Re: JSH: Was never a factor argument

marcus_bruckner@xxxxxxxxx a écrit :
On Dec 21, 4:48 pm, JSH <jst...@xxxxxxxxx> wrote:
On Dec 21, 2:43 pm, marcus_bruck...@xxxxxxxxx wrote:



On Dec 21, 4:23 pm, JSH <jst...@xxxxxxxxx> wrote:
On Dec 21, 2:03 pm, marcus_bruck...@xxxxxxxxx wrote:
On Dec 21, 3:36 pm, JSH <jst...@xxxxxxxxx> wrote:
On Dec 21, 12:28 pm, marcus_bruck...@xxxxxxxxx wrote:
On Dec 21, 2:18 pm, JSH <jst...@xxxxxxxxx> wrote:
On Dec 21, 12:08 pm, marcus_bruck...@xxxxxxxxx wrote:
On Dec 21, 1:50 pm, JSH <jst...@xxxxxxxxx> wrote:
On Dec 21, 11:45 am, marcus_bruck...@xxxxxxxxx wrote:
On Dec 21, 1:19 pm, JSH <jst...@xxxxxxxxx> wrote:
Now isn't that amazing, I go to the complex plane and can give an
example like
7(x^2 + 3x + 2) = (7x + 7)(x + 2)
and posters who arrogantly argued with me for years, convincing untold
numbers of people that I was wrong here fall apart just because I'm
forcing the issue that the equations are in the field of complex
numbers.
But factor arguments don't matter on the complex plane!!!
Correct.
Yup. You're right, as I never had a factor argument.
You may have never had a factor argument, but you have
claimed factor conclusions. For example, you have claimed
that one of a_1(x) or a_2(x) is divisible by 7, and the other
is coprime to 7, and you were not just making that claim in
the complex plane.
I've had a
distribution argument.
So you can see how the 7 *distributes* through the factorization of
x^2 + 3x + 2. That is about the ***distributive property***.
You are imputing more to the distributive property than it
actually says. All the distributive property says is
that for a, b, and c in a ring,
a*(b + c) = a*b + a*c.
That's ALL it says. It does NOT say, for example, that
the only way to "distribute" 'a' across a product of two
sums is the following:
a*(b + c)*(d + e) = (a*b + a*c)*(d + e).
Here is an exercise that may help you, in the complex plane how do you
*prove* mathematically that with
7(x^2 + 3x + 2) = (7x + 7)(x + 2)
the 7 distributed in only one way?
You DON'T prove it, because it is not true!
There are infinitely many ways to split up
7 in the complex numbers; in fact, infinitely
That is correct.
many ways to split up 7 in the ring of algebraic
integers. Choose one of those ways: 7 = a*b.
Then
7(x^2 + 3x + 2) = 7(x + 1)*(x + 2)
= a*b(x + 1)*(x + 2) = a*(x + 1)*b*(x + 2)
= (a*x + a)*(b*x + b*2).
That is correct, but then you do something weird: you ignore your own
result.
All of these are valid. In fact, to arrive
at the last expression, I used the distributive
property TWICE. If somewhere in the above, I have
done something that violates the distributive property,
please point it out. Everything is there, nothing
hidden. If I made a mistake in any of the steps
above, let me know what it is.
Nope. You made no mistake, but let a=7 and b=1, how can you not
accept that that is ONE WAY?
Of course it's ONE WAY. But what you keep claiming
is, it's the ONLY WAY.
Given
7(x^2 + 3x + 2) = (7x + 7)(x + 2)
the 7 distributed in only ONE WAY.
Of course in the equation above it is distributed
in only ONE WAY. Are you saying that therefore,
all other distributions are impossible????
Why?
Hmmm...you seem to be missing an intermediate step:
7(x^2 + 3x + 2) = 7(x+1)(x+2) = (7x + 7)(x + 2)
indicates that the 7 was distributed in only one way.
As I noted in another thread, I think I now think I
know how you are thinking of this.
You are saying
7(x + 1)(x + 2) = (7x + 7)(x + 2),
independent of x. That is, if you choose to distribute
7 across the two parentheses in this way for one x,
you must be doing the same for all other x.
Um, dude, don't you realize that in algebra the 'x' represents ALL x?

It's not just some x, or a few x's and then you can change your mind.

In algebra, the x means: any number that can fit in there.

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

is true for ALL x. There is no movement based on x, or any movement
whatsoever.

It's STATIC.


When you write 'x' as above, x is intended to
represent a VARIABLE. It is the exact opposite
of STATIC. x is itself a function of x. Note that in
the complex numbers, I can write 7 as a product of two
numbers,

7 = (7/(1 + x^2)) * (1 + x^2)

i.e., 7 = a(x) * b(x) where a(x) = 7/(1 + x^2),
b(x) = 1 + x^2. Then I can write

7*(x^2 + 3x + 2) = (x * a(x) + a(x))*(x * b(x) + 2*b(x))

True or false? And everything in sight is a complex
number. No problem there. But the way in which I split
up 7 is a FUNCTION OF X. As it happens, when x = 0,
a(x) = 7 and b(x) = 1. Hmm. But the equation above
works perfectly well for all values of x. And you
keep saying that a(x) HAS TO equal 7 for all x, and b(x)
HAS to equal 1 for all x. But there it is, the equation
above, staring right out at you and saying you are a liar.
Whattayagonnadoaboutit?


And of course, no one has begin to use 7= (2-i*sqrt3)(2+i*sqrt3) yet...

Marcus.

James Harris

.

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