With a quadratic polynomial, we can consider its factorization in general with:

P(x) = (g

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

_{2}(x) + 2)

where, g

_{1}(0) = g

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

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

Values for the g's are trivial for a polynomial factorization.

For example, if g

_{1}(x) = x, and g

_{2}(x) = x, then, of course P(x) = x

^{2}+ 3x + 2.

P(x) = (x+1)(x+2)

But if P(x) does not factor into polynomial factors, then the g's cannot be algebraic integers, which I've shown by multiplying by a constant integer where the result must be an algebraic integer.

For example,

11*P(x) = (f

_{1}(x) + 11)(f

_{2}(x) + 11), where g

_{1}(x) = f

_{1}(x)/11 and g

_{2}(x) = f

_{2}(x) + 9.

For algebraic integer x, the f's are in general within the ring of algebraic integers, which I've proven in detail in a post:

somemath.blogspot.com/2015/04/why-in-ring-of-algebraic-integers

But while the f's are algebraic integers for algebraic x, the g's can't both be, which is from the forced asymmetry. That's why I used 1 and 2, as I wanted something simple, and that's about as simple as I could get:

P(x) = (g

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

_{2}(x) + 2)

So consider some nonrational solutions for the g's, which I'll call o

_{1}and o

_{2}, which would be for some particular nonzero, algebraic integer x.

I have that one is an algebraic integer, while the other is not, and does not fit within any currently established ring.

But if I multiply it, for instance, by 11, then it is an algebraic integer. But if I multiply the same number by 7, then it is also an algebraic integer. And in general I can multiply the same number by any nonzero integer not 1 or -1, and it will be an algebraic integer.

That is to say, if o

_{1}is the non algebraic integer, then 7o

_{1}and 11o

_{1}are, and in general, ko

_{1}is an algebraic integer for k, nonzero and not 1 or -1, while o

_{1}is not:

k*P(x) = (f

_{1}(x) + k)(f

_{2}(x) + k)

where g

_{1}(x) = f

_{1}(x)/k and g

_{2}(x) = f

_{2}(x) + k-2

And then it is clear why fields are not impacted, as consider o

_{2}/o

_{1}, which might not seem to be in the field of algebraic numbers, but you just have to multiply top and bottom by some nonzero integer not 1 or -1, like:

2o

_{2}/2o

_{1}

and you have a ratio of algebraic integers, showing that it is indeed an algebraic number.

So this approach allows studying the underlying asymmetrical form by

*moving to a symmetrical form*using basic algebra.

And it's nice to mark over 10 years since I came up with a more robust classification scheme with the object ring.

That is the first post on this blog made over 10 years ago. Yeah, it's that important.

Probably is the reason this blog exists, because it was long enough ago, I'm not sure all of what I was thinking at the time. But since I made it the very first post, got a feeling it had much to do with my motivations.

So I'd also like to celebrate that special event.

And will note coming up on a 20 year celebration this month.

That there are numbers which behave like integers, which are themselves not rational, which are NOT algebraic integers is fascinating.

And since they are not algebraic integers it's hard to see them directly. What you can do is get a solution for the f's and divide off the constant multiplier from both solutions knowing it works for one, and for the other you have something that is fractional, but that doesn't tell you which is which.

I've been thinking about these numbers for a while now, over a decade.

So how were they missed before? If you step through the tests done on the ring of algebraic integers, it's easy to see how they don't catch these numbers.

So I came up with a more inclusive ring classification and call these numbers along with integers--objects.

Objects replace algebraic integers at the foundations of numbers. The object ring includes the ring of algebraic integers plus the additional numbers previously missed.

James Harris