proving the odd behavior of the ring of algebraic integer is easy. I > > just did it yet again above. > > > > Explaining it simply is easy as well, as I just did, yet again with the > > 2 and 6 analogy. > > > > But there are a lot of people around the world who learned mathematical > > ideas that have been shown to be flawed and so far that group has > > resisted publication, bizarre retraction of these ideas, and the death > > of the math journalto keep using the wrong ideas, and teach them to > > others. > > > > While posters have routinely lied about the details of the argument and > > objected for years in a steady attempt to deny what is mathematically > > correct. > > > > And meanwhile undergraduates have been taught mathematical ideas proven > > wrong years ago. > > > > I do not know what else to do here. If math people will not follow > > mathematical proof, and even second proof, then what can anyone do? > > > > If the will for error is too strong for rational argument to work, then > > it seems there are few options in this situation except to grieve over > > the fall of humanity and the failure of civilization in this area. > > > > > > James Harris

