Tuesday, January 27, 2004
JSH: Mathematical clarity
What I find interesting about this current situation is the willingness of some people on this newsgroup to question rather basic mathematics.
That's easy to show with basic mathematics using the Decker example.
Decker, a professor at Hamilton College, put forward a quadratic, and you can find his original post with the following headers:
Message-ID: <3FF47C4C.6080109@hamilton.edu>
>Date: Thu, 01 Jan 2004 15:00:12 -0500
>From: Rick Decker <rdecker@hamilton.edu>
Newsgroups: sci.math
>Subject: Re: Mathematical consistency, courage
Decker put forward the quadratic
(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)
where his a's are roots of
a^2 - (x - 1)a + 7(x^2 + x).
A simple linear transformation helps balance the factorization, as using
a_2(x) = b_2(x) - 1, and making the substitution gives
(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2).
The point I've made is that if you divide both sides by 7, and wish to try and remain the ring of algebraic integers then necessarily you have
(5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2
which follows rather easily from some basic algebra.
If follows because algebra gives a rigid way to multiply out.
That way of multiplying out means that on the left you have 1 and 2 multiplying times each other to get 2 on the right.
It's easy to demonstrate that multiplying out:
25 a_1(x) b_2(x)/7 + 10 a_1(x)/7 + 5 b_2(x) + 2 = 25x^2 + 30x + 2
and you can see the two 2's balanced to each side.
It's basic algebra. Now sure you have some people who let their egos get in the way, but why keep protecting them? What's in it for you?
Now I know, and you know that if you try to divide 7 out any other way then you can't remain in the ring of algebraic integers. Like if you have f_1 and f_2 where f_1 f_2 = 7, with non-unit f_1 and f_2, where
(5a_1(x)/f_1 + f_2)(5b_2(x)/f_2 + 2/f_2) = 25x^2 + 30x + 2
then the algebra tells you that the factors of 2 on the right are f_2 and 2/f_2, and if f_2 is a non-unit factor of 7, then you're not in the ring of algebraic integers.
Now then, you have a choice: ignore basic algebra assuming you can keep this fight up indefinitely for social reasons, or tell the truth.
It seems to me that some of you don't realize that even if you succeed here for a few days, or a few years, or even decades, some day the truth comes out.
It doesn't matter if you're dead. I've made certain that your name will live in infamy if it's known at all: Wiles, Ribet, Granville, or anyone else from this generation.
It doesn't matter, as you're all in it together.
It's kind of funny really. Sure Wiles can appreciate whatever approbation he gets today, but in the future, he's just another loser.
I'm the winner. It's about the legacy people. I'm destroying yours now and guaranteeing mine.
Pump each other up now. Tell yourselves how great you are. Congratulate yourself on your supposed accomplishments while you can.
History will hate you and love me.
I'm the misunderstood and persecuted genius.
You're the assholes.
That's easy to show with basic mathematics using the Decker example.
Decker, a professor at Hamilton College, put forward a quadratic, and you can find his original post with the following headers:
Message-ID: <3FF47C4C.6080109@hamilton.edu>
>Date: Thu, 01 Jan 2004 15:00:12 -0500
>From: Rick Decker <rdecker@hamilton.edu>
Newsgroups: sci.math
>Subject: Re: Mathematical consistency, courage
Decker put forward the quadratic
(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)
where his a's are roots of
a^2 - (x - 1)a + 7(x^2 + x).
A simple linear transformation helps balance the factorization, as using
a_2(x) = b_2(x) - 1, and making the substitution gives
(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2).
The point I've made is that if you divide both sides by 7, and wish to try and remain the ring of algebraic integers then necessarily you have
(5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2
which follows rather easily from some basic algebra.
If follows because algebra gives a rigid way to multiply out.
That way of multiplying out means that on the left you have 1 and 2 multiplying times each other to get 2 on the right.
It's easy to demonstrate that multiplying out:
25 a_1(x) b_2(x)/7 + 10 a_1(x)/7 + 5 b_2(x) + 2 = 25x^2 + 30x + 2
and you can see the two 2's balanced to each side.
It's basic algebra. Now sure you have some people who let their egos get in the way, but why keep protecting them? What's in it for you?
Now I know, and you know that if you try to divide 7 out any other way then you can't remain in the ring of algebraic integers. Like if you have f_1 and f_2 where f_1 f_2 = 7, with non-unit f_1 and f_2, where
(5a_1(x)/f_1 + f_2)(5b_2(x)/f_2 + 2/f_2) = 25x^2 + 30x + 2
then the algebra tells you that the factors of 2 on the right are f_2 and 2/f_2, and if f_2 is a non-unit factor of 7, then you're not in the ring of algebraic integers.
Now then, you have a choice: ignore basic algebra assuming you can keep this fight up indefinitely for social reasons, or tell the truth.
It seems to me that some of you don't realize that even if you succeed here for a few days, or a few years, or even decades, some day the truth comes out.
It doesn't matter if you're dead. I've made certain that your name will live in infamy if it's known at all: Wiles, Ribet, Granville, or anyone else from this generation.
It doesn't matter, as you're all in it together.
It's kind of funny really. Sure Wiles can appreciate whatever approbation he gets today, but in the future, he's just another loser.
I'm the winner. It's about the legacy people. I'm destroying yours now and guaranteeing mine.
Pump each other up now. Tell yourselves how great you are. Congratulate yourself on your supposed accomplishments while you can.
History will hate you and love me.
I'm the misunderstood and persecuted genius.
You're the assholes.
# posted by James Harris @ 08:38 
