Re: JSH: Simple logic, distributive property

On Oct 7, 6:13 pm, Rick Decker <rdec...@xxxxxxxxxxxx> wrote:
JSH wrote:
I may have hit on what is key to the problem with people understanding
my research, where you MUST hold a certain amount of information in
mind long enough to understand the argument, so it may be a problem of
what you could call human mental buffer space.

Like consider the distributive property:

a*(b+c) = a*b + a*c

which is fairly simple and should have been learned by most of you as
children, and let's make it just a little more complicated with the
introduction of the function f(x) where b = f(x):

a*(f(x) + c) = a*f(x) + a*c

and now note that the VALUE of f(x) does not change the distributive
property.

Now then, have all that in mind? Sure? As I'm going to up the ante
here with something even more complex:

7*C(x) = (f(x) + 7)*(g(x) + 2)

where f(0) = g(0) = 0, and C(x) is a polynomial, so you have

7*C(0) = (0 + 7)(0 + 2)

proving that the 7 distributes through just one factor, and by the
logical principle where I noted the distributive property doesn't care
about the value of what is being multiplied, it follows that 7
distributes the same way for all x.

That should not be hard to hold in mind.

Nope, it isn't hard to hold in mind--all it requires is the ability
to keep in mind the datum "false."

Suppose we use these functions defined on the integers:

f(x) = (7/2)(x^2 + x) if x is even and -6 if x is odd

g(x) = 0 if x is even and 7x^2 + 7x + 12 if x is odd

Now we have f(0) = g(0) = 0 and it's easy to see that

(f(x) + 7)(g(x) + 2) = 7(x^2 + x + 2) = 7*C(x)

But we also have (in Z)

x is even => 7 divides f(x) + 7 and 7 does not divide g(x) + 2
x is odd => 7 does not divide f(x) + 7 and 7 divides g(x) + 2


Division is NOT a ring operation.

The distributive property simply states that

a*(b+c) = a*b + a*c

and if b=f(x) then you have

a*(f(x) + c) = a*f(x) + a*c

and the value of f(x) is irrelevant.

The result I have is not a factor result at first, as I rely on the
distributive property and if you then look to see what the factor
result would be in a particular ring, like the ring of algebraic
integers, you get weird things happening.

Now it turns out that in certain instances you simply cannot be in the
ring of algebraic integers because you have a function that cannot be
in that ring, like if with

a*C(x) = f(x) + a*c

where a, c and x are integers, and f(x) has integer coefficients, and
C(x) is a polynomial with rational coefficients, if f(x)/a is NOT an
algebraic integer then, of course, it just isn't regardless of what
the distributive property forces, so in that instance you necessarily
are not within the ring of algebraic integers if you try to remove 'a'
from both sides.

The logic is not complicated. It's all rather basic math.

Running to division over and over again is just a sign of your
desperation or of limited intellect.

Readers need to note that this result has already killed a math
journal as do your search on SWJPAM.

A result big enough to kill a math journal, quietly, is a
revolutionary result.

It is the biggest result in number theory in over a hundred years.

Nothing else even comes close.


James Harris

.

Relevant Pages

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  • Re: JSH: Using composites to understand the arguments
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  • Re: JSH: Using composites to understand the arguments
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  • Re: JSH: Using composites to understand the arguments
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