Which to me is rather fascinating. And you can check if you have knowledge of basic number theory by downloading a paper I got published where here is link to post that has a link to it, and have talked story often on this blog.

What's cool is of course it stands on its own but for the curious can explain LOTS about how that's possible and implications for modern mathematics.

What allows that actually is that the paper starts by declaring the ring to be the ring of algebraic integers. And the Wikipedia has an article on the algebraic integer for those who need a reference.

While I'm NOT a mathematician, I stumbled across a problem with the ring of algebraic integers, where after YEARS can actually show very simply and won't give all details here, but is so short can enough.

So consider

**P(x) = (g**

_{1}(x) + 1)(g_{2}(x) + 2)where P(x) is a primitive quadratic with integer coefficients, g

_{1}(0) = g

_{2}(0) = 0, but g

_{1}(x) does not equal 0 for all x.

Of course a simple

*reducible*quadratic with those requirements is:

**x**

^{2}+ 3x + 2 = (x+1)(x+2)And I can force symmetry by introducing k, where k is a nonzero integer, and not 1 or -1, and new functions f

_{1}(x), and f

_{2}(x), where:

**g**and

_{1}(x) = f_{1}(x)/k**g**,

_{2}(x) = f_{2}(x) + k-2multiply both sides by k, and substitute for the g's, which gives me the now symmetrical form:

**k*P(x) = (f**

_{1}(x) + k)(f_{2}(x) + k)If you want to see more, I found a really in-depth post where talk much more here.

The forcing symmetry is what does it really. As I put an asymmetrical form at the BOTTOM as the base, and then force a symmetrical one later, and that does it! So cool to me.

Notice you have NO problems with our simple example:

k(x

^{2}+ 3x + 2) = (kx + k)((x+2-k) + k)

But all kinds of awesome happens when you have non-reducible quadratics for P(x). That's when the fun starts.

Turns out you can set things up so that it's easy to solve for the f's as roots of a monic polynomial with integer coefficients so they are DEFINITELY algebraic integers, but for the non-rational case you run into a problem with g

_{1}(x) = f

_{1}(x)/k as it forces one of the f's to have k as a factor.

So you can force that factor into algebraic integers and then NOT be able to see it directly. Knowing is there from mathematical logic. So absolute proof tells you is there but no way to see directly.

For example I often would show that 3+sqrt(-26) must have 7 as a factor for one of its two solutions which can seem strange if you follow human convention. But for example 5 + sqrt(4) is 5 ± 2 and equals 7 or 3. If that really bugs you not much I can do. Had wild arguments years ago where people would say that the convention is to take the positive so MUST be 7. And I'm like, what can you say to that?

So you find that in such cases both g's can NOT be algebraic integers. And I used a neat trick, where the f's can be, but I constructed with a throw-away factor of k, which means if you remove it, and reduce, you're forced to have:

P(x) = (g

_{1}(x) + 1)(g

_{2}(x) + 2)

So you can't run from the problem. The algebra says you

*can do it*.

Logically it works! Is fascinating to me though, how human beings can struggle with it, and may be one of the best pure examples of such.

There's no reason NOT to be able to do it, but our species didn't notice that over a hundred years ago in the late 1800's when coming up with algebraic integers, so it took me to show, in our time.

Yeah it is just NOT intuitive to multiply by that k, and math folks get taught to go in

*opposite direction*, removing factors if you can! Not adding them.

So I went against the usual flow.

And I showed how not knowing these things can let you use full mathematical rigor and come up with a wrong conclusion--in the ring of algebraic integers.

So why would I get a paper published showing that?

Well that was a great way to demonstrate, and may be the one key example known in human history to demonstrate the failure, where planned on a follow-up paper to explain the problem, thinking would have help from mathematicians, but that didn't happen.

What can someone do with this error?

Can appear to prove things with full mathematical rigor, which are NOT true.

And yes, I talked the question on this blog: Did mathematical rigor fail?

Answer will go ahead and note here is, no.

So yeah I had things not happen the way I expected. And was disappointed for years and acted out a bit online as well. But as time went on, I realized I had responsibilities and besides I could discover more things!

Eventually I focused more on mathematical industry as healthy, which includes lots more people than mathematicians, as lots of people all over the world use mathematics. The problem I found is important but rather esoteric. It helps to be a number theory expert to fully understand it and its implications.

And I am NOT a mathematician.

James Harris

## No comments:

Post a Comment