Rings and valid expressions

There have been these discussions about my non-polynomial factorization research where the replies go on and on, when it can be about dumb things like square roots having two values (others arguing against, me arguing for), and considering the big picture, I think quite a few of them may be clipped by talking some about rings and expressions valid within them.

There seems to be confusion about the distributive property when you have an expression not valid in a particular ring, like consider the simple example:

In the ring of algebraic integers let

7*(f(x) + 1) = x + 7

—and you have a contradiction as the expression is NOT valid in that ring, as it requires

f(x) = x/7

so notice that SAYING you're in the ring of algebraic integers, doesn't mean you actually are.

Here you're pushed out of the ring.

Notice though that the distributive property still works so you have

7*f(x) + 7 = x + 7

and it's just that f(x) = x/7, so the value you get from a function does not impact the distributive property.

So 7 still multiplied through.

With more complex factorizations you can get more complex behavior so that when you move to

7*C(x) = (f(x) + 7)(g(x) + 1)

where f(0) = g(0) and C(x) is a polynomial

you can start in the ring of algebraic integers, have the equations valid in the ring of algebraic integers, but still get a result not valid in that ring.

That's where things get subtle and where it's possible for malicious people to confuse the issue.

Now the argument that shows that you're forced out of a particular ring, like the ring of algebraic integers, for certain expressions isn't complicated, and relies on the distributive property still.

Just like with

7*(f(x) + 1) = x + 7

where you can SAY you're in the ring of algebraic integers, and find something that may look to a naive person like you defied the distributive property, it's still true that 7 multiplied through, as the distributive property simply says that if you multiply a group, you multiply each of the elements within that group, so f(x) and 1 get multiplied by 7.

But you don't see 7 times x on the right side.

So? It doesn't change the distributive property.

With more complicated expressions you get more complicated behavior, so that with

7*C(x) = (f(x) + 7)(g(x) + 1)

where f(0)=g(0) = 0

you can find C(x), f(x) and g(x) such that you are pushed out of the ring if you follow logically where the 7 must have multiplied through.

What makes this situation frustrating is that the mathematics is simple to the point of triviality, and I think most of you can comprehend that a function can be outside of a ring, so that someone arguing against the distributive property who jumped up and down about

7*(f(x) + 1) = x + 7

because you can't SEE the 7 multiplied times x, would not be very convincing.

But with slightly more complicated expressions where the proof is simple, people seem willing to try and SEE the 7, and when they don't see it, they are convinced by posters like Rick Decker or W. Dale Hall, who make it their business to be confusing on this issue.

The value of the functions does not change the distributive property.

That is just immutable.

If you are mathematicians, given a proof, you will follow the proof.

When posters claim examples that contradict a proof, you will look to find what's wrong with their claims versus simply tossing out the proof as if mathematical proof meant nothing to you.

Doubt is a good thing. It should lead you to depend on what is absolute, and in mathematics that is mathematical proof.

The proof is simple with

7*C(x) = (f(x) + 7)(g(x) + 1)

on the complex plane, where f(0) = g(0) and C(x) is a polynomial

if you accept that the distributive property does not care about the value of what is being multiplied, then the value of the function does not change the distributive property.

That's the main point. Accept it or challenge.

If you accept it, then it must be true that if the value of the function does not matter then ANY VALUE can be used, and therefore, it follows logically that x=0 is valid as a useful value.

Accept that logical step, or challenge.

Now then, if it does not matter what value is used, so x=0 is valid, then it is just a matter of noting that at x=0 you have

7*C(0) = (0 + 7)(0 + 1)

showing that 7 multiplied through only one, and the proof is done.

Examples equivalent to

7*(f(x) + 1) = x + 1

do NOT REFUTE A MATHEMATICAL PROOF!!!

I think part of the problem here is that many of you do not comprehend what a mathematical proof actually is, so you think that proofs can be broken, or created, or they are fragile things that can shake with arguing.

But what proofs are, are points of truth that do not require faith.

You don't have to believe in some person or entity. You don't have to trust or worry about what education level one person has versus another.

All those are points irrelevant to a proof, so when people question my not having a degree in mathematics as my degree is in physics, you do not have to worry about that if you know what mathematics is about, as what you really need is the proof.

With the proof social crap is irrelevant. It doesn't matter how many people argue with me. It doesn't matter what your gut feeling tells you.

Trace the logical steps from a truth, and the conclusion that follows MUST BE TRUE.

See my definition of mathematical proof:

http://mymath.blogspot.com/2005/07/definition-of-mathematical-proof.html

The arguing would have been over years ago if mathematicians followed proofs as I have the proofs, people just lie about them.

Given a proof, argument is just a waste of time, because a proof cannot be refuted.

You may have heard of proofs being refuted. You may have heard of proofs failing, but then, they were not proofs!!!

If someone has proof that you are dead, but it turns out you are not dead (as none of you are) then did they really have proof?

Was it failed proof? Did their proof collapse? Can someone create a proof that you are dead and then someone else refute it?

NO!!! They just didn't have proof.