Sunday, January 11, 2009
JSH: Algebraic integer problem, another try at explaining
I find it hard to believe that even physicists would not get very interested in this result if they believed it true, even though it seems on the surface to only matter about some abstract number theory, so here's yet another try at explaining this esoteric math issue that takes out Galois Theory.
On the complex plane consider an unknown factorization:
175x^2 - 15x + 2 = (5c_1(x) + 1)(5c_2(x)+ 2)
Notice with just that information there are an infinity of functions that will work for the c's.
If you're wondering why THAT particular quadratic which I keep using over and over again, it's for historical reasons.
Now I also multiply by 7 for historical reasons but if it would help to multiply times some other number like say, 27, because it's not prime I can do that as well, and then the quadratic will change as well. But for now on with 7:
7*(175x^2 - 15x + 2) = 7*(5c_1(x) + 1)(5c_2(x)+ 2)
That is the OPPOSITE of what's normally done of course, as why multiply on an extraneous factor?
Answer is, because by doing this major "out of the box" step, I can reveal a massive error.
But next I have to make a choice!
7*(175x^2 - 15x + 2) = (5*7*c_1(x) + 7)(5c_2(x)+ 2)
and now I need to hide things a bit, which I can do with new functions:
b_1(x) = 7*c_1(x), and c_2(x) = b_2(x)
so
7*(175x^2 - 15x + 2) = (5b_1(x) + 7)(5b_2(x)+ 2)
with still as of yet undetermined functions, but I have a problem now with the 7 hidden, so I need a condition to start determining my functions:
Let b_1(0) = b_2(0) = 0.
But there are still an infinite number of possible functions at this point.
And now I need symmetry, as the issue is actually a symmetry problem, so I need one last set of new functions:
a_1(x) = b_1(x), and a_2(x) = b_2(x) - 1
and making those substitutions gives me:
7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)
The symmetry with regards to the 7's is required so that the a's can be roots of the same quadratic, which is NOT possible until that fix is made, so it's not an option here.
Now, of course, I have that ONE SOLUTION—as it is one solution out of infinity—for the a's is, as roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0.
and now you can solve for the a's using the quadratic formula:
a_1(x) = ((7x-1) +/- sqrt((7x-1)^2 - 4(49x^2 - 14x)))/2
a_2(x) = ((7x-1) -/+ sqrt((7x-1)^2 - 4(49x^2 - 14x)))/2
I need to emphasize at this point that the entire point of this exercise was to get to what I call a quadratic generator, which is just an expression that gives you quadratics, so for example, with x=1, you have
a^2 - 6a + 35 = 0
which is, of course, a quadratic. But also I needed a monic quadratic, which just means the leading coefficient is 1, which is satisfied, because then the a's will always be things called algebraic integers, when x is an integer.
But so what? Why bother?
Well, because mathematicians do not teach and prior to my research did not believe (I guess mostly they still don't) that given
a^2 - 6a + 35 = 0
only one of the roots is a product of 7, but my substitutions above mean that
a_1(x) = 7c_1(x) and a_2(x) = c_2(x) - 1
so you can SEE the 7 multiplying times the function c_1(x), where I remind that the start was:
175x^2 - 15x + 2 = (5c_1(x) + 1)(5c_2(x)+ 2)
Now the preferred ring for considering factors in such a situation for mathematicians is something called the ring of algebraic integers, and given all of the above you might suppose then that
a^2 - 6a + 35 = 0
has 7 as a factor for just one root, but instead you can mathematically prove the OPPOSITE result!!!
Well, you may wonder, maybe there is something weird about the algebraic operations I used to get to the a's, so I suggest that you go back through this post and consider each mathematical step carefully.
You will find that NONE of them are invalid on the complex plane, or apparently even on the ring of algebraic integers!!!
But that CANNOT be true, as if all the steps are valid on the ring of algebraic integers, then the result would hold in that ring, but we already have at x=1 that it DOES NOT HOLD IN THAT RING.
The mystery should resolve with a leap for some of you that of course the one thing that must be invalid on the ring of algebraic integers, as all the steps look like ok algebra is the start!!!
175x^2 - 15x + 2 = (5c_1(x) + 1)(5c_2(x)+ 2)
That is the only conclusion left for you. If that start is ok, then everything else follows on the ring of algebraic integers and the conclusion logically should hold, so it must not be possible for the c's to exist on the ring of algebraic integers for x=1, where you can get to the a's that are roots of a quadratic generator as given.
The more general conclusion though is that the c's cannot exist for ANY situation where you can get to a's that are roots of a quadratic generator!!!
So my using
a^2 - (7x-1)a + (49x^2 - 14x) = 0
is not just some special pick that matters as you have the conclusion that NO quadratic generator will work because all the algebraic steps I used are ok, and the conclusion for any that could would be that
a_1(x) = 7c_1(x)
which is the factor result being blocked by the ring of algebraic integers for certain cases.
So the ring of algebraic integers is BLOCKING the existence of
175x^2 - 15x + 2 = (5c_1(x) + 1)(5c_2(x)+ 2)
whereas, of course, the complex plane does not care, so everything was valid on the complex plane!
The ring of algebraic integer is saying, no way to a particular factorization. Now isn't that kind of odd? I think so.
So you have two conclusions:
The math appears to be fighting itself.
But why even care about what the result is on the ring of algebraic integers? Why should that matter?
Why can't I just say, screw that ring, I like the original result—as it makes sense—and I believe now that
a^2 - 6a + 35 = 0
has 7 as a factor for one root?
Turns out, you can.
And THAT is the reason math people get up in arms as traditionally they trust the ring of algebraic integers, but cannot in this instance tell you why you can't just say, forget it!!! It's bonkers!!!
Pull the thread though, and things get kind of nasty if you wish to toss out the ring of algebraic integers, as down the line you also lose Galois Theory as a useful tool.
Now this post is still an overview. I hope that maybe you now have a better grasp of what all the arguing is about, and maybe have some sense of why I say the issue is so huge, as I have shown you a first in mathematics:
direct demonstration of apparent contradiction
There is no other case in known mathematics where you can get something like the field of complex numbers disagreeing with something like the ring of algebraic integers.
Posters claiming that factors don't matter on the complex plane at all are being disingenuous as that is not the issue.
Like with y = 7x. On the complex plane y has 7 as a factor, but it also has 11, so what?
The equation however is valid in the ring of integers, where it is a factor result.
If
175x^2 - 15x + 2 = (5c_1(x) + 1)(5c_2(x)+ 2)
were valid on the ring of algebraic integers then a_1(x) = 7c_1(x), would be a factor result, in the same way.
Now to me it's kind of interesting that the c's CANNOT EXIST in the ring of algebraic integers if they can lead to roots of the same quadratic, for certain special cases, but I'm a person with a lot of curiosity.
And I've been taught at one of the best universities in the world to pull at threads, to see where that leads.
Math people fighting me on this issue are always coming to one conclusion: nothing to see here, not important, don't worry about it, just go with what you were taught, trust the textbooks.
Who to you sounds like the scientist, and who like people fighting to hold on to dogma?
Like I said, this post is an overview. But I have worked out the mathematical issues in much greater depth. I can delve very deeply into the esoteric aspects of the problem. Explain in extraordinary detail exactly how all the mathematics operates, bring in the wrapper theorem, talk about the ring of objects and show you how just advanced the mathematics is once this problem is properly considered.
It gets simpler but far more powerful.
But the first hurdle is your BELIEF.
And I have years of seeing how little I can get done when you just don't believe.
I cannot make progress if you do not have curiosity to care about the world's first case of apparent contradiction in established mathematics.
6 years have already passed here…
You being curious enough to force an answer from the mathematical community is the only hope that 6 more or more don't pass, and what then?
What if decades from now some physicists come to grips with this issue? Pull the thread and go where curious minds go?
And figure it out that I am right?
Consider the decades lost to the human species just because you and others like you were not curious enough to just find out the answers.
Real researchers wouldn't tell you to just let it all go and forget any of what I've shown here as nothing but trivial musings of a deranged mind.
What kind of people would? Think about it.
On the complex plane consider an unknown factorization:
175x^2 - 15x + 2 = (5c_1(x) + 1)(5c_2(x)+ 2)
Notice with just that information there are an infinity of functions that will work for the c's.
If you're wondering why THAT particular quadratic which I keep using over and over again, it's for historical reasons.
Now I also multiply by 7 for historical reasons but if it would help to multiply times some other number like say, 27, because it's not prime I can do that as well, and then the quadratic will change as well. But for now on with 7:
7*(175x^2 - 15x + 2) = 7*(5c_1(x) + 1)(5c_2(x)+ 2)
That is the OPPOSITE of what's normally done of course, as why multiply on an extraneous factor?
Answer is, because by doing this major "out of the box" step, I can reveal a massive error.
But next I have to make a choice!
7*(175x^2 - 15x + 2) = (5*7*c_1(x) + 7)(5c_2(x)+ 2)
and now I need to hide things a bit, which I can do with new functions:
b_1(x) = 7*c_1(x), and c_2(x) = b_2(x)
so
7*(175x^2 - 15x + 2) = (5b_1(x) + 7)(5b_2(x)+ 2)
with still as of yet undetermined functions, but I have a problem now with the 7 hidden, so I need a condition to start determining my functions:
Let b_1(0) = b_2(0) = 0.
But there are still an infinite number of possible functions at this point.
And now I need symmetry, as the issue is actually a symmetry problem, so I need one last set of new functions:
a_1(x) = b_1(x), and a_2(x) = b_2(x) - 1
and making those substitutions gives me:
7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)
The symmetry with regards to the 7's is required so that the a's can be roots of the same quadratic, which is NOT possible until that fix is made, so it's not an option here.
Now, of course, I have that ONE SOLUTION—as it is one solution out of infinity—for the a's is, as roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0.
and now you can solve for the a's using the quadratic formula:
a_1(x) = ((7x-1) +/- sqrt((7x-1)^2 - 4(49x^2 - 14x)))/2
a_2(x) = ((7x-1) -/+ sqrt((7x-1)^2 - 4(49x^2 - 14x)))/2
I need to emphasize at this point that the entire point of this exercise was to get to what I call a quadratic generator, which is just an expression that gives you quadratics, so for example, with x=1, you have
a^2 - 6a + 35 = 0
which is, of course, a quadratic. But also I needed a monic quadratic, which just means the leading coefficient is 1, which is satisfied, because then the a's will always be things called algebraic integers, when x is an integer.
But so what? Why bother?
Well, because mathematicians do not teach and prior to my research did not believe (I guess mostly they still don't) that given
a^2 - 6a + 35 = 0
only one of the roots is a product of 7, but my substitutions above mean that
a_1(x) = 7c_1(x) and a_2(x) = c_2(x) - 1
so you can SEE the 7 multiplying times the function c_1(x), where I remind that the start was:
175x^2 - 15x + 2 = (5c_1(x) + 1)(5c_2(x)+ 2)
Now the preferred ring for considering factors in such a situation for mathematicians is something called the ring of algebraic integers, and given all of the above you might suppose then that
a^2 - 6a + 35 = 0
has 7 as a factor for just one root, but instead you can mathematically prove the OPPOSITE result!!!
Well, you may wonder, maybe there is something weird about the algebraic operations I used to get to the a's, so I suggest that you go back through this post and consider each mathematical step carefully.
You will find that NONE of them are invalid on the complex plane, or apparently even on the ring of algebraic integers!!!
But that CANNOT be true, as if all the steps are valid on the ring of algebraic integers, then the result would hold in that ring, but we already have at x=1 that it DOES NOT HOLD IN THAT RING.
The mystery should resolve with a leap for some of you that of course the one thing that must be invalid on the ring of algebraic integers, as all the steps look like ok algebra is the start!!!
175x^2 - 15x + 2 = (5c_1(x) + 1)(5c_2(x)+ 2)
That is the only conclusion left for you. If that start is ok, then everything else follows on the ring of algebraic integers and the conclusion logically should hold, so it must not be possible for the c's to exist on the ring of algebraic integers for x=1, where you can get to the a's that are roots of a quadratic generator as given.
The more general conclusion though is that the c's cannot exist for ANY situation where you can get to a's that are roots of a quadratic generator!!!
So my using
a^2 - (7x-1)a + (49x^2 - 14x) = 0
is not just some special pick that matters as you have the conclusion that NO quadratic generator will work because all the algebraic steps I used are ok, and the conclusion for any that could would be that
a_1(x) = 7c_1(x)
which is the factor result being blocked by the ring of algebraic integers for certain cases.
So the ring of algebraic integers is BLOCKING the existence of
175x^2 - 15x + 2 = (5c_1(x) + 1)(5c_2(x)+ 2)
whereas, of course, the complex plane does not care, so everything was valid on the complex plane!
The ring of algebraic integer is saying, no way to a particular factorization. Now isn't that kind of odd? I think so.
So you have two conclusions:
- a_1(x) = 7c_1(x) found on the complex plane.
- Neither root of a^2 - 6a + 35 = 0 can have 7 as a factor in the ring of algebraic integers.
The math appears to be fighting itself.
But why even care about what the result is on the ring of algebraic integers? Why should that matter?
Why can't I just say, screw that ring, I like the original result—as it makes sense—and I believe now that
a^2 - 6a + 35 = 0
has 7 as a factor for one root?
Turns out, you can.
And THAT is the reason math people get up in arms as traditionally they trust the ring of algebraic integers, but cannot in this instance tell you why you can't just say, forget it!!! It's bonkers!!!
Pull the thread though, and things get kind of nasty if you wish to toss out the ring of algebraic integers, as down the line you also lose Galois Theory as a useful tool.
Now this post is still an overview. I hope that maybe you now have a better grasp of what all the arguing is about, and maybe have some sense of why I say the issue is so huge, as I have shown you a first in mathematics:
direct demonstration of apparent contradiction
There is no other case in known mathematics where you can get something like the field of complex numbers disagreeing with something like the ring of algebraic integers.
Posters claiming that factors don't matter on the complex plane at all are being disingenuous as that is not the issue.
Like with y = 7x. On the complex plane y has 7 as a factor, but it also has 11, so what?
The equation however is valid in the ring of integers, where it is a factor result.
If
175x^2 - 15x + 2 = (5c_1(x) + 1)(5c_2(x)+ 2)
were valid on the ring of algebraic integers then a_1(x) = 7c_1(x), would be a factor result, in the same way.
Now to me it's kind of interesting that the c's CANNOT EXIST in the ring of algebraic integers if they can lead to roots of the same quadratic, for certain special cases, but I'm a person with a lot of curiosity.
And I've been taught at one of the best universities in the world to pull at threads, to see where that leads.
Math people fighting me on this issue are always coming to one conclusion: nothing to see here, not important, don't worry about it, just go with what you were taught, trust the textbooks.
Who to you sounds like the scientist, and who like people fighting to hold on to dogma?
Like I said, this post is an overview. But I have worked out the mathematical issues in much greater depth. I can delve very deeply into the esoteric aspects of the problem. Explain in extraordinary detail exactly how all the mathematics operates, bring in the wrapper theorem, talk about the ring of objects and show you how just advanced the mathematics is once this problem is properly considered.
It gets simpler but far more powerful.
But the first hurdle is your BELIEF.
And I have years of seeing how little I can get done when you just don't believe.
I cannot make progress if you do not have curiosity to care about the world's first case of apparent contradiction in established mathematics.
6 years have already passed here…
You being curious enough to force an answer from the mathematical community is the only hope that 6 more or more don't pass, and what then?
What if decades from now some physicists come to grips with this issue? Pull the thread and go where curious minds go?
And figure it out that I am right?
Consider the decades lost to the human species just because you and others like you were not curious enough to just find out the answers.
Real researchers wouldn't tell you to just let it all go and forget any of what I've shown here as nothing but trivial musings of a deranged mind.
What kind of people would? Think about it.
# posted by James Harris @ 04:08 
