Re: JSH: Stolen dreams and the distributive property

jstevh@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.

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.

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).

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.

Here's an example. Suppose P(x) = 9x^2 + 21x - 1.
Then we could write (with c = 2)

2P(x) = (3f_1(x) + 2)(3f_2(x) + 2) [*]

where

f_1(x) = (x-1 + sqrt(-7x^2 - 18x +1)/2
f_2(x) = (x-1 - sqrt(-7x^2 - 18x +1)/2

Now observe that f_1(0) = 0, so 2 divides f_1(0) and
so the factor of 2 on the left is accounted for by
the factor of 2 in the first term on the right. However,
the fact that you have this pattern for x = 0 does
not imply that you'll have this splitting for other values
of x.

For instance, if x = 1 [*] takes the values

2(29) = 58 = (3 sqrt(-6) + 2)(-3 sqrt(-6) + 2)

but 3 sqrt(-6) is not divisible by 2. In this case,
the 2 on the left is distributed as sqrt(2).sqrt(2)
since, obviously

2(29) = [sqrt(2)(3sqrt(-3)+sqrt(2))][sqrt(2)(-3sqrt(-3)+sqrt(2)]
= [sqrt(2)sqrt(2)][3sqrt(-3)+sqrt(2)][-3sqrt(-3)+sqrt(2)]
= [2][29]

In summary, the problem you face is not with the distributive
property, which is true in any ring, but with your assumption
that if c divides AB then c divides A or c divides B. While
this is true in the integers of c is a prime, it is certainly
not true in the algebraic integers if c is a rational prime.

If this is a "failure" of the algebraic integers, so be it.
You have no recourse but to accept it.


Regards,

Rick
.

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