### Saturday, December 12, 2009

## JSH: Mathematical impossibility and Galois Theory

Showing an important result in complex numbers allows me to prove absolutely a problem with some well-established number theory which underpins Galois Theory, which removes use of Galois groups, at least, useful use. But being able to prove something absolutely is not as big a convincer as you might suppose, as, I've done it before.

What's different now is I'm pointing out to people that this result CAN be used to humble any, oh, string theorist you wish to torture with it. Human nature is odd. People might not care about some massive error in the abstract, but knowing they can blow up another human being's world with it, now that gets their attention.

So here's the result in the field of complex numbers yet again, where notice it's easy to show absolute proof.

I use a special construction:

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 a's can be found using the quadratic formula, but are so ugly I usually don't bother to give them, but maybe it'll help for those who wonder what they look like:

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 or vice versa.

At x=0, one of the a's is 0, while the other is -1, as a^2 + a = 0, has those solutions (you can check with the explicit equations as well if you wish). So it pays to get functions where both are 0 at x=0, as that trivially finishes the exercise.

Letting b_1(x) = a_1(x) + 1, assuming a_2(0) = 0, I have a_1(x) = b_1(x) - 1, and substituting gives:

7(175x^2 - 15x + 2) = (5b_1(x) + 2)(5a_2(x) + 7)

So now in the field of complex numbers you have the interesting result that the 7 was multiplied in only one way, which is the way I picked as fundamentally the example is no different from 7(x^2 + 3x + 2) = (x + 2)(7x + 7).

If I'd picked another way, you might have:

7(175x^2 - 15x + 2) = (5b_1(x) + 14)(5a_2(x) + 1)

If I'd wanted to split the 7 with square roots, you could have:

7(175x^2 - 15x + 2) = (5b_1(x) + 2sqrt(7))(5a_2(x) + sqrt(7))

So over the complex plane you have the answer as to what I wanted and the only way the 7 could have multiplied.

That is a mathematical absolute. It's mathematically impossible for the 7 to have multiplied any different way with:

7(175x^2 - 15x + 2) = (5b_1(x) + 2)(5a_2(x) + 7)

where b_1(0) = a_2(0) = 0.

But that says that with

a^2 - (7x-1)a + (49x^2 - 14x) = 0

only one of the roots is a product of 7, and I say it that weird way as I'm on the complex plane and "factor" isn't really meaningful.

And here's where a lot of algebraic number theory blows up which brings into question the usefulness of Galois Theory, and rather than go into that a lot, I'll point out that you're looking at a really easy absolute proof.

After all, if the 7 is being multiplied times one factor, so that only one root is a product of 7 and something else, then how in the hell could its factors end up in the other root?

But mathematicians teach that it DOES if with integer x

a^2 - (7x-1)a + (49x^2 - 14x) = 0

has non-rational roots.

Their mistake was easy to make, and the mistake was made over a hundred years ago, and it just took a while for someone (me) to come along and put it out there boldly what that mistake was, and predictably with an over one hundred year error there has been resistance.

It's actually kind of fascinating to watch posters try to dance around what you can determine in the field of complex numbers!

So what's different now? I've explained that the result is valid over the complex field before.

Again, it's that simple thing: the ability to humiliate others.

I'm guessing there are people in this world who will love the opportunity to screw with academics, to laugh at them, to show them up.

String theorists are a sad lot, if you know of this error, or they're a funny group of people whom you can watch with endless humor.

You can laugh and laugh and laugh as they go on about what the universe might be like, or claim that this or that notion of theirs actually works in the REAL WORLD.

When mathematicians claim some BIG RESULT, you can chuckle. I know I do.

I can dismiss awards yearly. I see the dark humor in it.

They're funny people.

What I've given is the key to seeing people with no clothes on, as the saying goes.

You can laugh at department heads. An undergrad can smile knowingly at her math professor. She may do the homework assigned. Take the tests. Do the bogus math because what's her choice? But deep down she knows she's dealing with a fool.

One time I got so in a mood I actually made a post about how the village idiots took over number theory.

Am I being too mean?

Nope. They had their chance. They've had YEARS. But they seem to believe that they can run.

Ok, now they can see the consequences. Humiliation.

They will be the world's next big laugh, and cry. Especially for the parents of the students taught the bogus research over the last six years. Those parents may just cry. So much for a Harvard education, eh?

Through the humiliation of academics, the truth can come out and that tragedy can end.

The betrayal of young minds by their professors.

Yeah, I have full justification to wreck the pride of arrogant idiots who betrayed so much for so little, who in trying to run from an absolute error, only set themselves up to be shown for what they are.

What they truly are.

They had a duty to tell the truth. Now it's your duty to make them pay.

What's different now is I'm pointing out to people that this result CAN be used to humble any, oh, string theorist you wish to torture with it. Human nature is odd. People might not care about some massive error in the abstract, but knowing they can blow up another human being's world with it, now that gets their attention.

So here's the result in the field of complex numbers yet again, where notice it's easy to show absolute proof.

I use a special construction:

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 a's can be found using the quadratic formula, but are so ugly I usually don't bother to give them, but maybe it'll help for those who wonder what they look like:

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 or vice versa.

At x=0, one of the a's is 0, while the other is -1, as a^2 + a = 0, has those solutions (you can check with the explicit equations as well if you wish). So it pays to get functions where both are 0 at x=0, as that trivially finishes the exercise.

Letting b_1(x) = a_1(x) + 1, assuming a_2(0) = 0, I have a_1(x) = b_1(x) - 1, and substituting gives:

7(175x^2 - 15x + 2) = (5b_1(x) + 2)(5a_2(x) + 7)

So now in the field of complex numbers you have the interesting result that the 7 was multiplied in only one way, which is the way I picked as fundamentally the example is no different from 7(x^2 + 3x + 2) = (x + 2)(7x + 7).

If I'd picked another way, you might have:

7(175x^2 - 15x + 2) = (5b_1(x) + 14)(5a_2(x) + 1)

If I'd wanted to split the 7 with square roots, you could have:

7(175x^2 - 15x + 2) = (5b_1(x) + 2sqrt(7))(5a_2(x) + sqrt(7))

So over the complex plane you have the answer as to what I wanted and the only way the 7 could have multiplied.

That is a mathematical absolute. It's mathematically impossible for the 7 to have multiplied any different way with:

7(175x^2 - 15x + 2) = (5b_1(x) + 2)(5a_2(x) + 7)

where b_1(0) = a_2(0) = 0.

But that says that with

a^2 - (7x-1)a + (49x^2 - 14x) = 0

only one of the roots is a product of 7, and I say it that weird way as I'm on the complex plane and "factor" isn't really meaningful.

And here's where a lot of algebraic number theory blows up which brings into question the usefulness of Galois Theory, and rather than go into that a lot, I'll point out that you're looking at a really easy absolute proof.

After all, if the 7 is being multiplied times one factor, so that only one root is a product of 7 and something else, then how in the hell could its factors end up in the other root?

But mathematicians teach that it DOES if with integer x

a^2 - (7x-1)a + (49x^2 - 14x) = 0

has non-rational roots.

Their mistake was easy to make, and the mistake was made over a hundred years ago, and it just took a while for someone (me) to come along and put it out there boldly what that mistake was, and predictably with an over one hundred year error there has been resistance.

It's actually kind of fascinating to watch posters try to dance around what you can determine in the field of complex numbers!

So what's different now? I've explained that the result is valid over the complex field before.

Again, it's that simple thing: the ability to humiliate others.

I'm guessing there are people in this world who will love the opportunity to screw with academics, to laugh at them, to show them up.

String theorists are a sad lot, if you know of this error, or they're a funny group of people whom you can watch with endless humor.

You can laugh and laugh and laugh as they go on about what the universe might be like, or claim that this or that notion of theirs actually works in the REAL WORLD.

When mathematicians claim some BIG RESULT, you can chuckle. I know I do.

I can dismiss awards yearly. I see the dark humor in it.

They're funny people.

What I've given is the key to seeing people with no clothes on, as the saying goes.

You can laugh at department heads. An undergrad can smile knowingly at her math professor. She may do the homework assigned. Take the tests. Do the bogus math because what's her choice? But deep down she knows she's dealing with a fool.

One time I got so in a mood I actually made a post about how the village idiots took over number theory.

Am I being too mean?

Nope. They had their chance. They've had YEARS. But they seem to believe that they can run.

Ok, now they can see the consequences. Humiliation.

They will be the world's next big laugh, and cry. Especially for the parents of the students taught the bogus research over the last six years. Those parents may just cry. So much for a Harvard education, eh?

Through the humiliation of academics, the truth can come out and that tragedy can end.

The betrayal of young minds by their professors.

Yeah, I have full justification to wreck the pride of arrogant idiots who betrayed so much for so little, who in trying to run from an absolute error, only set themselves up to be shown for what they are.

What they truly are.

They had a duty to tell the truth. Now it's your duty to make them pay.