Harristotelian Logic

I am going to rather easily explain how you can show a remarkable but devastating problem with the coverage of the ring of algebraic integers, but what makes the interesting twist here is that I didn't think of the base equations for this approach, or find a crucial infinite series as a poster who goes by Gene Ward Smith did:

"Consider the polynomial x^2 - (3+t)x + 2. The two roots of this can be expanded in series as

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

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

If t is 2-adically a nonunit, so that |t|_2 < 1, then these converge in the 2-adic integers. One root, r2, is even, the other, r1, is odd. The 2 does not split up. If t is a rational integer, this means that when t is even, the roots split into an even root and an odd root 2-adically."

He goes on about adics, but in another post he said he just got the series out of Maple.

Now there's the issue of convergence of the series, but before getting into lots of technical stuff, notice that with his r2(t) in EACH TERM you can clearly see 2 is a factor.

I'll assume that goes out to infinity, which I'm sure is easily verifiable.

Now then, if t is an algebraic integer, and the series converges, then necessarily it converges to the roots of x^2 - (3+t)x + 2. Some of you may be confused on this convergence thing, but the gist of it is that yes, you can expand out a solution with an infinite series and it only apply in certain ranges, as otherwise the series blows up, as in goes infinite.

Here you have

x^2 - (3+t)x + 2 = (x - r1(t))*(x - r2(t))

with

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

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

when t is in a range where the infinite series do not blow up.

So, like, you know that r1(t)*r2(t) = 2.

And it's easy to see that r2(t) has 2 as a factor in EACH TERM so it makes sense that r1(t) will give a unit, when you have t an algebraic integer for which the series is valid.

And in fact, you can see that at the trivial t=0.

So now if you wish to disbelieve the result you have to believe that with every term multiplied by 2, the sum does not have 2 as a factor—when the actuality is that it may not in the ring of algebraic integers.

What's interesting here is that you have

r1(t) = 3+ t - r2(t)

so the infinite series is the same for each function, BUT r1(t) and r2(t) must be algebraic integers if t is an algebraic integer because they are roots of a monic polynomial with algebraic integer coefficients.

So the infinite series if it converges will converge to algebraic integers.

Oh yeah, you can get answers easily enough since

x^2 - (3+t)x + 2 = (x - r1(t))*(x - r2(t))

with

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

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

when t is in a range where the infinite series do not blow up.

as you can just plug in some t, like t=sqrt(5) - 2, and get to a monic polynomial with integer coefficients to verify for yourself that you have algebraic integers. Proving the series then converges is a different matter, but that can be done as well.

To ignore this result and continue with what you think you knew about mathematics you have to accept that 2 can be a factor of EVERY TERM but somehow, mysteriously and against any logical position, just fade away as a factor of the full sum just because it might not be in the ring of algebraic integers.

Note: Standard teaching is that BOTH r1(t) and r2(t) would have non-unit factors of 2 because they would in the ring of algebraic integers if 2 is not a factor of r2(t) in that ring.

But despite each term having 2 as a factor in the ring of algebraic integers, the sum may not have 2 as a factor—because that sum may not be able to be the root of a monic polynomial with integer coefficients!!!

The rule limits whether or not you get an algebraic integer, but notice algebraically it has no impact, so with each term having 2 as a factor, the sum must have 2 as a factor, in a ring that algebraically makes sense and does not have the coverage problem.

Now I have proven a problem with the ring of algebraic integers multiple ways over the years since I found it, and even had a paper published, but mathematicians have decided to go against proof in this area because of the fallout from the coverage problem.

Part of my point though to undergrads is that the graduate students and professors who grew up with the coverage problem and built their careers on flawed foundations are screwed.

You are not.

But they are not going to help you as they ARE screwed.

So math undergrads can go along if they like to protect people who thought they were brilliant mathematicians but missed an error that destroys their "proofs" and maybe their careers, but why?

Why keep working hard at your studies to learn crap? Why fight for knowledge that doesn't work? Why take tests to show your mastery of wrong ideas?

And make no mistake, I've proven this problem multiple ways and even got a paper published—before some sci.math'ers with Ph.D's who are screwed by this problem emailed the editors attacking the paper and managed to get it yanked.

They are SCREWED. This error shatters their illusions of mastery in mathematics, so they fight it as if they admit the truth then they are ordinary people, and not brilliant mathematicians.

So they will sell out the undergrads who have a chance for a fresh start, to bring you in, so that when you have spent years invested in bad ideas, you will fight for them too, and humanity loses out as mathematicians turn against mathematical proof, while hiding the truth from everyone.