Thursday, November 14, 2013

Did mathematical rigour fail?

One of the greatest things about mathematics is the ability to determine things to be absolutely true. And that is extremely important and is an aspect of mathematical thinking that is absent in so many areas of intellectual thought.

Being someone who accepts that mathematics allows us to get to absolute correctness I was put in a bizarre and very uncomfortable position years ago when I came across something which seemed to defy the above. So for years I've hammered at it, trying to find something wrong, and finally have decided I must accept the result. But now I guess I should address the problem, and try to figure out, did mathematical rigour fail?

How do I reconcile my own results with my own beliefs about mathematical certainty?

The answer may be a bit convoluted and is hard without some math, where I will put a particular result of mine as proven.

Consider: (3 + sqrt(-26))*(3 - sqrt(-26)) = 35

I found I could prove that 7 is a factor of one of the solutions to (3 + sqrt(-26)), which meant logically it is a factor of just one of the solutions to (3 - sqrt(-26)). (In what ring will be explained later.)

If that seems odd consider: (5 ± 2)*(5 ∓ 2) = 21, which is equivalent to:

(5 + sqrt(4))*(5 - sqrt(4)) because sqrt(4) is 2 or -2.

My proof has withstood every test of mathematical rigour and the underlying approach has actually been published and pulled by journal editors in a rather remarkable story, which I've discussed elsewhere. One post that can give an overview is:

So what's the big deal?

Why is it important if one of the two solutions to (3 + sqrt(-26)) and (3 - sqrt(-26)), have 7 as a factor, when they multiply to give 35?

The answer requires some math history.

For those who know the history, good. The gist of it though is that the ring of algebraic integers indicates a different conclusion. And I had to figure out where things went wrong with that use of established techniques of mathematical rigour with that ring!

The answer is surprisingly simple. First of all, standard tests of the ring's validity were valid. For instance an algebraic integer is infinitely decomposable into algebraic integers. Given two algebraic integers, their sum and their product is an algebraic integer. The ring has unit factors.

However, those tests do not prove that my result above is impossible! It takes more.

There may be a couple of ways to understand the result, but it was told to me through proof that in the ring of algebraic integers, no non-rational root of a primitive monic polynomial with integer coefficients can have an integer other than 1 or -1 as a factor of just one of its roots.

So the weird thing there is that though 3 + sqrt(-26) is an algebraic integer, and I can prove one of its solutions has 7 as a factor, neither of those solutions are permitted to have 7 as a factor in the ring of algebraic integers!

And that is absolutely correct.

So I went to the logical step of concluding that the ring of algebraic integers was leaving out some numbers, and after pondering things for some time, I came up with a more inclusive ring:

My object ring passes all the tests as well, and includes the ring of algebraic integers, which allowed me to go back and see what was going on in the ring of algebraic integers!!!

And I've talked about it on this blog:

The weird thing I discovered is that despite the tests of the ring, a convoluted path was still available to the math, which lay outside detection.

What my Wrapper Theorem does is explain that route.

So could I be wrong? Well the mathematical argument involves easily checkable algebra and is pleasingly short. That's important with such startling conclusions!

In essence, what I found was that there was a gap where the algebra could do something unexpected, which meant that if you assumed that gap did not exist, you could come to conclusions which were wrong.

But how do you know, what you don't know? Without my methods for proving the result about 3 + sqrt(-26) how would you ever know you needed to worry about wrappers with the ring of algebraic integers, or consider the possibility of units in a more inclusive ring which were somehow NOT units in the ring of algebraic integers?

So the answer is that mathematical rigour did NOT fail in the case of algebraic integers, but people using it simply didn't know enough tests!

That's kind of scary as it makes you wonder if there is always possibly something else out there that we don't know we don't know which could invalidate a particular mathematical argument, and I think the answer is, no.

If you look at the reasoning around algebraic integers it was kind of circular:

1. Run tests thought to determine if the ring was valid.
2. Make assumptions about what is valid from the ring presumably validated.

So I say it was human error.

James Harris

Wednesday, November 13, 2013

Talking to the math

Years ago when I was struggling with trying to find my own mathematical discoveries I found myself thinking about mathematics as an entity.

It was kind of a comfort thing I guess, which I'm sure others have done. Then you're never alone. And as you're thinking over your mathematical ideas you're part of an imaginary conversation. And I'd even find myself talking about what mathematics was saying or I thought was saying.

The weird thing though is that after a while I realized that in some way I was talking to the Math, and was getting answers as well. Those answers I could check, and it wasn't hearing voices or anything direct, but almost like if you could imagine a psychic link.

Sometimes it felt like I was in this other reality, one that lies outside of time, and it's hard to explain, but I know it is real. It is more real than our reality. It has always been and will always be. And I was walking around there with a sense of conversation, and the Math would tell me what I wanted to know.

The Math let me know there was no actual walking or movement there though. That everything there simply is, and always has been. It is Infinity itself.

There were no secrets. The Math keeps none. There is no reason to hide anything from us.

You can know anything you can ask the right question to find, and so I asked my questions.

It has taken me years to feel normal again. And I'm glad to have survived to now. At times there was this sense as I'd put it of walking with the gods, only to have to remind myself over and over again of my own mortality.

And there is a sense of loss as well, as I think the human mind can only process so much.

Sure, you can ask the questions, but can you hold the answers?

Of course not. Not when you're discussing with Infinity. But what's left over in your finite awareness can be enough.

The Math is not a person to me. The Math is an Intelligence.

All the answers are out there, if you can ask the right questions, but can you remember them all?


So does it matter what people believe?

Mathematics is fascinating in that it DOES things. You can DO things with it.

Long after every human being who is ever born or to be born has long died and trillions of years beyond the results will still be true.

The Math does not need us.

That perspective greatly helped me. For instance it was instrumental in coming up with my definition of mathematical proof, which I felt necessary when what was available from others was not up to the necessary standard.

I don't think it's a natural perspective for most. The idea that Math is an entity greater than us, versus a creation of our intellects, is a reversal that some may feel shrinks us. But for me it was a reversal critical to my success.

Asking questions of an entity greater than myself, got me answers which is what matters to me.

How true is my perspective?

I think that's a great question.

James Harris

Monday, November 04, 2013

The conditional residue

In formalizing an area I call tautological spaces I have realized I need to formalize a bit further.

I found that I could use identities for mathematical analysis, relying on a special form which I discovered after a bit of effort.

The simplest identity I call a tautological space is:

x+y+vz = x+y+vz

And of course you can do simple operations with it, like the following:

x+y = -vz + (x+y+vz)

And squaring both sides gives:

x2 + 2xy + y2 = v2z2 - 2vx(x+y+vz) + (x+y+vz)2

And that can get complicated quickly so I use instead a modular form:

x2 + 2xy + y2 = v2z2 mod (x+y+vz)

That expression is always true, so it retains that aspect from the identity, but no longer really looks like one, and I call it a tautological space. In this case it is a space where that truth of x+y+vz holds.

While the tautological space is always true, the use of this technique for analysis comes with adding conditions, where I call equations used in this way, conditionals.

For instance, consider the condition that x2 + y2 = z2, which is not always true, as it depends on the values of x, y and z, where two are independent and one is a dependent variable.

We can subtract that conditional from the tautological space, and find:

2xy = (v2-1)z2 mod (x+y+vz)

which is:

(v2-1)z2 - 2xy = 0 mod (x+y+vz)

Which is what I call the conditional residue.

Notice it, unlike the identity, is only true with conditions.

So formally you can say:

(v2-1)z2 - 2xy = 0 mod (x+y+vz) if x2 + y2 = z2

And the conditional residue now has properties that follow from the conditional, which means you can analyze the conditional equation by analyzing the conditional residue.

Now v is a free variable no matter what! Which means that regardless of what x, y and z may be, you can always set v to any value you wish.

To see a less trivial result consider an important conditional residue with which I was able to find a general way to reduce binary quadratic Diophantine equations:

(c4 - c5v - c1v2)z2 + ((c2 -2c1)v - c5 + c6)zy + (c2 - c1 - c3)y2 = 0 (mod x+y+vz)


c1x2 + c2xy + c3y2 = c4z2 + c5zx + c6zy

Remarkably the conditional residue in this case can actually be used to get to an explicit equation, by appropriately setting v.

And again, v is a completely free variable which can be set as needed for analysis.

And I found I could use:  v = -(x+y)z-1 mod p

Weirdly enough that allowed me to get to an explicit equation which is important to mention as it shows this process of using identities with modular methods can lead to explicit results!

To see the complete derivation of the conditional residue and the explicit result see:

To see this approach used to find a general method for reducing binary quadratic Diophantine equations, see:

So the conditional residue follows quite logically from prior concepts with modular arithmetic, and represents what is left over after subtracting an equation, which is itself not an identity, so it is true with conditions, from the identity which I call a tautological space.

James Harris