marcus_b
Posts:
1,138
Registered:
2/5/06
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Re: JSH: End of an error
Posted:
Dec 8, 2006 3:18 PM
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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)
Why on earth are you factoring P(x) in this way? What
kind of functions are f(x) and g(x)? Are you assuming that if
x is an algebraic integer, then f(x) and g(x) are algebraic
integers?
>
> where f(0) = g(0) = 0, where the functions are otherwise to be
> determined.
>
OK, that's more specific. But it's not enough.
> 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 re-order in a special way on the left side and pick one way to
> multiply by 7 on the right:
>
> (49x^2 - 14x)5^2 + (7x-1)(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
Why do you need to "balance the right side" ???
> so let
>
> f(x) = 5a_1(x) + 5
>
> and
>
> 7g(x) =5a_2(x)
>
OK. So you haven't defined what kind of function
g(x) is. Now you introduce a_2(x), which is evidently
defined as
a_2(x) = 7 * g(x) / 5,
and since the nature of g(x) is not specified, we also
don't know anything about a_2(x) either, except that
a_2(0) = 0.
> which gives
>
> (49x^2 - 14x)5^2 + (7x-1)(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 - (7x-1)a + (49x^2 - 14x) = 0.
>
Oh, you're back to that. You are now suddenly thinking of the
polynomial on the left as if it were a quadratic in the number "5".
> The problem though is in getting to this point--using some VERY simple
> algebra--I multiplied g(x) times 7 to get 5a_2(x), so it makes sense
> that a_2(x) has 7 as a factor.
>
> 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.
>
No, that doesn't follow. This goes back to the equation
a_2(x) = 7 * g(x) / 5,
as I noted above. If g(x) were an algebraic integer, you
could say that a_2(x) has 7 as a factor. But what if
g(x) is simply an algebraic number which can be written
as a quotient
g(x) = h(x)/(7*r(x)),
where h(x) is an algebraic integer, and r(x) is a rational
integer? Why can't that happen? After all, you never
specified in what ring g(x) lives. You have not assumed
it is an algebraic integer. You have given no reason to
conclude that it does NOT have a "7" in its
denominator, or a non-unit factor of 7.
> But provably neither do in the ring of algebraic integers.
>
Correct. But you don't have a contradiction, because
you have not succeeded in showing that a_2(x) has 7
as a factor.
> 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)
>
But you have not shown that g(x) is an algebraic integer.
It may well be an algebraic *number* which has either 7 or
a factor of 7 in its denominator. If that is the case, then
the first "7" partially cancels out, and you cannot conclude
a_2(x) has 7 as a factor.
> if a_2(x) does NOT have 7 itself as a factor then g(x) must divide off
> some factors in common with 7.
>
I think what you mean to say here is, if a_2(x) does not have
7 itself as a factor, then g(x) can be written as
an algebraic integer divided by an ordinary integer, and the
denominator is divisible by 7. But since you never defined
what kind of thing g(x) is, you cannot rule this out. So you
have no contradiction.
> 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.
>
True enough.
> But now let
>
> f(x) = 2*g'(x) and g(x) = f'(x)/2
>
So, are you now implying that f(x) is divisible by 2
in some ring? Why should that be? As far as I can
tell, f(x) is just some algebraic number (although you
haven't even said ***that***. f(x) could be just some
function that takes values in the complex numbers.).
> 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,
> 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.
>
Same problem as before with g(x). You have not specified what
kind of number f(x) is. There is no reason to assume it is an
algebraic integer, for example.
> 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 non-zero integer x as noted above.
>
> So what is the resolution to this weirdness?
>
You have not yet defined any "weirdness", because you have not
said what kind of functions f(x) and g(x) are.
> 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.
>
Since you have never said that f(x) and g(x) are algebraic
integers, this is just irrelevant to your present discussion.
> 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.
>
You haven't done it "again", because (1) you have never given
a valid previous argument to this effect, and (2) your present
argument is not valid either.
> 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 journal--to keep using the wrong ideas, and teach them to
> others.
>
No - you have been consistently wrong on this and you
are continuing in the same vein. What you have done here
adds zero what you had before, and what you had before
was zero to start with.
> 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?
>
Try listening. Your mistakes have been carefully explained and you
are either incapable of understanding them, or refusing to believe
what you read. You have no proofs at all.
> 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.
>
Such melodrama. Stick to the math and keep away from all
the value judgements and the temper tantrums. They just make
you look stupider.
Marcus.
>
> James Harris
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