Re: Non-polynomial factorization, critical point
- From: amy666 <tommy1729@xxxxxxxxxxx>
- Date: Thu, 30 Oct 2008 19:26:27 EDT
jsh wrote:
Some easy algebra casts doubt on core algebraic
number theory but that
conclusion is so hard to accept despite the ease of
the mathematics.
Consider a simple polynomial P(x) = 175x^2 - 15x + 2.
Multiply it
times 7, to get
7*P(x) = 1225x^2 - 105x + 14. Cleverly re-group
terms:
1225x^2 - 105x + 14 = (49x^2 - 14x)5^2 + (7x-1)(7)(5)
+ 7^2
and now factor into non-polynomials:
7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)
where the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0.
But now go backwards and divide the 7 back off, or
try:
175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)
where now the b's are unknown functions. If I
multiply by 7 again,
can I do this?
7*(175x^2 - 15x + 2) = (sqrt(7)*5b_1(x) +
2*sqrt(7))(sqrt(7)*5b_2(x)+
sqrt(7))?
Yes. Of course I can multiply 7 through ANY way I
want, or I could
multiply through by 13 or some other number.
So how many ways can 7 divide off from
7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)
where the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0?
People arguing with me say, one way per x. So one
way for x=0, and
another way for x=1, and another way for x=2.
BUT if you just divide 7 off and get
175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)
how many ways can you multiply 7 BACK onto the
expression? An
infinity is the answer.
So guess what? The mathematics will not let you
start with
175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)
where the b's are any possible algebraic integer
function, multiply by
7 and get
7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)
where the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0.
You KNOW the math won't allow that because if you
could do that then
you could just divide the 7 off and get b_1(x) and
b_2(x), as
algebraic integer functions.
So the 7 cannot be divided off in general, so math
people claim you
have to figure it out at EACH x, so like with x=1,
you have
7*(175 - 15 + 2) = 7*162 = (5a_1(1) + 7)(5a_2(1)+ 7)
where the a's are roots of
a^2 - 6a + 35 = 0, so you have a = (6 +/-
sqrt(-104))/2,
so you'd look for factors of 7 to divide through that
in a different
way according to them than when x=2, but how does the
math know how to
choose?
After all, if ANY particular factorization is used, I
can just
multiply back through with another as
175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)
is bare. There is no way mathematically to force me,
a quirky human
being with free will, to choose to multiply that 7 in
a particular way
just to get
7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)
where the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0
so how does the math know? Mathies in arguing
against the obvious,
claim that the tail wags the dog in that the
FUNCTIONS tell the math
how the factors of 7 are distributed, but that is
circular as who
tells the functions?
Remember, with something like
175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)
you can multiply times 7 ANY WAY you like and
distribute its factors
in an INFINITY of ways, so how does the mathematics
choose a way?
How about with
x^2 + 3x + 2 = (x+2)(7x+7)?
But you say, no fair, you can SEE how a choice was
made out of
infinity. How does that help anyone with
7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)
where the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0?
How can the math pick and choose how 7 multiplies
through based on the
value of x, when
175x^2 - 15x + 2
is being multiplied BY 7?
How? How is that possible? It doesn't work that way
with x^2 + 3x +
2. If I multiply it times 7, and get
x^2 + 3x + 2 = (x+2)(7x+7)
that is done for all x, because x is being
multiplied. The tail does
not wag the dog.
There is no way for 7 to multiply times a
factorization in a way
determined by the very thing it is multiplied
against, so with
7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)
where the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0
there is NO WAY the value of x can control how 7
multiplies. It's not
possible. It's stupendously nonsensical to claim
that it does, but a
dead math journal SWJPAM is dead because some people
argued against
the possible, convincingly.
There is NO WAY that 7 on the outside is being told
which of an
infinity of ways to multiply times the factorization
175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)
by the value of x. That is impossible. There is no
mathematical way
that can happen.
But then you can just pick any value for x, like x=0,
and find that
with
7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)
where the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0
the 7 is a factor of (5a_1(x) + 7) or (5a_2(x)+ 7)
and know that is true for all x.
And then know that with x=1, only one of the roots of
a^2 - 6a + 35 = 0, so a = 3 +/- sqrt(-26)
is 7 a factor, and I remember when I was talking
about these kinds of
results years ago, posters would promptly ask, but
which one has 7 as
a factor? 3 + sqrt(-26) or 3 - sqrt(26)? And the
answer is, there is
no way to know.
let me translate that:
jsh makes a fuss about 7 having to be a factor of one of the factors.
heres a trivial parody example that explains it all.
7x = ( sqrt(49) sqrt(x) ) * ( sqrt(1/7) sqrt(x) )
so how do we know which one has 7 as a factor :
sqrt(49) * sqrt(x) or sqrt(1/7) sqrt(x)
we cant !!!
and mathematicians have been lying about that since pythagoras
and they pulled a math magazine
they did not understood the genius jsh
whine whine whine ...
The frustrating thing for me in being able to explain
so simply why
there must be this huge issue is that posters get
away with claiming
that given
175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)
there is a way to multiply by 7, on the outside, so
that it is
controlled by what is being multiplied on the inside,
which is
impossible. There is no way, with say 7*(x^2 + 3x +
2) = (7x + 14)(x
+1) for the thing being multiplied to inform me how
it is to be done,
so that occurs different ways with different values
of x.
So like, if x=2, then it's
7*(x^2 + 3x + 2) = (7x + 14)(x+1)
but if x=3, then it's 7*(x^2 + 3x + 2) = (x +
2)(7x+7)?
The tail does not wag the dog. The value of x cannot
force 7 to
multiply times
175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)
one way versus another.
Now math people are not stupid. It's not hard to see
what must be
true mathematically and accept that 7 divides off for
all x the same
way, so why did they instead kill an entire math
journal?
That dead math journal should tell you something.
How huge this
result is, and how powerful the resistance to it.
They killed a math journal that was a decade old.
Trashed it in
public. Berated the editors. Its hosting university
scrubbed all
mention of it from their site.
THEY KNOW I AM RIGHT AND STILL TELL YOU I AM WRONG.
What could be big enough for that kind of reaction?
What in the physics field? Maybe if someone
disproved quantum
mechanics. But hey, quantum mechanics works in the
real world, no one
CAN disprove it as it works. But in "pure math"
areas there is no
theory to test in the real world!
The mathematical community not only could kill a math
journal to
protect against knowledge of this result, wage a
smear campaign
against me, lie to the public repeatedly in replies
and in its
behavior in not accepting this result, but also keep
teaching the
flawed mathematical ideas to NEW STUDENTS.
Which forces a responsibility on others in the
academic field to do
the right thing as the experts within the field have
refused to do it
for years now and clearly have no intention of ever
doing the right
thing here.
Does the tail wag the dog? Can the thing being
multiplied dictate to
what is multiplying times it?
How could they destroy an entire math journal like
that? Lie like
that? Keep teaching their students wrong
information?
How could they? How DO they. As we speak those
students are still
being taught around the world.
James Harris
regards
tommy1729
.
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