Thursday, October 30, 2008
How could they? How DO they. As we speak those students are still
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.
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.
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.
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.
# posted by James Harris @ 14:53 
