JSH: Dividing off 7
JSH
special example which shows how bizarre (insane) the ring of algebraic
integers is, but can it be divided off in a saner ring? YES!
Here's the example again:
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.
If you arbitrarily pick a_1(x) = 7b(x), you get:
7(175x^2 - 15x + 2) = (5(7)b_1(x) + 7)(5a_2(x)+ 7)
and you can trivially divide off the 7:
175x^2 - 15x + 2 = (5b_1(x) + 1)(5a_2(x)+ 7)
Easy as pie. Some may wonder about the apparent 7 still remaining,
especially as I talked of ghosts with 7 in a previous reply, but the
situation is easily figured out with x=0, as then:
175(0)^2 - 15(0) + 2 = (5b_1(0) + 1)(5a_2(0)+ 7)
which is
2 = (1)(-5 + 7) = (1)(2)
no problems and no 7. You can remove the final appearance that 7 is
still there with a_2(x) = b_2(x) - 1, giving:
175x^2 - 15x + 2 = (5b_1(x) + 1)(5b_2(x) + 2)
where you'll note the b's are now normalized, in that with x=0, b_1(0)
= b_2(0) = 0.
(Isn't that neat? Now it looks a LOT like x^2 + 3x + 2 = (x+1)(x+2)
and YES that was DELIBERATE! Isn't that clever?)
And now I can multiply back by 7 if I wish to get my original example,
or I can multiply 8, or 27, or any number I wish.
See? Constants multiplied ARE easy to handle, just like you've been
taught! So what's the big deal?
The big deal is that option is not available in the ring of algebraic
integers, as it says one of the roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0
has 7 as a factor, while the other does not.
But NEITHER of the roots have 7 as a factor, if that gives you a
quadratic with integer coefficients and the a's are non-rational, in
the ring of algebraic integers! The ring is insane!!!
Example: with x=1, you have a^2 - 6a + 35 = 0, so
a_1 = (6 +/- sqrt(36 - 4(35)))/2, a_2 = (6 -/+ sqrt(36 - 4(35)))/2
So you have simply enough a_1 = 3 +/- sqrt(-26), and a_2 = 3 -/+ sqrt
(-26) where I arbitrarily picked that order.
And I don't see that either of them has 7 as a factor, but you have
mathematical proof that one of them does, but not in the ring of
algebraic integers! So what gives?
The mathematics is telling you that there can be situations where with
combinations of radicals and integers, you can have an integer factor
that is not visible to the naked eye, but provably there or
mathematics contradicts!
Yup. Without that conclusion you run into mathematical contradiction.
So in a sense the people who keep arguing with me are arguing for the
death of mathematics, by claiming it is fundamentally illogical.
They are then: anti-mathematicians
Yes, they can call themselves mathematicians. They can be called
mathematicians by the world, but if you take a position that requires
that mathematics itself be illogical, then you cannot in truth BE a
mathematician.
Go back over it. Go back over everything in this post as much as you
need. And realize the logical necessity or find an error in the
reasoning.
I constructed a powerful little example that blows apart some human
social structures. It reveals that some pretenders are claiming to be
mathematicians when they are not. That's all. It's not complicated.
These things happen. Otherwise, human history would be rather boring.
James Harris
Mark Murray
> Ok, I've gone on and on about a problem with dividing off 7 with a
Blah blah blah
Your denial is beautiful to behold, particularly in the context of
your own essays on the subject.
M
Bacle
And you could not be any more boring. One-trick pony: boohoo, the ring of integers, boohoo, there is a plot against me.
Take a nickel and go buy yourself a life; it will certainly afford you more than you have now, and becoming a loser will be a step up in your case.
>
>
> James Harris
>
>
>