Tuesday, December 01, 2009


JSH: Dividing off 7

Ok, I've gone on and on about a problem with dividing off 7 with a 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.

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