### Wednesday, February 08, 2006

## JSH: Generalized Decker example, examples

Decker's example can be generalized to help in finding rational solutions, where you will find that the mathematics follows the distributive property:

f Q(x) = f((x^2 + x)(5^2) + (-1 + x)(5) + f) = f(25 x^2 + 30 x + 2)

and

f Q(x) = (5a_1(x) + f)(5a_2(x) + f)

where the a's are defined by

a^2 - (x - 1)a + f(x^2 + x) = 0

and you can let f be a rational number. Now there will be posters who will loudly declare coprimeness means nothing with rationals, but consider

f(x^2 + 3x + 2) = (fx + f)(x + 2)

and solutions with rationals, as guess what? You can STILL see that one factor is multiplied by f, even with rationals. If you don't think so, play with that example with some rational f's and rational x and see if the factor multiplied by f doesn't betray that it was.

Now then, if I am wrong, some rational f can be found with a rational x that shows it.

Like let f=32 with the generalized Decker example, and find some rational solutions and see if that f gets split up. Or let f=1024 or anything you want!!!

You see, no counterexample exists, as the distributive property is right!!!

So posters here at best can loudly proclaim that Galois Theory doesn't work with rationals but only with non-rationals, which is the dodge because it actually doesn't work, but you can't see that with non-rationals.

The proof I've given relies on the distributive property.

Even lower rung mathematicians cannot be incapable of quickly seeing it MUST be correct, but clearly as this impasse continues they are running.

And them running means they are hoping that none of you who are not already established in careers, who are just learning as you're still in school, will stop protecting them by ignoring this result.

They are in the weak positon of needing your protection so that they can teach you wrong mathematical ideas, as if some students start protesting, they will collapse like the cowards they are, running the other way, selling each other out to protect themselves.

First mathematicians on the block will be the ones who are posters on sci.math, and their own will destroy their careers. People like Magidin and Ullrich will be out of their universities so fast your head will spin, as they are tossed to the wolves.

So they sit and wait, checking each day to see if any of you are breaking out of the wall of silent acceptance, or irrational denial in the face of a simple proof that relies on the distributive property at the point of dispute.

That check may be the best evidence against those mathematicians as they leave cyber clues to prove they knew, but were checking to see if they could keep getting away with lying.

f Q(x) = f((x^2 + x)(5^2) + (-1 + x)(5) + f) = f(25 x^2 + 30 x + 2)

and

f Q(x) = (5a_1(x) + f)(5a_2(x) + f)

where the a's are defined by

a^2 - (x - 1)a + f(x^2 + x) = 0

and you can let f be a rational number. Now there will be posters who will loudly declare coprimeness means nothing with rationals, but consider

f(x^2 + 3x + 2) = (fx + f)(x + 2)

and solutions with rationals, as guess what? You can STILL see that one factor is multiplied by f, even with rationals. If you don't think so, play with that example with some rational f's and rational x and see if the factor multiplied by f doesn't betray that it was.

Now then, if I am wrong, some rational f can be found with a rational x that shows it.

Like let f=32 with the generalized Decker example, and find some rational solutions and see if that f gets split up. Or let f=1024 or anything you want!!!

You see, no counterexample exists, as the distributive property is right!!!

So posters here at best can loudly proclaim that Galois Theory doesn't work with rationals but only with non-rationals, which is the dodge because it actually doesn't work, but you can't see that with non-rationals.

The proof I've given relies on the distributive property.

Even lower rung mathematicians cannot be incapable of quickly seeing it MUST be correct, but clearly as this impasse continues they are running.

And them running means they are hoping that none of you who are not already established in careers, who are just learning as you're still in school, will stop protecting them by ignoring this result.

They are in the weak positon of needing your protection so that they can teach you wrong mathematical ideas, as if some students start protesting, they will collapse like the cowards they are, running the other way, selling each other out to protect themselves.

First mathematicians on the block will be the ones who are posters on sci.math, and their own will destroy their careers. People like Magidin and Ullrich will be out of their universities so fast your head will spin, as they are tossed to the wolves.

So they sit and wait, checking each day to see if any of you are breaking out of the wall of silent acceptance, or irrational denial in the face of a simple proof that relies on the distributive property at the point of dispute.

That check may be the best evidence against those mathematicians as they leave cyber clues to prove they knew, but were checking to see if they could keep getting away with lying.