Re: JSH: Stolen dreams and the distributive property

jstevh@xxxxxxxxx wrote:
On Jul 25, 6:20 pm, Rick Decker <rdec...@xxxxxxxxxxxx> wrote:
jst...@xxxxxxxxx wrote:
On Jul 25, 10:02 am, marcus_b <marcus_bruck...@xxxxxxxxx> wrote:
On Jul 24, 10:32 pm, jst...@xxxxxxxxx wrote:
<snip>





More generally, you were claiming that the factorization
of c out of the product
(f_1(x) + c)*(f_2(x) + c)
does not depend on the value of x. This proves that in
No I did not so claim.
What I did say is that the distributive property is independent of the
value of x.
For instance
a*(f(x) + b) = a*f(x) + a*b
without regard to the value of x [...]
Of course. No one denies that.



Why don't you go back to the starting post in this thread and read
what I did actually claim.
The confusion here seems to come from reading more into the
distributive property than is actually there.

Seems instead you read more into my post than is actually there. I'll
show below.

You have posited an equality

cP(x) = (f_1(x) + c)(f_2(x) + c)

for some functions f_1 and f_2. No problem there, for
suitable choices of P. Then you observe that since c
divides the left hand side, it must be the case that
c divides the right. Also no problem, even though you
haven't specified the ring in which you're working.


That's because what I've shown is valid in any integral domain.

Importantly I specify that P(x) is a polynomial with integer
coefficients and c is a non-zero integer which is coprime to P(0).

That forces the conclusion that both f_1(0) and f_2(0) cannot equal 0,
which leads you down the path of normalizing to functions that BOTH
equal 0 when x=0, and the proper conclusion naturally jumps out at you
which is why I like this example and said so little in my original
post.

But intriguingly you claim I said more, so let's continue.

Then, however, you go off the rails. Since c divides
the right hand side, you then assume that c must divide
one of the two factors, say f_1(x) + c, from which it
correctly follows that c must divide f_1(x).


Nope. What if c factors in whatever ring you are in? Like if you are
in Z, can't c be a composite?

You're either being disingenuous or illiterate here. Immediately
after the paragraph above I wrote:

This would be correct if everything in sight was an integer
and c was a prime. The problem is that we're working in
the ring of algebraic integers, where none of the "ordinary"
(read rational) primes are primes. Simply said, the c
you use might very well split into other factors.

And continued with yet another example of why your assertion
failed.


Regards,

Rick


James Harris

.

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