To me what makes mathematical discovery really hard is the pressure to get to a point of no error. That pressure to be perfect is so daunting but so absolute! As with a single error an entire framework of thought can collapse.
Pursuing perfection defines mathematical discovery unlike any other human endeavor, except I guess logic, because there are so few areas where a perfect argument can be known!
For other areas there is an approximation, or finding of things that work where it may not be clear why.
But in mathematics for a mathematical proof every single connecting element must be known, and every single one must be absolutely perfect. And every single piece of the mathematical argument must be absolutely correct.
Or you have nothing.
James Harris
Monday, November 23, 2015
Friday, November 20, 2015
Push beyond with simple relations
Playing with simple mathematical relations you can push the instinctive limits of human thought in key mathematical areas.
For example, if you dare, in the complex plane consider:
k*P(x) = k*(g1(x) + 1)(g2(x) + 2)
where P(x) is a quadratic with integer coefficients, g1(0) = g2(0) = 0, but g1(x) does not equal 0 for all x. And k is any nonzero non-unit integer.
Yes, I know the k can be removed, but now see what I do as instead I use it to allow me to force symmetry on the right.
k*P(x) = (f1(x) + k)(f2(x) + k)
where: f1(0) = 0, f2(0) = -k+2
And now the k is wrapped up! How do you remove now?
There ARE simple solutions, for example, try: g1(x) = x and g2(x) = x
That forces P(x) = x2 + 3x + 2
Which means, switchable by indices of course:
f1(x) = kx and f2(x) = x - (k-2)
Which is the simplest case of it. Polynomial cases are easy. But can you go beyond?
If you DO try and get stuck I have answers. And guess what? The absolute truth requires only algebra. And understanding that truth requires that you can step through a mathematical argument and care fiercely about wanting the truth. Because the typical human mind may balk at trying to go beyond polynomials.
I figured out a way to get a handle on the equations with a function H(x), where yeah the H is about "handle", and I capitalize because I like the look, though actually it can be an infinity of functions in the complex plane:
f1(x) + f2(x) = H(x)
So now I can substitute out for one of the f's, and choose easily:
f12(x) - H(x)f1(x) - kH(x) - k2 + k*P(x) = 0
Which covers an infinite class of equations now. Easy.
Now we can see the obvious, but also see how you step off into beyond. And notice that all I did was do something counter to habit. That is easier said than done.
So many talk about stepping outside the box, but the point is that thinking a certain way is easier than not.
That I have a quadratic isn't a surprise of course, but now can use the quadratic formula which has a square root! And that has HUGE consequences.
And you can solve for f1(x) using the quadratic formula:
f1(x) = (H(x) +/- sqrt[(H(x) + 2k)2 - 4k*P(x)])/2
Because H(x) is actually a function with x as a variable, turns out that to get that square root to go away, you need a quadratic. And that means that then the f's are just simple polynomials themselves, linear.
There is no way around it.
Also if H(x) is itself an algebraic integer function, that is, give algebraic integers for algebraic integer x, then know that the f's are as well.
Weirdly enough, H(x) can cover ALL possible solutions for the g's and f's, but is limited with my proof by: H(0) = -k + 2
That just amazes me. It in essence tells the math all it needs to know about what I want, as I picked the initial conditions. And wanted 1 and 2 in there for simplicity.
Here's something with actual numbers:
7*162 = (5(3+sqrt(-26)) + 7)(5(3-sqrt(-26)) + 7)
Ask yourself, what are the values for the g's here?
And I've talked about this subject quite a bit including a recent mathematical proof. While I've also noted that the first post on this blog back in 2005 relates to this subject as I've been considering mathematics in this area now for over a decade.
LOTS of controversy in this subject area as well, including a dead mathematical journal no less.
These kinds of things are fun, weird, and fascinating as in human history they don't happen that often.
People living through such things? Rarely have a clue how huge it all is. Which probably has its benefits I guess. Maybe I shouldn't tell you then, eh? But I'm not worried, only a special group of you will believe me early anyway, I'm sure.
At least at first, while eventually of course will just be settled mathematical history.
That weird feeling though that some may have with the result and its consequences is all about limitations of the normal. We think a certain way! It can be SO HARD to shift even with absolute proof.
Am guessing the human brain wasn't built for such mathematics as I've discovered, as even for me has taken some time to get comfortable with it. But the great thing is with effort our minds can comprehend the math anyway.
Makes me question how any people could think that human beings create mathematics rather than discover it, which is why an ego check like this result may have such a hard way to go! After all, for some self-important person that may be a cherished belief.
So maybe for some people it's a welcome to humility if they accept it.
Welcome to mathematical reality where you're small. The Math is infinite.
And guess what? The Math does not need you. It does not need you to understand.
The Math does not care.
James Harris
For example, if you dare, in the complex plane consider:
k*P(x) = k*(g1(x) + 1)(g2(x) + 2)
where P(x) is a quadratic with integer coefficients, g1(0) = g2(0) = 0, but g1(x) does not equal 0 for all x. And k is any nonzero non-unit integer.
Yes, I know the k can be removed, but now see what I do as instead I use it to allow me to force symmetry on the right.
k*P(x) = (f1(x) + k)(f2(x) + k)
where: f1(0) = 0, f2(0) = -k+2
And now the k is wrapped up! How do you remove now?
There ARE simple solutions, for example, try: g1(x) = x and g2(x) = x
That forces P(x) = x2 + 3x + 2
Which means, switchable by indices of course:
f1(x) = kx and f2(x) = x - (k-2)
Which is the simplest case of it. Polynomial cases are easy. But can you go beyond?
If you DO try and get stuck I have answers. And guess what? The absolute truth requires only algebra. And understanding that truth requires that you can step through a mathematical argument and care fiercely about wanting the truth. Because the typical human mind may balk at trying to go beyond polynomials.
I figured out a way to get a handle on the equations with a function H(x), where yeah the H is about "handle", and I capitalize because I like the look, though actually it can be an infinity of functions in the complex plane:
f1(x) + f2(x) = H(x)
So now I can substitute out for one of the f's, and choose easily:
f12(x) - H(x)f1(x) - kH(x) - k2 + k*P(x) = 0
Which covers an infinite class of equations now. Easy.
Now we can see the obvious, but also see how you step off into beyond. And notice that all I did was do something counter to habit. That is easier said than done.
So many talk about stepping outside the box, but the point is that thinking a certain way is easier than not.
That I have a quadratic isn't a surprise of course, but now can use the quadratic formula which has a square root! And that has HUGE consequences.
And you can solve for f1(x) using the quadratic formula:
f1(x) = (H(x) +/- sqrt[(H(x) + 2k)2 - 4k*P(x)])/2
Because H(x) is actually a function with x as a variable, turns out that to get that square root to go away, you need a quadratic. And that means that then the f's are just simple polynomials themselves, linear.
There is no way around it.
Also if H(x) is itself an algebraic integer function, that is, give algebraic integers for algebraic integer x, then know that the f's are as well.
Weirdly enough, H(x) can cover ALL possible solutions for the g's and f's, but is limited with my proof by: H(0) = -k + 2
That just amazes me. It in essence tells the math all it needs to know about what I want, as I picked the initial conditions. And wanted 1 and 2 in there for simplicity.
Here's something with actual numbers:
7*162 = (5(3+sqrt(-26)) + 7)(5(3-sqrt(-26)) + 7)
Ask yourself, what are the values for the g's here?
And I've talked about this subject quite a bit including a recent mathematical proof. While I've also noted that the first post on this blog back in 2005 relates to this subject as I've been considering mathematics in this area now for over a decade.
LOTS of controversy in this subject area as well, including a dead mathematical journal no less.
These kinds of things are fun, weird, and fascinating as in human history they don't happen that often.
People living through such things? Rarely have a clue how huge it all is. Which probably has its benefits I guess. Maybe I shouldn't tell you then, eh? But I'm not worried, only a special group of you will believe me early anyway, I'm sure.
At least at first, while eventually of course will just be settled mathematical history.
That weird feeling though that some may have with the result and its consequences is all about limitations of the normal. We think a certain way! It can be SO HARD to shift even with absolute proof.
Am guessing the human brain wasn't built for such mathematics as I've discovered, as even for me has taken some time to get comfortable with it. But the great thing is with effort our minds can comprehend the math anyway.
Makes me question how any people could think that human beings create mathematics rather than discover it, which is why an ego check like this result may have such a hard way to go! After all, for some self-important person that may be a cherished belief.
So maybe for some people it's a welcome to humility if they accept it.
Welcome to mathematical reality where you're small. The Math is infinite.
And guess what? The Math does not need you. It does not need you to understand.
The Math does not care.
James Harris
Tuesday, November 17, 2015
Non polynomial factorization short argument
It can be shown that there must exist additional numbers besides algebraic integers, which are also integer-like numbers.
Proof:
In the complex plane given:
P(x) = (g1(x) + 1)(g2(x) + 2)
where P(x) is a primitive quadratic with integer coefficients, g1(0) = g2(0) = 0, but g1(x) does not equal 0 for all x.
Introduce k, where k is a nonzero integer, and not 1 or -1, and new functions f1(x), and f2(x), where:
g2(x) = f2(x) + k-2 and g1(x) = f1(x)/k,
multiply both sides by k, and substitute for the g's, which gives me the now symmetrical form:
k*P(x) = (f1(x) + k)(f2(x) + k)
And introduce H(x), where I like using the capital letter here for visual reasons, but it is not to signify H(x) must be a polynomial, where:
f1(x) + f2(x) = H(x)
So I can find:
f12(x) - H(x)f1(x) - kH(x) - k2 + k*P(x) = 0
And you can solve for f1(x) using the quadratic formula:
f1(x) = (H(x) +/- sqrt[(H(x) + 2k)2 - 4k*P(x)])/2
And sqrt[(H(x) + 2k)2 - 4k*P(x)] will only resolve if it is to a linear function, which proves that the f's are only polynomials when H(x) is, which means the g's are then as well.
And H(x) is a handle for every possible factorization with the g's. While it has a key constraint, H(0) = -k + 2.
By smoothly transitioning H(x) through all possible equations with the given constraint can encompass all possible equations for the g's with the given requirements at x=0, and in no case can they both be equations that produce algebraic integers with algebraic integer x, except the polynomial case, when k is not zero or a unit.
Not really part of the main argument, but notice if we relax the restriction on k and let it be a unit, like try k=1, then necessarily the g's give algebraic integers. And using k =1 or -1 and using all possible H(x) in that ring covers ALL algebraic integer solutions for the g's.
But with the condition that k not be zero or a unit, there can be values of x for which you get algebraic integers, for instance x=0 will always work. But such exceptions don't change the typical case.
But regardless of whether k is a unit or not, the f's must be in the ring of algebraic integers if H(x) is. So it is shown that in those cases ANY k with the requirements given that it be a nonzero integer, not 1 or -1, will give algebraic integer f's.
Therefore there must exist numbers which I will call objects o, such that k*o must be an algebraic integer, when o is not an algebraic integer, for any k with the requirements.
Proof complete.
And it is easy to show why fields are not impacted, as consider o2/o1, which might not seem to be in the field of algebraic numbers, but you just have to multiply top and bottom by some nonzero integer not 1 or -1, like:
2o2/2o1
and you have a ratio of algebraic integers, showing that it is indeed an algebraic number.
These numbers are such that k*o will give an algebraic integer, regardless of k as long as not 0, 1 or -1, with it an integer. So for a particular o not an algebraic integer, 3o and 5o will both be algebraic integers.
With these new numbers needed a ring to contain them as well as algebraic integers.
And after pondering them for some time I came up with a more robust classification scheme with what I decided to call the object ring.
That link goes to the first post on this blog. Posted back in 2005.
These numbers were the biggest reason for the existence of this blog.
But I hadn't discovered this simple approach back then. Wonder if I had, would it have made a difference? Will it make one now?
These results change so much in mathematics. Resistance has been so fierce for over a decade now.
James Harris
Proof:
In the complex plane given:
P(x) = (g1(x) + 1)(g2(x) + 2)
where P(x) is a primitive quadratic with integer coefficients, g1(0) = g2(0) = 0, but g1(x) does not equal 0 for all x.
Introduce k, where k is a nonzero integer, and not 1 or -1, and new functions f1(x), and f2(x), where:
g2(x) = f2(x) + k-2 and g1(x) = f1(x)/k,
multiply both sides by k, and substitute for the g's, which gives me the now symmetrical form:
k*P(x) = (f1(x) + k)(f2(x) + k)
And introduce H(x), where I like using the capital letter here for visual reasons, but it is not to signify H(x) must be a polynomial, where:
f1(x) + f2(x) = H(x)
So I can find:
f12(x) - H(x)f1(x) - kH(x) - k2 + k*P(x) = 0
And you can solve for f1(x) using the quadratic formula:
f1(x) = (H(x) +/- sqrt[(H(x) + 2k)2 - 4k*P(x)])/2
And sqrt[(H(x) + 2k)2 - 4k*P(x)] will only resolve if it is to a linear function, which proves that the f's are only polynomials when H(x) is, which means the g's are then as well.
And H(x) is a handle for every possible factorization with the g's. While it has a key constraint, H(0) = -k + 2.
By smoothly transitioning H(x) through all possible equations with the given constraint can encompass all possible equations for the g's with the given requirements at x=0, and in no case can they both be equations that produce algebraic integers with algebraic integer x, except the polynomial case, when k is not zero or a unit.
Not really part of the main argument, but notice if we relax the restriction on k and let it be a unit, like try k=1, then necessarily the g's give algebraic integers. And using k =1 or -1 and using all possible H(x) in that ring covers ALL algebraic integer solutions for the g's.
But with the condition that k not be zero or a unit, there can be values of x for which you get algebraic integers, for instance x=0 will always work. But such exceptions don't change the typical case.
But regardless of whether k is a unit or not, the f's must be in the ring of algebraic integers if H(x) is. So it is shown that in those cases ANY k with the requirements given that it be a nonzero integer, not 1 or -1, will give algebraic integer f's.
Therefore there must exist numbers which I will call objects o, such that k*o must be an algebraic integer, when o is not an algebraic integer, for any k with the requirements.
Proof complete.
But what about fields? Does this impact them?
And it is easy to show why fields are not impacted, as consider o2/o1, which might not seem to be in the field of algebraic numbers, but you just have to multiply top and bottom by some nonzero integer not 1 or -1, like:
2o2/2o1
and you have a ratio of algebraic integers, showing that it is indeed an algebraic number.
These numbers are such that k*o will give an algebraic integer, regardless of k as long as not 0, 1 or -1, with it an integer. So for a particular o not an algebraic integer, 3o and 5o will both be algebraic integers.
With these new numbers needed a ring to contain them as well as algebraic integers.
And after pondering them for some time I came up with a more robust classification scheme with what I decided to call the object ring.
That link goes to the first post on this blog. Posted back in 2005.
These numbers were the biggest reason for the existence of this blog.
But I hadn't discovered this simple approach back then. Wonder if I had, would it have made a difference? Will it make one now?
These results change so much in mathematics. Resistance has been so fierce for over a decade now.
James Harris
Monday, November 16, 2015
Beyond polynomial mental barrier
Thinking about ways of looking at problems intrigues me, where I recently posted about something rather simple to try, which has remarkable consequences:
k*P(x) = k*(g1(x) + 1)(g2(x) + 2)
k*P(x) = k*(g1(x) + 1)(g2(x) + 2)
where P(x) is a quadratic with integer coefficients, g1(0) = g2(0) = 0, but g1(x) does not equal 0 for all x. And k is any nonzero non-unit integer.
And:
k*P(x) = (f1(x) + k)(f2(x) + k)
These two expressions force you in certain directions while including the polynomial case which I think is the one the human brain by design prefers. So then:
g1(x) = x and g2(x) = x
That forces P(x) = x2 + 3x + 2
Which means, switchable by indices of course:
f1(x) = kx and f2(x) = x - (k-2)
Which is the simplest case of it. Polynomial cases are easy. But I suggest to you there is a mental barrier to accepting that the g's are not limited to what is familiar to a human mind! But if you push beyond polynomials you break what most people think they know about mathematics and numbers.
Which is why I like this example.
Test yourself--find another set of f's for a different set of g's.
Polynomials are boring. There are more of those of course, an infinity more but that pattern is easy.
If you're stuck, welcome to the limitations of your brain. Some may be incapable of seeing anything other than polynomials here. You may have an inherent limitation in the wiring of your brain. And if so that's not my fault! Don't get mad at me because of it.
Notice I didn't specify a ring. See how that matters. Try different rings, or even the complex field! The f's and g's will still exist there of course.
Push beyond your mental barriers--if you can.
Are you smart enough to find any other answer beyond the basic your mind already knows?
But some of you don't need to be so enticed, as you've been looking for a door into something beyond, maybe your entire life. For those people, you're welcome to the fun!
James Harris
These two expressions force you in certain directions while including the polynomial case which I think is the one the human brain by design prefers. So then:
g1(x) = x and g2(x) = x
That forces P(x) = x2 + 3x + 2
Which means, switchable by indices of course:
f1(x) = kx and f2(x) = x - (k-2)
Which is the simplest case of it. Polynomial cases are easy. But I suggest to you there is a mental barrier to accepting that the g's are not limited to what is familiar to a human mind! But if you push beyond polynomials you break what most people think they know about mathematics and numbers.
Which is why I like this example.
Test yourself--find another set of f's for a different set of g's.
Polynomials are boring. There are more of those of course, an infinity more but that pattern is easy.
If you're stuck, welcome to the limitations of your brain. Some may be incapable of seeing anything other than polynomials here. You may have an inherent limitation in the wiring of your brain. And if so that's not my fault! Don't get mad at me because of it.
Notice I didn't specify a ring. See how that matters. Try different rings, or even the complex field! The f's and g's will still exist there of course.
Push beyond your mental barriers--if you can.
Are you smart enough to find any other answer beyond the basic your mind already knows?
But some of you don't need to be so enticed, as you've been looking for a door into something beyond, maybe your entire life. For those people, you're welcome to the fun!
James Harris
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