Tuesday, December 19, 2006
JSH: Clarified disproof
Remarkably a simple refutation of claims against my research is easy enough to obtain, where I noticed that I could just directly give a solution that ends the debate. After brainstorming through it a bit I'll go ahead and give the clarified disproof.
Consider
175x^2 - 15x + 2 = 2(f(x) + 1)*(g(x) + 1)
in an integral domain, where f(x) and g(x) are functions I'll soon give.
With
a_1(x) = ((7x-1) +/- sqrt((7x-1)^2 - 4(49x^2 - 14x))/2
it can be shown that
2f(x) = 5a_1(x) + 5
so 2f(x) is an algebraic integer.
Solving then I have
2f(x) = 5((7x-1) +/- sqrt((7x-1)^2 - 4(49x^2 - 14x))/2 + 5
and it follows that
2g(x) = 5((7x-1) -/+ sqrt((7x-1)^2 - 4(49x^2 - 14x))/2 + 5
proving that both 2f(x) and 2g(x) are algebraic integers.
The solution for a_1(x) follows from it being a root of
a^2 - (7x-1)a + (49x^2 - 14x) = 0.
Using x=1, I have that the quadratic that follows is
a^2 - 6a + 35 = 0
which gives
a = 3 +/- sqrt(-26)
as the solutions that follow from the quadratic formula, where now it is clear that one solution has 7 as a factor, while the other does not, in the ring of algebraic integers.
But let a = 7z, and you have
49z^2 - 42a + 35 = 0
which is
7z^2 - 6a + 5 = 0
which is a non-monic primitive polynomial irreducible over Q, which has been shown to have on root that must be an algebraic integer.
This disproof of a previously widely held belief—if sci.math'ers are to be believed and you may notice I tested them on this all day—is remarkable for its conciseness and reliance on very elementary algebra.
Previous to this disproof, it seems some people wrongly believed that an algebraic integer could not be the root of a non-monic primitive polynomial irreducible over Q.
[A reply to someone who said that James should get a grip on his life an put it back on track.]
You people have gotten away way too long with babbling in posts versus doing mathematics.
You show no respect whatsoever for proof and instead are lead around like dumb cows by ANY reply from people like Rupert or Magidin without paying attention to their errors or avoidance of the obvious.
If Rupert posted 1+1=3, you'd probably be right there cheering him on talking about how great it was, and respond with derision if I pointed out that 1+1=2.
Oh and yes, in case you people are too dumb to get it, your behavior with absolute proof is useful for political operators who yes, do wish to know how to get away with not telling the truth or ignoring the obvious in their better understanding of how to manipulate large groups of supposedly highly intelligent people.
So the rest of your crime is in helping people like George W. Bush understand how to keep power no matter what the truth is.
And you people give everything away, including your credibility as mathematicians, for what?
Nothing. You're giving it away for absolutely nothing in return, which is the lesson for real political players.
The best way to really win is to make sure you give your patsies nothing in return, as rather than acknowledge they've been completely taken for nothing, they will make up their own rewards.
They'll do all the work to convince themselves they are getting something.
NONE of you noticed that Bush's rise has been coincident with my arguments with you knuckleheads?
[A reply to someone who told James that he and Bush “are both symptoms of the sorry state of the nation”.]
Bush is even brainstorming now, which at least is a good thing.
Some of you must have noticed similarities between what you could see on the newsgroup and political machinations different than at any other time in US history.
But of course you will simply explain it all away versus considering that by being recalcitrant and resistant to mathematical proof you provide a roadmap for how politicians can, well, rule the world.
[A reply of James to his own assertion that “both 2f(x) and 2g(x) are algebraic integers”.]
Well I've thought about that and it should be correct but some posters are saying it's not, so I'll switch from brainstorming to critiquing.
For those who don't know I practice what I call extreme mathematics, which involves a creative brainstorming phase followed by a critiquing phase.
During brainstorming ideas are tossed out with little consideration for correctness as idea generation is the aim.
Now then I have asserted that with
175x^2 - 15x + 2 = 2(f(x) + 1)*(g(x) + 1)
it works to use
2f(x) = 5((7x-1) +/- sqrt((7x-1)^2 - 4(49x^2 - 14x))/2 + 5
and
2g(x) = 5((7x-1) -/+ sqrt((7x-1)^2 - 4(49x^2 - 14x))/2 + 5
So I guess I'll multiply that out. To make it a little easier I'll shift to
350x^2 - 30x + 6 = (2f(x) + 2)*(2g(x) + 2)
so
350x^2 - 30x + 6 = (5((7x-1) +/- sqrt((7x-1)^2 - 4(49x^2 - 14x))/2 + 5 + 2)*(5((7x-1) -/+ sqrt((7x-1)^2 - 4(49x^2 - 14x))/2 + 5 + 2)
Which is
1400x^2 - 120x + 24 = (5((7x-1) +/- sqrt((7x-1)^2 - 4(49x^2 - 14x)) + 14)*(5((7x-1) -/+ sqrt((7x-1)^2 - 4(49x^2 - 14x))+14)
which is
1400x^2 - 120x + 24 = (35x + 9 +/- 5sqrt((7x-1)^2 - 4(49x^2 - 14x)))*(35x + 9 -/+ 5sqrt((7x-1)^2 - 4(49x^2 - 14x)))
which is
1400x^2 - 120x + 24 = ((35x + 9)^2 - 25((7x-1)^2 - 4(49x^2 - 14x)))
which is
1400x^2 - 120x + 24 = ((35x + 9)^2 - 25(49x^2 - 14x + 1 - 196x^2 + 56x))
which is
1400x^2 - 120x + 24 = ((35x + 9)^2 - 25(-147x^2 + 42x + 1))
which is
1400x^2 - 120x + 24 = ((35x + 9)^2 - 25(-147x^2 + 42x + 1))
and I can already see they don't match, but continuing on the right:
Q(x) = (1225x^2 + 630x + 81 + 3675x^2 - 1050x - 25)
so I finally have
Q(x) = 4900x^2 - 420x + 56 = 7(700x^2 - 60x + 8)
so that idea failed. One thing is interesting though, as the two solutions for f(x) plus one multiply together to give 7 itself as a factor.
So using f(x) and f*(x) as the conjugate, I have that with integer x,
(f(x)+1)*(f*(x)+1)
is an integer with 7 itself as a factor.
But the original idea failed so it's back to brainstorming!
Some may wonder about a delay in my checking ideas, which is about me letting other people check them so I save time. During brainstorming I do little if any checking but toss the ideas out there and often some other person will go and check for me.
[Another reply of James to himself, this time to his assertion that “350x^2 - 30x + 6 = (2f(x) + 2)*(2g(x) + 2)”.]
Oh, inexplicably at this point I put 6 instead of 4.
Fix that then there are only extra factors of 2, as 2f(x) + 2 is just 5a_1(x) + 7, so I just went in a Big Freaking Circle.
Oh well, at least now it all makes sense. So though
2f(x) = 5((7x-1) +/- sqrt((7x-1)^2 - 4(49x^2 - 14x))/2 + 5
I can't just say that 2g(x) is the conjugate, as that gives a wrong answer with the sides not matching up because of factors of 2.
But it's still true that g(x) and f(x) should be indistinguishable, unless I'm missing something.
More brainstorming needed!!!
Consider
175x^2 - 15x + 2 = 2(f(x) + 1)*(g(x) + 1)
in an integral domain, where f(x) and g(x) are functions I'll soon give.
With
a_1(x) = ((7x-1) +/- sqrt((7x-1)^2 - 4(49x^2 - 14x))/2
it can be shown that
2f(x) = 5a_1(x) + 5
so 2f(x) is an algebraic integer.
Solving then I have
2f(x) = 5((7x-1) +/- sqrt((7x-1)^2 - 4(49x^2 - 14x))/2 + 5
and it follows that
2g(x) = 5((7x-1) -/+ sqrt((7x-1)^2 - 4(49x^2 - 14x))/2 + 5
proving that both 2f(x) and 2g(x) are algebraic integers.
The solution for a_1(x) follows from it being a root of
a^2 - (7x-1)a + (49x^2 - 14x) = 0.
Using x=1, I have that the quadratic that follows is
a^2 - 6a + 35 = 0
which gives
a = 3 +/- sqrt(-26)
as the solutions that follow from the quadratic formula, where now it is clear that one solution has 7 as a factor, while the other does not, in the ring of algebraic integers.
But let a = 7z, and you have
49z^2 - 42a + 35 = 0
which is
7z^2 - 6a + 5 = 0
which is a non-monic primitive polynomial irreducible over Q, which has been shown to have on root that must be an algebraic integer.
This disproof of a previously widely held belief—if sci.math'ers are to be believed and you may notice I tested them on this all day—is remarkable for its conciseness and reliance on very elementary algebra.
Previous to this disproof, it seems some people wrongly believed that an algebraic integer could not be the root of a non-monic primitive polynomial irreducible over Q.
[A reply to someone who said that James should get a grip on his life an put it back on track.]
You people have gotten away way too long with babbling in posts versus doing mathematics.
You show no respect whatsoever for proof and instead are lead around like dumb cows by ANY reply from people like Rupert or Magidin without paying attention to their errors or avoidance of the obvious.
If Rupert posted 1+1=3, you'd probably be right there cheering him on talking about how great it was, and respond with derision if I pointed out that 1+1=2.
Oh and yes, in case you people are too dumb to get it, your behavior with absolute proof is useful for political operators who yes, do wish to know how to get away with not telling the truth or ignoring the obvious in their better understanding of how to manipulate large groups of supposedly highly intelligent people.
So the rest of your crime is in helping people like George W. Bush understand how to keep power no matter what the truth is.
And you people give everything away, including your credibility as mathematicians, for what?
Nothing. You're giving it away for absolutely nothing in return, which is the lesson for real political players.
The best way to really win is to make sure you give your patsies nothing in return, as rather than acknowledge they've been completely taken for nothing, they will make up their own rewards.
They'll do all the work to convince themselves they are getting something.
NONE of you noticed that Bush's rise has been coincident with my arguments with you knuckleheads?
[A reply to someone who told James that he and Bush “are both symptoms of the sorry state of the nation”.]
Bush is even brainstorming now, which at least is a good thing.
Some of you must have noticed similarities between what you could see on the newsgroup and political machinations different than at any other time in US history.
But of course you will simply explain it all away versus considering that by being recalcitrant and resistant to mathematical proof you provide a roadmap for how politicians can, well, rule the world.
[A reply of James to his own assertion that “both 2f(x) and 2g(x) are algebraic integers”.]
Well I've thought about that and it should be correct but some posters are saying it's not, so I'll switch from brainstorming to critiquing.
For those who don't know I practice what I call extreme mathematics, which involves a creative brainstorming phase followed by a critiquing phase.
During brainstorming ideas are tossed out with little consideration for correctness as idea generation is the aim.
Now then I have asserted that with
175x^2 - 15x + 2 = 2(f(x) + 1)*(g(x) + 1)
it works to use
2f(x) = 5((7x-1) +/- sqrt((7x-1)^2 - 4(49x^2 - 14x))/2 + 5
and
2g(x) = 5((7x-1) -/+ sqrt((7x-1)^2 - 4(49x^2 - 14x))/2 + 5
So I guess I'll multiply that out. To make it a little easier I'll shift to
350x^2 - 30x + 6 = (2f(x) + 2)*(2g(x) + 2)
so
350x^2 - 30x + 6 = (5((7x-1) +/- sqrt((7x-1)^2 - 4(49x^2 - 14x))/2 + 5 + 2)*(5((7x-1) -/+ sqrt((7x-1)^2 - 4(49x^2 - 14x))/2 + 5 + 2)
Which is
1400x^2 - 120x + 24 = (5((7x-1) +/- sqrt((7x-1)^2 - 4(49x^2 - 14x)) + 14)*(5((7x-1) -/+ sqrt((7x-1)^2 - 4(49x^2 - 14x))+14)
which is
1400x^2 - 120x + 24 = (35x + 9 +/- 5sqrt((7x-1)^2 - 4(49x^2 - 14x)))*(35x + 9 -/+ 5sqrt((7x-1)^2 - 4(49x^2 - 14x)))
which is
1400x^2 - 120x + 24 = ((35x + 9)^2 - 25((7x-1)^2 - 4(49x^2 - 14x)))
which is
1400x^2 - 120x + 24 = ((35x + 9)^2 - 25(49x^2 - 14x + 1 - 196x^2 + 56x))
which is
1400x^2 - 120x + 24 = ((35x + 9)^2 - 25(-147x^2 + 42x + 1))
which is
1400x^2 - 120x + 24 = ((35x + 9)^2 - 25(-147x^2 + 42x + 1))
and I can already see they don't match, but continuing on the right:
Q(x) = (1225x^2 + 630x + 81 + 3675x^2 - 1050x - 25)
so I finally have
Q(x) = 4900x^2 - 420x + 56 = 7(700x^2 - 60x + 8)
so that idea failed. One thing is interesting though, as the two solutions for f(x) plus one multiply together to give 7 itself as a factor.
So using f(x) and f*(x) as the conjugate, I have that with integer x,
(f(x)+1)*(f*(x)+1)
is an integer with 7 itself as a factor.
But the original idea failed so it's back to brainstorming!
Some may wonder about a delay in my checking ideas, which is about me letting other people check them so I save time. During brainstorming I do little if any checking but toss the ideas out there and often some other person will go and check for me.
[Another reply of James to himself, this time to his assertion that “350x^2 - 30x + 6 = (2f(x) + 2)*(2g(x) + 2)”.]
Oh, inexplicably at this point I put 6 instead of 4.
Fix that then there are only extra factors of 2, as 2f(x) + 2 is just 5a_1(x) + 7, so I just went in a Big Freaking Circle.
Oh well, at least now it all makes sense. So though
2f(x) = 5((7x-1) +/- sqrt((7x-1)^2 - 4(49x^2 - 14x))/2 + 5
I can't just say that 2g(x) is the conjugate, as that gives a wrong answer with the sides not matching up because of factors of 2.
But it's still true that g(x) and f(x) should be indistinguishable, unless I'm missing something.
More brainstorming needed!!!
# posted by James Harris @ 02:37 
