Sunday, October 07, 2007
JSH: Simple logic, distributive property
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.
Let me recap, because the distributive property does not care what value f(x) has, you can note at a convenient value, like x=0 as then f(0) = 0, how the 7 distributed and it cannot now matter if x changes as, why?
Because a*(f(x) + c) = a*f(x) + a*c without regard to the value of x and I emphasize this point as it is CRUCIAL.
The distributive property does not care what value f(x) has, as it just multiplies through—distributes the multiplication—without regard to the value of what is being multiplied, so
a*(b+c) = a*b + a*c
without regard to the value of b or c.
So logically, the conclusion must be true, but now I have this general result!
What is the result?
Given
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)
it must be true that the 7 distributes through just the first factor (f(x) + 7).
It is a HUGE result in terms of consequences and it is one that mathematicians have argued against with me for years. Literally years. And if you understand it, you may believe that is impossible as it seems so trivial, but mathematics can be remarkable in that seemingly trivial results can have a huge impact.
I'm about to up the complexity yet again, so see if you can hold all the information needed so far in mind, as now consider
7*C(x) = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 49
where the odd form is to show the factorization which is
7*C(x) = (5a_1(x) + 7)(5a_2(x)+ 7)
where the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0
and I haven't reached
7*C(x) = (f(x) + 7)*(g(x) + 2)
yet as you have two 7's visible in the factorization, but notice at x=0 I have for the a's
a^2 + a = 0
so one of the a's equals 0 at that point, while the other equals -1.
So let f(x) equal the one that goes to 0, while g(x) - 1 equals the other, as its indeterminate which one does, and I have worked my way back to
7*C(x) = (f(x) + 7)*(g(x) + 2)
and the conclusion that the 7 multiplies times C(x) so that it distributes in only one way, where by the distributive property you can see what that way is directly at x=0.
But the distributive property doesn't care about the value of f(x), so it distributes one way for all x.
Trouble is, you can prove that in a ring mathematicians call the ring of algebraic integers, 7 cannot be a factor of one of the roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0
if you have integer x and that gives a polynomial irreducible over Q.
Now then, could you hold all the information in mind or did some dribble out of your mental buffer?
If you could then you know that it follows from the distributive property that only one of the roots of that expression should have 7 as a factor, so if they don't in the ring of algebraic integers under certain conditions then something must be wrong with that ring.
And you know that follows from the distributive property.
But if you cannot hold all that information in mind, some poster can come back and claim it doesn't follow from the distributive property and because information leaked out of your brain, you can believe them from a failure of your mental circuitry.
So failing to understand such a trivial argument is evidence of a fundamental lack in your mental wiring.
[A reply to someone who asked what “distributes through” means.]
The clue is in "distributive property" as in, why is "distributive" in there?
Do you know? I will try to answer your question, so can you try to answer why you think "distributive" is part of the distributive property?
With
a*(b+c) you have that 'a' distributes through the term (b+c) so that you have a*b+a*c
and the fully mathematicized form you have then is
a*(b+c) = a*b+a*c
and it seems so simple until you end up in a heated dispute where people challenge the very basics of mathematics down to the most trivial because so much is at stake.
What is at stake here is that if you accept what "distributes" means, and acknowledge that the value of b and c does not matter to how 'a' multiplies times (b+c), then of course, with
a*(f(x) + c) you still have a*f(x) + a*c
and then I can go to a more complex expression and prove a remarkable result using what many think is trivial mathematics.
[A reply to someone who wrote that how 7 divides 71 − x7x depends on x.]
Division is not a ring operation.
And so many of the arguments around this issue are all about division when posters should know that division is not a ring operation.
Let's look again at
a*(f(x) + c) = a*f(x) + a*c
and say a=4, and f(x) = x/2, so you have
4(x/2 + c) = 2x + 4c
and now I say that the value of x does not matter but if you counter with x=2 to say it DOES matter, can you see the logical flaw in your thinking?
Obviously the ring above is one that contains 1/2, so let's say it is the ring of algebraic numbers which is also a field, does that matter?
Crucially the distributive property is not about division, but about multiplication, which is a ring operation, and the value of x does not affect it at all.
The problem for modern mathematics though is that I can go from there to proving a problem with the ring of algebraic integers using a rather complex form as shown in my original post in this thread, and that THAT is why there is a continuing argument against the distributive property and blatant attempts at distraction like tossing
in division, when it's not a ring operation.
Readers ready to confront the full issue with the ring of algebraic integers can go to my web page at my Extreme Mathematics group:
http://groups.google.com/group/extrememathematics/web/non-polynomial-factorization-paper
There you can get my most modern paper in pdf format as well as see the original paper that was published—and then pulled—by the now defunct journal SWJPAM.
See what kind of mathematical result was powerful enough to blow away a modern math journal with the rest of the math world pretending like it didn't happen.
This result is that big.
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.
Let me recap, because the distributive property does not care what value f(x) has, you can note at a convenient value, like x=0 as then f(0) = 0, how the 7 distributed and it cannot now matter if x changes as, why?
Because a*(f(x) + c) = a*f(x) + a*c without regard to the value of x and I emphasize this point as it is CRUCIAL.
The distributive property does not care what value f(x) has, as it just multiplies through—distributes the multiplication—without regard to the value of what is being multiplied, so
a*(b+c) = a*b + a*c
without regard to the value of b or c.
So logically, the conclusion must be true, but now I have this general result!
What is the result?
Given
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)
it must be true that the 7 distributes through just the first factor (f(x) + 7).
It is a HUGE result in terms of consequences and it is one that mathematicians have argued against with me for years. Literally years. And if you understand it, you may believe that is impossible as it seems so trivial, but mathematics can be remarkable in that seemingly trivial results can have a huge impact.
I'm about to up the complexity yet again, so see if you can hold all the information needed so far in mind, as now consider
7*C(x) = (49x^2 - 14x)5^2 + (7x-1)(7)(5) + 49
where the odd form is to show the factorization which is
7*C(x) = (5a_1(x) + 7)(5a_2(x)+ 7)
where the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0
and I haven't reached
7*C(x) = (f(x) + 7)*(g(x) + 2)
yet as you have two 7's visible in the factorization, but notice at x=0 I have for the a's
a^2 + a = 0
so one of the a's equals 0 at that point, while the other equals -1.
So let f(x) equal the one that goes to 0, while g(x) - 1 equals the other, as its indeterminate which one does, and I have worked my way back to
7*C(x) = (f(x) + 7)*(g(x) + 2)
and the conclusion that the 7 multiplies times C(x) so that it distributes in only one way, where by the distributive property you can see what that way is directly at x=0.
But the distributive property doesn't care about the value of f(x), so it distributes one way for all x.
Trouble is, you can prove that in a ring mathematicians call the ring of algebraic integers, 7 cannot be a factor of one of the roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0
if you have integer x and that gives a polynomial irreducible over Q.
Now then, could you hold all the information in mind or did some dribble out of your mental buffer?
If you could then you know that it follows from the distributive property that only one of the roots of that expression should have 7 as a factor, so if they don't in the ring of algebraic integers under certain conditions then something must be wrong with that ring.
And you know that follows from the distributive property.
But if you cannot hold all that information in mind, some poster can come back and claim it doesn't follow from the distributive property and because information leaked out of your brain, you can believe them from a failure of your mental circuitry.
So failing to understand such a trivial argument is evidence of a fundamental lack in your mental wiring.
[A reply to someone who asked what “distributes through” means.]
The clue is in "distributive property" as in, why is "distributive" in there?
Do you know? I will try to answer your question, so can you try to answer why you think "distributive" is part of the distributive property?
With
a*(b+c) you have that 'a' distributes through the term (b+c) so that you have a*b+a*c
and the fully mathematicized form you have then is
a*(b+c) = a*b+a*c
and it seems so simple until you end up in a heated dispute where people challenge the very basics of mathematics down to the most trivial because so much is at stake.
What is at stake here is that if you accept what "distributes" means, and acknowledge that the value of b and c does not matter to how 'a' multiplies times (b+c), then of course, with
a*(f(x) + c) you still have a*f(x) + a*c
and then I can go to a more complex expression and prove a remarkable result using what many think is trivial mathematics.
[A reply to someone who wrote that how 7 divides 71 − x7x depends on x.]
Division is not a ring operation.
And so many of the arguments around this issue are all about division when posters should know that division is not a ring operation.
Let's look again at
a*(f(x) + c) = a*f(x) + a*c
and say a=4, and f(x) = x/2, so you have
4(x/2 + c) = 2x + 4c
and now I say that the value of x does not matter but if you counter with x=2 to say it DOES matter, can you see the logical flaw in your thinking?
Obviously the ring above is one that contains 1/2, so let's say it is the ring of algebraic numbers which is also a field, does that matter?
Crucially the distributive property is not about division, but about multiplication, which is a ring operation, and the value of x does not affect it at all.
The problem for modern mathematics though is that I can go from there to proving a problem with the ring of algebraic integers using a rather complex form as shown in my original post in this thread, and that THAT is why there is a continuing argument against the distributive property and blatant attempts at distraction like tossing
in division, when it's not a ring operation.
Readers ready to confront the full issue with the ring of algebraic integers can go to my web page at my Extreme Mathematics group:
http://groups.google.com/group/extrememathematics/web/non-polynomial-factorization-paper
There you can get my most modern paper in pdf format as well as see the original paper that was published—and then pulled—by the now defunct journal SWJPAM.
See what kind of mathematical result was powerful enough to blow away a modern math journal with the rest of the math world pretending like it didn't happen.
This result is that big.
# posted by James Harris @ 02:12 
