### Tuesday, January 06, 2009

## JSH: Demonstrating the distributive property

The problem with proof is, people can try to deny it, so here's yet another explanation of a remarkable error in number theory, where the proof only requires you accept the distributive property.

In the complex plane consider

7(x+1) = 7x + 7

which is a very simple demonstration of the distributive property. Now multiply both sides by x+2:

7(x+1)(x+2) = (7x + 7)(x + 2)

and on the left hand side simplify (x+1)(x+2) to get x^2 + 3x + 2, and you have

7(x^2 + 3x + 2) = (7x + 7)(x + 2)

and the stepped out demonstration that yes, indeed, it is the distributive property, where the only addition to the distributive property is multiplying both sides by x+2 and simplifying the left hand side, slightly. (Yes, my proof really is this simple.)

Moving on to the more complicated argument consider

7(f_1(x) + 1) = 7f_1(x) + 7

where f_1(0) = 0, so it's a normalized function. Now multiply both sides by f_2(x) + 2, where f_2(0)=0, and you have

7(f_1(x) + 1)(f_2(x) + 2) = (7f_1(x) + 7)(f_2(x) + 2)

and now let (f_1(x) + 1)(f_2(x) + 2) be a polynomial P(x), and you have have:

7P(x) = (7f_1(x) + 7)(f_2(x) + 2)

as a demonstration yet again of the distributive property, which you'll notice encompasses the prior result with

P(x) = x^2 + 3x + 2, and f_1(x) = x, and f_2(x) = x.

Notice it is a proof on the complex plane that given

7P(x) = (7f_1(x) + 7)(f_2(x) + 2), where P(x) is a polynomial and f_1(0) = f_2(0) = 0,

you have a demonstration of the distributive property.

So that is true for ALL such cases. ALL of them, over infinity.

THAT is the result being fought by the posters arguing with me, as now let:

P(x) = 175x^2 - 15x + 2

7f_1(x) + 7 = 5a_1(x) + 7, and f_2(x) + 2 = 5a_2(x)+ 7

where the a's are roots of

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

And you only need prove that f_1(0) = f_2(0) = 0, and you have the same distribution result on the complex plane, as before.

So that proves a general result encompasses both:

7(x^2 + 3x + 2) = (7x + 7)(x + 2)

and

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

where the a's are roots of

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

in a simple and elegant proof by direct demonstration of the distributive property.

Mathematical proof is irrefutable. What you have is an absolute argument which is not refutable mathematically.

In the complex plane consider

7(x+1) = 7x + 7

which is a very simple demonstration of the distributive property. Now multiply both sides by x+2:

7(x+1)(x+2) = (7x + 7)(x + 2)

and on the left hand side simplify (x+1)(x+2) to get x^2 + 3x + 2, and you have

7(x^2 + 3x + 2) = (7x + 7)(x + 2)

and the stepped out demonstration that yes, indeed, it is the distributive property, where the only addition to the distributive property is multiplying both sides by x+2 and simplifying the left hand side, slightly. (Yes, my proof really is this simple.)

Moving on to the more complicated argument consider

7(f_1(x) + 1) = 7f_1(x) + 7

where f_1(0) = 0, so it's a normalized function. Now multiply both sides by f_2(x) + 2, where f_2(0)=0, and you have

7(f_1(x) + 1)(f_2(x) + 2) = (7f_1(x) + 7)(f_2(x) + 2)

and now let (f_1(x) + 1)(f_2(x) + 2) be a polynomial P(x), and you have have:

7P(x) = (7f_1(x) + 7)(f_2(x) + 2)

as a demonstration yet again of the distributive property, which you'll notice encompasses the prior result with

P(x) = x^2 + 3x + 2, and f_1(x) = x, and f_2(x) = x.

Notice it is a proof on the complex plane that given

7P(x) = (7f_1(x) + 7)(f_2(x) + 2), where P(x) is a polynomial and f_1(0) = f_2(0) = 0,

you have a demonstration of the distributive property.

So that is true for ALL such cases. ALL of them, over infinity.

THAT is the result being fought by the posters arguing with me, as now let:

P(x) = 175x^2 - 15x + 2

7f_1(x) + 7 = 5a_1(x) + 7, and f_2(x) + 2 = 5a_2(x)+ 7

where the a's are roots of

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

And you only need prove that f_1(0) = f_2(0) = 0, and you have the same distribution result on the complex plane, as before.

So that proves a general result encompasses both:

7(x^2 + 3x + 2) = (7x + 7)(x + 2)

and

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

where the a's are roots of

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

in a simple and elegant proof by direct demonstration of the distributive property.

Mathematical proof is irrefutable. What you have is an absolute argument which is not refutable mathematically.