Sunday, April 15, 2007
JSH: Newsgroups mistakes, updated explanation and proof
One of the more horrible additions to mathematical history will be the role of the sci.math newsgroup in acting to block the world's knowledge of important new mathematical research, where I post now to point out a simplified mathematical argument that proves my point about a problem with the ring of algebraic integers in a way that removes the ability of others to make seemingly intelligent objections.
I start with
2975x^2 - 15x + 2 = 2(f(x) + 1)*(g(x) + 1)
where x is a non-zero algebraic integer and f(x) and g(x) are functions of x determined by the following derivations. I've done something like this before in posts, so it's important to note upfront that these demonstrations are different in that I do two versus one.
Multiplying through the first factor by 2 and multiplying both sides by 7, picked because it is a prime factor of 2975, as 2975 = 7(17)(25) I have
7(2975x^2 - 15x + 2) = 7*(2f(x) + 2)*(g(x) + 1)
now re-order in a special way on the left side and now pick one way to multiply by 7 on the right:
7(7(17)(5^2)x^2 - (3)(5)x + 2) = (2f(x) + 2)*(7g(x) + 7)
Now let
2f(x) = 5a_1(x) + 5
and
7g(x) = 5a_2(x)
so I have:
(7(7(17)(5^2)x^2 - (3)(5)x + 2) = (5a_1(x) + 7)*(5a_2(x) + 7)
Now multiply through by 7 on the left, add and subtract 35 and introduce the polynomial Q(x), where I add and subtract it so everything still balances, and now I have
(49(17)x^2+7(17)Q(x))5^2 - (3x+1+5(17)Q(x))(5)(7) + 7^2 = (5a_1(x)+7)*(5a_2(x)+7)
where now a general solution can be seen by multiplying out on the right side as that gives
a_1(x)*a_2(x) = 49(17)x^2 + 7(17)Q(x)
and
a_1(x) + a_2(x) = -(3x + 1 + 5(17)Q(x))
which determines the following of which the a's must be roots:
a^2 + (3x + 1 + 5(17)Q(x))a + (17)(49x^2 + 7Q(x)) = 0
Then with Q(x) = -2x, the a's are roots of
a^2 - (168x-1)a + (17)(49x^2 - 14x) = 0
And letting x=1 gives as a solution:
a^2 - 167a + 595 = 0
where I say the simple argument above shows that ONLY ONE ROOT can have 7 as a factor, where of course in the ring of algebraic integers neither has 7 itself as a factor, and objectors have pointed at solving
7g(x) = 5a_2(x)
to get
g(x) = 5a_2(x)/7
and claim that the 7 does not divide through a_2(x), so now let's consider the second derivation.
Starting yet again with
2975x^2 - 15x + 2 = 2(f(x) + 1)*(g(x) + 1)
Still multiplying through the first factor by 2, but this time multiplying both sides by 17:
17(2975x^2 - 15x + 2) = 17*(2f(x) + 2)*(g(x) + 1)
and again re-order in a special way on the left side and now pick one way to multiply by 17 on the right:
17(7(17)(5^2)x^2 - (3)(5)x + 2) = (2f(x) + 2)*(17g(x) + 17)
Now let
2f(x) = 5b_1(x) + 15
and
17g(x) = 5b_2(x)
so I have:
(17(7(17)(5^2)x^2 - (3)(5)x + 2) = (5b_1(x) + 17)*(5b_2(x) + 17)
Now multiply through by 17 on the left, add and subtract 255 and introduce the polynomial R(x) so I have
(289(7)x^2+119R(x))5^2-(3x+3+5(7)R(x))(5)(17)+17^2= (5b_1(x)+17)*(5b_2(x)+17)
where now a general solution can be seen by multiplying out on the right side as that gives
b_1(x)*b_2(x) = (7)(289x^2 + 17R(x)
and
b_1(x) + b_2(x) = -(3x + 3 + 5(7)R(x))
which determines the following of which the b's must be roots:
b^2 + (3x + 3 + 5(7)R(x))b + ((7)(289x^2 + 17R(x)) = 0
but now notice I have 17g(x) = 5b_2(x), so
g(x) = 5b_2(x)/17
and the prior objection MUST go away as 17 was not in the denominator with the first derivation as instead you had 7.
Where of course I started EACH DERIVATION with
2975x^2 - 15x + 2 = 2(f(x) + 1)*(g(x) + 1)
so, you have the SAME g(x) in BOTH DERIVATIONS.
Readers should note that you can solve for a_2(x) and b_2(x), as the first is a root of
a^2 + (3x + 1 + 5(17)Q(x))a + ((17)(49x^2 + 7Q(x)) = 0
while the second is a root of
b^2 + (3x + 3 + 5(7)R(x))b + ((7)(289x^2 + 17R(x)) = 0
and solving the first using the quadratic formula gives
a_2(x) = (-(3x + 1 + 5(17)Q(x)) +/- sqrt((3x + 1 + 5(17)Q(x))^2 - 4((17)(49x^2 + 7Q(x)) )/2
so g(x) = 5a_2(x)/7 means that
g(x) = 5(-(3x + 1 + 5(17)Q(x)) +/- sqrt((3x + 1 + 5(17)Q(x))^2 - 4((17)(49x^2 + 7Q(x)) )/14
but from the SECOND solution, you have g(x) = 5b_2(x)/17, so
g(x) = 5(-(3x + 3 + 5(7)R(x))) +/- sqrt((3x + 3 + 5(7)R(x)))^2 - 4((7)(289x^2 + 17R(x)) )/34
so with one case you have 7 as a factor in the denominator and with the other you have 17, so 7 must divide though with the first and 17 must divide through with the second for at least one solution.
Some might wonder about Q(x) and R(x), but those are polynomials which you select, as remember above I selected Q(x) = -2x to show that only one root of
a^2 - 167a + 595 = 0
has 7 as a factor. And importantly note that if x is an algebraic integer and Q(x) is as well then
a_2(x) = (-(3x + 1 + 5(17)Q(x)) +/- sqrt((3x + 1 + 5(17)Q(x))^2 - 4((17)(49x^2 + 7Q(x)) )/2
shows you that the a's are algebraic integers as that is the solution for both a's while I put a_2(x) as I was focusing on it.
So if you try to fudge in factors of 17 in the denominator with g(x), you run into the wall of
g(x) = 5a_2(x)/7
as there is just the 7, and also you can't say that factors of 7 are in the denominator because of
g(x) = 5b_2(x)/17
so the simply brilliant idea I had to remove even the semblance of an intelligent objection to this approach was to use TWO PRIME NUMBERS in concert with each other.
I will note though that mathematicians have steadfastly refused to accept what is mathematically true in this area for years now, creating the necessity for clever explanations to get around emphatic denials of what is mathematically correct.
The reason for those denials is that now you know that a lot of Galois Theory is not quite right, and you know that the ring of algebraic integers is used in a flawed way, as in that ring NEITHER of the roots of
a^2 - 167a + 595 = 0
can have 7 as a factor!!!
This error has been explained by me many times, and the refusal of the mathematical community to acknowledge it is best explained by the importance of the wrong math for research done by quite a few mathematicians who care more about their careers and belief that they are right than on what is mathematically true.
And they are willing to teach error to trusting students coming to them for a mathematical education.
That must stop. These students are no longer to be victims.
Remember at this point the error is willful. And it betrays a hatred of mathematical truth.
Mathematicians who tolerate this error and teach it are like arsonists who become firefighters, not because they actually hate error, but because they love it, and hate mathematical proof.
[A reply to someone who explained James that there is more that one pair of functions f(x) and g(x) that satisfy his equation.]
You are correct as I don't necessarily have the same g(x) in both cases.
Excellent reply.
Thinking it over I realized that Q(x) and R(x) are the functions that handle the full space of possible solutions for f(x) and g(x), and what I want is for Q(x) and R(x) to give the SAME quadratic defining the a's and b's, with an integer x, where both are integer functions in that they themselves give an integer value.
That is, I want the solution for the a's and the b's to be the same, with all integers.
Turns out that because I have two equations with three degrees of freedom since I have x, Q(x) and R(x), I can solve out and find congruence relationships that should define values that will work!!!
The nice thing about that is that then you can also directly see it work by plugging in actual values, noting that you have the same solutions for the a's and b's and therefore for g(x), which completes the demonstration I need.
I am going now to update the paper, but any of you who are serious mathematicians can step through the exercise on your own as it should be truly sublime mathematics.
It is just a remarkable proof, and it leave no shadow of doubt.
Have fun. I'll be checking in later to see if you any of you bothered.
[A reply to someone who told James not to worry because this time he would be right.]
William Hughes gave a cogent objection that was RIGHT ON TARGET.
And in doing so he showed at least some mathematical sophistication.
I gave a direct reply acknowledging the issue and giving what I thought was a resolution.
So what's your problem?
As an update I went ahead to try that approach and found it didn't work!!!
Turns out that I can't force the a's and b's to be equal, but I CAN force them to be equal by absolute value, so it's the same difference.
To me hanger's on like you are just such a waste of the time of people who actually care about doing some mathematics.
I may be loud and obnoxious often, and get things wrong a lot, but I am working, and it takes me years at times to figure things out, but I keep at it until I do.
It's done people. The direct demonstration now allows one to use numerical methods to prove that with the defined congruences g(x) is the SAME by absolute value for both the a's and b's though they are opposite in sign, which is why I talk about the absolute value.
That is, I figured out a route where you get g(x) and g'(x) where g(x) = -g'(x), and one is given by the a's and the other by the b's and the sign doesn't matter, but I had to do that as the other way didn't work.
If you figure it out you will be floored. I mean freaking floored as it is just so beautiful.
Worth all the crap. Definitely worth all of it.
One of the most beautiful proofs I've ever seen or imagined.
Maybe life does have meaning after all. Such a beautiful, beautiful, beautiful proof.
I am simply floored.
[A reply to someone who explained James that he is repeatedly rude, obnoxious, abusive, and that he believes that others have to accept that.]
It's not about abuse.
Figuring out mathematics can just simply be hard beyond belief.
It stresses people out. Pushes them outside of normal limits.
And most importantly, it makes you realize just how small you are.
Just to keep your head above water, yes, you may come over the top, and sound very arrogant and full of yourself, but that's mostly bravado.
Mathematics is so much bigger than any of us.
It is overwhelming and if you ever really get a good feel for just how big it is, and accept that, then you just might stop, and never try to figure anything out.
I do not fully understand why I do what I do.
But this latest proof gives me a greater sense of so much that is so far beyond me.
I don't know what people like you think is so important about other people liking you, or telling you that you are whatever it is that you think they should tell you.
But what people tell me means little to me.
What I can prove means so much more to me.
And what I can prove takes me so far beyond what you can get from people liking you, or telling you whatever it is that is so important for you to hear.
Mathematical proof is certainty. And it's knowing the truth without wondering about motivations, or wondering what's in it for some other person, or wondering why this or that person likes you or doesn't like you.
Mathematical proof is truth.
You will die someday, as will I. And when you die, will you be running through your mind all the good things people said about you? Or worrying about all the nasty things?
Either way, will it matter? When you die, you'll still be dead.
You will die and when you die what will any of this matter?
You will still be dead and everyone you know now will die, and everything you build will be torn down.
I know you will die. I know that no matter what you say now billions of years will pass and none of that will be around to matter any more.
Nothing any of you say will still be around billions of years from now.
But this proof that I discovered will still be true.
And as I comprehend absolutes, I understand certainty, and appreciate that which will still be here long after we are all dead.
Take your social nothings with you to your grave which will come soon enough.
You will die. Everyone you know now will die. Everything you build will be torn down.
What do you think is so permanent about your silly social crap?
I take mathematical proof.
As with mathematical proof I hear what God hears, and understand just a little bit, as much as I can, and that is the best that I can do.
[A reply to someone who wrote James usually starts his arguments with statements that are true.]
I have a record of actually trying versus acting like I am trying.
One of the more profound things I realized some years ago was that sci.math'ers were deliberately acting to confuse versus caring about what was mathematically true and that happened when I wrote a very detailed rebuttal to "Nora Baron" only to have that poster delete out everything and just repeat flawed claims.
It's easier to claim you are honest than to actually be honest.
I screw up. I make lots of mistakes in my search for truth and quite a few times over the years I have been rude, very wrong, and regretted things I've posted.
But I at least admit my humanity.
And in the world I thought I knew when I was growing up, human beings do those things and then some, because they work harder at being right, than just trying to look right.
If mathematicians proved me wrong I'd accept that because I know fantasy, and I prefer knowing truth.
But to the extent that they claim they prefer truth and rip on me as if the truth were the opposite, when I admit to being a very flawed human being just trying to get some things straight while these monsters pretend to be perfect—or very damn close to perfect—while they lie, and lie, and lie then I say we have a serious problem people.
The monsters have taken the high ground, and made it illegal to be human.
They spit on mistakes made in the search for truth.
And they piss on failures made in the pursuit of truth, to tell you lies, when they do not actually try to know what is mathematically correct—preferring only to rely on what they can convince you is correct.
I fail. And I cheer my ability to not only fail in trying, but to admit that I fail, and I fail, and I fail.
And through failing honestly, I find success, honestly.
While the demons tell you lies.
I start with
2975x^2 - 15x + 2 = 2(f(x) + 1)*(g(x) + 1)
where x is a non-zero algebraic integer and f(x) and g(x) are functions of x determined by the following derivations. I've done something like this before in posts, so it's important to note upfront that these demonstrations are different in that I do two versus one.
Multiplying through the first factor by 2 and multiplying both sides by 7, picked because it is a prime factor of 2975, as 2975 = 7(17)(25) I have
7(2975x^2 - 15x + 2) = 7*(2f(x) + 2)*(g(x) + 1)
now re-order in a special way on the left side and now pick one way to multiply by 7 on the right:
7(7(17)(5^2)x^2 - (3)(5)x + 2) = (2f(x) + 2)*(7g(x) + 7)
Now let
2f(x) = 5a_1(x) + 5
and
7g(x) = 5a_2(x)
so I have:
(7(7(17)(5^2)x^2 - (3)(5)x + 2) = (5a_1(x) + 7)*(5a_2(x) + 7)
Now multiply through by 7 on the left, add and subtract 35 and introduce the polynomial Q(x), where I add and subtract it so everything still balances, and now I have
(49(17)x^2+7(17)Q(x))5^2 - (3x+1+5(17)Q(x))(5)(7) + 7^2 = (5a_1(x)+7)*(5a_2(x)+7)
where now a general solution can be seen by multiplying out on the right side as that gives
a_1(x)*a_2(x) = 49(17)x^2 + 7(17)Q(x)
and
a_1(x) + a_2(x) = -(3x + 1 + 5(17)Q(x))
which determines the following of which the a's must be roots:
a^2 + (3x + 1 + 5(17)Q(x))a + (17)(49x^2 + 7Q(x)) = 0
Then with Q(x) = -2x, the a's are roots of
a^2 - (168x-1)a + (17)(49x^2 - 14x) = 0
And letting x=1 gives as a solution:
a^2 - 167a + 595 = 0
where I say the simple argument above shows that ONLY ONE ROOT can have 7 as a factor, where of course in the ring of algebraic integers neither has 7 itself as a factor, and objectors have pointed at solving
7g(x) = 5a_2(x)
to get
g(x) = 5a_2(x)/7
and claim that the 7 does not divide through a_2(x), so now let's consider the second derivation.
Starting yet again with
2975x^2 - 15x + 2 = 2(f(x) + 1)*(g(x) + 1)
Still multiplying through the first factor by 2, but this time multiplying both sides by 17:
17(2975x^2 - 15x + 2) = 17*(2f(x) + 2)*(g(x) + 1)
and again re-order in a special way on the left side and now pick one way to multiply by 17 on the right:
17(7(17)(5^2)x^2 - (3)(5)x + 2) = (2f(x) + 2)*(17g(x) + 17)
Now let
2f(x) = 5b_1(x) + 15
and
17g(x) = 5b_2(x)
so I have:
(17(7(17)(5^2)x^2 - (3)(5)x + 2) = (5b_1(x) + 17)*(5b_2(x) + 17)
Now multiply through by 17 on the left, add and subtract 255 and introduce the polynomial R(x) so I have
(289(7)x^2+119R(x))5^2-(3x+3+5(7)R(x))(5)(17)+17^2= (5b_1(x)+17)*(5b_2(x)+17)
where now a general solution can be seen by multiplying out on the right side as that gives
b_1(x)*b_2(x) = (7)(289x^2 + 17R(x)
and
b_1(x) + b_2(x) = -(3x + 3 + 5(7)R(x))
which determines the following of which the b's must be roots:
b^2 + (3x + 3 + 5(7)R(x))b + ((7)(289x^2 + 17R(x)) = 0
but now notice I have 17g(x) = 5b_2(x), so
g(x) = 5b_2(x)/17
and the prior objection MUST go away as 17 was not in the denominator with the first derivation as instead you had 7.
Where of course I started EACH DERIVATION with
2975x^2 - 15x + 2 = 2(f(x) + 1)*(g(x) + 1)
so, you have the SAME g(x) in BOTH DERIVATIONS.
Readers should note that you can solve for a_2(x) and b_2(x), as the first is a root of
a^2 + (3x + 1 + 5(17)Q(x))a + ((17)(49x^2 + 7Q(x)) = 0
while the second is a root of
b^2 + (3x + 3 + 5(7)R(x))b + ((7)(289x^2 + 17R(x)) = 0
and solving the first using the quadratic formula gives
a_2(x) = (-(3x + 1 + 5(17)Q(x)) +/- sqrt((3x + 1 + 5(17)Q(x))^2 - 4((17)(49x^2 + 7Q(x)) )/2
so g(x) = 5a_2(x)/7 means that
g(x) = 5(-(3x + 1 + 5(17)Q(x)) +/- sqrt((3x + 1 + 5(17)Q(x))^2 - 4((17)(49x^2 + 7Q(x)) )/14
but from the SECOND solution, you have g(x) = 5b_2(x)/17, so
g(x) = 5(-(3x + 3 + 5(7)R(x))) +/- sqrt((3x + 3 + 5(7)R(x)))^2 - 4((7)(289x^2 + 17R(x)) )/34
so with one case you have 7 as a factor in the denominator and with the other you have 17, so 7 must divide though with the first and 17 must divide through with the second for at least one solution.
Some might wonder about Q(x) and R(x), but those are polynomials which you select, as remember above I selected Q(x) = -2x to show that only one root of
a^2 - 167a + 595 = 0
has 7 as a factor. And importantly note that if x is an algebraic integer and Q(x) is as well then
a_2(x) = (-(3x + 1 + 5(17)Q(x)) +/- sqrt((3x + 1 + 5(17)Q(x))^2 - 4((17)(49x^2 + 7Q(x)) )/2
shows you that the a's are algebraic integers as that is the solution for both a's while I put a_2(x) as I was focusing on it.
So if you try to fudge in factors of 17 in the denominator with g(x), you run into the wall of
g(x) = 5a_2(x)/7
as there is just the 7, and also you can't say that factors of 7 are in the denominator because of
g(x) = 5b_2(x)/17
so the simply brilliant idea I had to remove even the semblance of an intelligent objection to this approach was to use TWO PRIME NUMBERS in concert with each other.
I will note though that mathematicians have steadfastly refused to accept what is mathematically true in this area for years now, creating the necessity for clever explanations to get around emphatic denials of what is mathematically correct.
The reason for those denials is that now you know that a lot of Galois Theory is not quite right, and you know that the ring of algebraic integers is used in a flawed way, as in that ring NEITHER of the roots of
a^2 - 167a + 595 = 0
can have 7 as a factor!!!
This error has been explained by me many times, and the refusal of the mathematical community to acknowledge it is best explained by the importance of the wrong math for research done by quite a few mathematicians who care more about their careers and belief that they are right than on what is mathematically true.
And they are willing to teach error to trusting students coming to them for a mathematical education.
That must stop. These students are no longer to be victims.
Remember at this point the error is willful. And it betrays a hatred of mathematical truth.
Mathematicians who tolerate this error and teach it are like arsonists who become firefighters, not because they actually hate error, but because they love it, and hate mathematical proof.
[A reply to someone who explained James that there is more that one pair of functions f(x) and g(x) that satisfy his equation.]
You are correct as I don't necessarily have the same g(x) in both cases.
Excellent reply.
Thinking it over I realized that Q(x) and R(x) are the functions that handle the full space of possible solutions for f(x) and g(x), and what I want is for Q(x) and R(x) to give the SAME quadratic defining the a's and b's, with an integer x, where both are integer functions in that they themselves give an integer value.
That is, I want the solution for the a's and the b's to be the same, with all integers.
Turns out that because I have two equations with three degrees of freedom since I have x, Q(x) and R(x), I can solve out and find congruence relationships that should define values that will work!!!
The nice thing about that is that then you can also directly see it work by plugging in actual values, noting that you have the same solutions for the a's and b's and therefore for g(x), which completes the demonstration I need.
I am going now to update the paper, but any of you who are serious mathematicians can step through the exercise on your own as it should be truly sublime mathematics.
It is just a remarkable proof, and it leave no shadow of doubt.
Have fun. I'll be checking in later to see if you any of you bothered.
[A reply to someone who told James not to worry because this time he would be right.]
William Hughes gave a cogent objection that was RIGHT ON TARGET.
And in doing so he showed at least some mathematical sophistication.
I gave a direct reply acknowledging the issue and giving what I thought was a resolution.
So what's your problem?
As an update I went ahead to try that approach and found it didn't work!!!
Turns out that I can't force the a's and b's to be equal, but I CAN force them to be equal by absolute value, so it's the same difference.
To me hanger's on like you are just such a waste of the time of people who actually care about doing some mathematics.
I may be loud and obnoxious often, and get things wrong a lot, but I am working, and it takes me years at times to figure things out, but I keep at it until I do.
It's done people. The direct demonstration now allows one to use numerical methods to prove that with the defined congruences g(x) is the SAME by absolute value for both the a's and b's though they are opposite in sign, which is why I talk about the absolute value.
That is, I figured out a route where you get g(x) and g'(x) where g(x) = -g'(x), and one is given by the a's and the other by the b's and the sign doesn't matter, but I had to do that as the other way didn't work.
If you figure it out you will be floored. I mean freaking floored as it is just so beautiful.
Worth all the crap. Definitely worth all of it.
One of the most beautiful proofs I've ever seen or imagined.
Maybe life does have meaning after all. Such a beautiful, beautiful, beautiful proof.
I am simply floored.
[A reply to someone who explained James that he is repeatedly rude, obnoxious, abusive, and that he believes that others have to accept that.]
It's not about abuse.
Figuring out mathematics can just simply be hard beyond belief.
It stresses people out. Pushes them outside of normal limits.
And most importantly, it makes you realize just how small you are.
Just to keep your head above water, yes, you may come over the top, and sound very arrogant and full of yourself, but that's mostly bravado.
Mathematics is so much bigger than any of us.
It is overwhelming and if you ever really get a good feel for just how big it is, and accept that, then you just might stop, and never try to figure anything out.
I do not fully understand why I do what I do.
But this latest proof gives me a greater sense of so much that is so far beyond me.
I don't know what people like you think is so important about other people liking you, or telling you that you are whatever it is that you think they should tell you.
But what people tell me means little to me.
What I can prove means so much more to me.
And what I can prove takes me so far beyond what you can get from people liking you, or telling you whatever it is that is so important for you to hear.
Mathematical proof is certainty. And it's knowing the truth without wondering about motivations, or wondering what's in it for some other person, or wondering why this or that person likes you or doesn't like you.
Mathematical proof is truth.
You will die someday, as will I. And when you die, will you be running through your mind all the good things people said about you? Or worrying about all the nasty things?
Either way, will it matter? When you die, you'll still be dead.
You will die and when you die what will any of this matter?
You will still be dead and everyone you know now will die, and everything you build will be torn down.
I know you will die. I know that no matter what you say now billions of years will pass and none of that will be around to matter any more.
Nothing any of you say will still be around billions of years from now.
But this proof that I discovered will still be true.
And as I comprehend absolutes, I understand certainty, and appreciate that which will still be here long after we are all dead.
Take your social nothings with you to your grave which will come soon enough.
You will die. Everyone you know now will die. Everything you build will be torn down.
What do you think is so permanent about your silly social crap?
I take mathematical proof.
As with mathematical proof I hear what God hears, and understand just a little bit, as much as I can, and that is the best that I can do.
[A reply to someone who wrote James usually starts his arguments with statements that are true.]
I have a record of actually trying versus acting like I am trying.
One of the more profound things I realized some years ago was that sci.math'ers were deliberately acting to confuse versus caring about what was mathematically true and that happened when I wrote a very detailed rebuttal to "Nora Baron" only to have that poster delete out everything and just repeat flawed claims.
It's easier to claim you are honest than to actually be honest.
I screw up. I make lots of mistakes in my search for truth and quite a few times over the years I have been rude, very wrong, and regretted things I've posted.
But I at least admit my humanity.
And in the world I thought I knew when I was growing up, human beings do those things and then some, because they work harder at being right, than just trying to look right.
If mathematicians proved me wrong I'd accept that because I know fantasy, and I prefer knowing truth.
But to the extent that they claim they prefer truth and rip on me as if the truth were the opposite, when I admit to being a very flawed human being just trying to get some things straight while these monsters pretend to be perfect—or very damn close to perfect—while they lie, and lie, and lie then I say we have a serious problem people.
The monsters have taken the high ground, and made it illegal to be human.
They spit on mistakes made in the search for truth.
And they piss on failures made in the pursuit of truth, to tell you lies, when they do not actually try to know what is mathematically correct—preferring only to rely on what they can convince you is correct.
I fail. And I cheer my ability to not only fail in trying, but to admit that I fail, and I fail, and I fail.
And through failing honestly, I find success, honestly.
While the demons tell you lies.
# posted by James Harris @ 11:22 
