### Saturday, January 10, 2009

## Direct proof of algebraic integer problem

When you stop accepting mathematical proof then you are not mathematicians, no matter if you keep claiming you are.

In what follows I walk through a simple proof on the complex plane, and then show how the result on the complex plane gives a conclusion in other rings, and how in one ring, the ring of algebraic integers, there is a problem shown.

The argument is as easy as a, b, c so the issue is not correctness, but whether or not you and people like you really believe in mathematics.

On the complex plane given the expression

7*(175x^2 - 15x + 2) = 7*(5c_1(x) + 1)(5c_2(x)+ 2)

where the c's are not yet determined function of x, there is NOTHING in algebra that prevents me from choosing:

7*(175x^2 - 15x + 2) = (5*7*c_1(x) + 7)(5c_2(x)+ 2).

That distribution IS ALLOWED. It is one distribution but it is allowed as a valid step.

If you accept that one point you are 99% of the way to being able to cut through the noise when others work desperately to challenge the conclusion of this proof. Now I introduce new functions:

b_1(x) = 7*c_1(x), and b_2(x) = c_2(x)

and now require c_1(0) = 0, and c_2(0) = 0, so

b_1(0) = b_2(0) = 0, and substitute to get

7*(175x^2 - 15x + 2) = (5b_1(x) + 7)(5b_2(x)+ 2).

and again there is NOTHING that says I cannot do that on the complex plane.

The b's are still not determined functions of x.

I will introduce a final set of functions:

a_1(x) = b_1(x), and a_2(x) = b_1(x) - 1

Now I'll make those substitutions and get:

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

and now note that a solution—no I'm not saying it's the only solution!!!—for the a's is, as roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0.

You can now solve for the a's easily using the quadratic formula:

a_1(x) = ((7x-1) +/- sqrt((7x-1)^2 - 4(49x^2 - 14x)))/2

a_2(x) = ((7x-1) -/+ sqrt((7x-1)^2 - 4(49x^2 - 14x)))/2

and that is a valid solution set as can be verified easily enough by substitution.

But now you have the conclusion that only one of the a's was actually a product of 7, as you know the start:

7*(175x^2 - 15x + 2) = (5*7*c_1(x) + 7)(5c_2(x)+ 2).

The result that only one of the roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0

was actually multiplied by 7 becomes a factor result in other rings, but is contradicted by the ring of algebraic integers in specific cases.

THAT brings into question the naive use of that ring, as unfortunately it can be shown that the problem allows you to appear to prove things that are not true, and it also brings into question the usefulness of Galois Theory.

In what follows I walk through a simple proof on the complex plane, and then show how the result on the complex plane gives a conclusion in other rings, and how in one ring, the ring of algebraic integers, there is a problem shown.

The argument is as easy as a, b, c so the issue is not correctness, but whether or not you and people like you really believe in mathematics.

On the complex plane given the expression

7*(175x^2 - 15x + 2) = 7*(5c_1(x) + 1)(5c_2(x)+ 2)

where the c's are not yet determined function of x, there is NOTHING in algebra that prevents me from choosing:

7*(175x^2 - 15x + 2) = (5*7*c_1(x) + 7)(5c_2(x)+ 2).

That distribution IS ALLOWED. It is one distribution but it is allowed as a valid step.

If you accept that one point you are 99% of the way to being able to cut through the noise when others work desperately to challenge the conclusion of this proof. Now I introduce new functions:

b_1(x) = 7*c_1(x), and b_2(x) = c_2(x)

and now require c_1(0) = 0, and c_2(0) = 0, so

b_1(0) = b_2(0) = 0, and substitute to get

7*(175x^2 - 15x + 2) = (5b_1(x) + 7)(5b_2(x)+ 2).

and again there is NOTHING that says I cannot do that on the complex plane.

The b's are still not determined functions of x.

I will introduce a final set of functions:

a_1(x) = b_1(x), and a_2(x) = b_1(x) - 1

Now I'll make those substitutions and get:

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

and now note that a solution—no I'm not saying it's the only solution!!!—for the a's is, as roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0.

You can now solve for the a's easily using the quadratic formula:

a_1(x) = ((7x-1) +/- sqrt((7x-1)^2 - 4(49x^2 - 14x)))/2

a_2(x) = ((7x-1) -/+ sqrt((7x-1)^2 - 4(49x^2 - 14x)))/2

and that is a valid solution set as can be verified easily enough by substitution.

But now you have the conclusion that only one of the a's was actually a product of 7, as you know the start:

7*(175x^2 - 15x + 2) = (5*7*c_1(x) + 7)(5c_2(x)+ 2).

The result that only one of the roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0

was actually multiplied by 7 becomes a factor result in other rings, but is contradicted by the ring of algebraic integers in specific cases.

THAT brings into question the naive use of that ring, as unfortunately it can be shown that the problem allows you to appear to prove things that are not true, and it also brings into question the usefulness of Galois Theory.