Harristotelian Logic

I noted that a poster going by the name Gene Ward Smith claimed that

r1(t) = 1 - t + 2t^2 - 6t^3 + 22t^4 - 90t^5 + …

r2(t) = 2 + 2t - 2t^2 + 6t^3 - 22t^4 + 90t^5 - …

are the two roots of x^2 - (3+t)x + 2, and I've explained how you can easily use that to show the coverage problem I talk about—if he's right.

But you can make a fairly easy argument regardless of whether or not that series is correct, as consider the solution when t=0, as then the roots are definitely 1 and 2.

So you should have something like those two series no matter what, as everything else has to go away when t=0.

Given that the roots of

x^2 - (3+t)x + 2 = 0

must be algebraic integers if t is an algebraic integer, then you can use the fact that

r_1 + r_2 = 3+t

and use something like t = sqrt(6) - 2, which forces r_1 and r_2 to be coprime to each other.

Since t has sqrt(2) itself as a factor then, and given the earlier situation where I noted that every term must have t itself as a factor in any expansion because when t=0, you just have 1 and 2, it's then trivial that one root will then not have 2 as a factor.

But now you just substitute

x^2 - (3 + sqrt(6) - 2)x + 2 = 0

and get to a polynomial with rational coefficients, and see if it's reducible over Q.

I think the problem here is partly that many number theorists lack the proper intuition in this area to realize that there was something wacky about how Galois Theory supposedly worked, as why would there be this weirdness only for non-rationals?

But worse than lacking the proper intuition modern number theorists clearly have a lot of political know-how, as I've been blocked repeatedly, where even publication hasn't mattered, while they've kept teaching the wrong information, protecting their careers against the interests of humanity.

After all, mathematics is an important subject. The lack of linking between "pure math" and physics in most areas is probably a result of the "pure math" being wrong.

So there was never ever going to be a time where the ideas using ideal theory would be practical as they were crap, but with non-experts being able to masquerade as experts and block people with mathematical know-how, there has been no way to get the truth out.

So the real story is bizarrely the opposite as I'm not the crackpot here—modern number theorists are.

I am the actual number theory expert trying to handle a situation where the crackpots took over a field.

[A reply to someone who told James to prove that modern number theorists are crackpots.]

Easily done. The polynomial again is

x2 - (3+t)x + 2 = 0

where when t is an algebraic integer both roots must be algebraic integers, and also when t = 0, the roots are 1 and 2.

So express the roots r_1 and r_2 as

r_1 = 1 + t^k*f_1(t)

r_2 = 2 + t^k*f_2(t)

where k is some non-zero rational number, and f_1(t) and f_2(t) are algebraic integer functions of t.

Notice then that

r_1 + r_2 = 3+t = 3 + t^k*(f_1(t) + f_2(t)

so

t^(1-k) = f_1(t) + f_2(t)

so abs(k) is less than or equal to1.

Multiplying the root together you have

r_1 * r_2 = 2 = 2 + t^k*f_2(t) + 2*t^k*f_1(t) + t^{2k}*f_1(t)*f_2(t)

so

t^k*f_2(t) + 2*t^k*f_1(t) + t^{2k}*f_1(t)*f_2(t) = 0

so

f_2(t) + 2*f_1(t) + t^k*f_1(t)*f_2(t) = 0

which as a sidenote tells you that f_1(t) must give unit results.

But, of course, it can't in the ring of algebraic integers.

Notice then the coverage problem is trivially shown by using a t that has 2 as a factor:

r_1 = 1 + t^k*f_1(t)

r_2 = 2 + t^k*f_2(t)

because then one root must have 2 as a factor while the other then is a unit.

But now I can let t just equal 2, and note that

x2 - 5x + 2 = 0

and you have this interesting result because then

r_1 = 1 + 2^k*f_1(t)

r_2 = 2 + 2^k*f_2(t)

as one root has 2 as a factor, but provably does NOT have 2 as a factor in the ring of algebraic integers.

Readers wishing to attack the result run into the problem of what happens when t=0, as how else can you get the zeroing out of all the extra terms unless you have t^k as a factor with k a non-rational number?

I like this way of explaining as then I don't have to worry about whether a particular infinite series converges or not, as the complexity is wrapped up in the function f_1(t) and f_2(t).

Easy proof. Trouble is, it kills ideal theory.

Mathematicians though who refuse mathematical proof are not actually mathematicians but must then be crackpots!

This result in and of itself does not give crackpot status to posters arguing with me, but refusing to acknowledge mathematical proof does.