Re: JSH: Stolen dreams and the distributive property

On Sat, 21 Jul 2007 23:30:15 -0000, jstevh@xxxxxxxxx wrote:

To understand succinctly how low mathematicians can go,

Skip the polemic, let's see the math.

just consider the following simple expression:

c*P(x)=(f_1(x) + c)*(f_2(x) + c)

It's not an expression -- it's an equation.

where c is a non-zero integer--so it is a constant--and P(x) is a
polynomial with integer coefficients, while f_1(x) and f_2(x) are
functions of x.

c is no problem, but f_1, f_2 and P are problems since you haven't
defined them except by an implicit equation.

Where are the _definitions_ of f_1, f_2 and P?

You see, you've been at this for years and you still can't write
rigorous math for even just the first few lines of your argument.

Ok, so let's defer the objection for now and temporarily accept that
f_1, f_2 and P are such that your equation is satisfied.

Now you indicate that P(x) is a polynomial. You never bothered to
mention the type of coefficients. Did you want integer coefficients?
If so, why didn't you declare it? You complain that your proofs
constantly get shot down as nonsense, but then you don't bother to
specify things that could be unclear.

Also, you say that f_1(x) and f_2(x) are functions. Did you mean
polynomials? If so, say so, and indicate the type of coefficients. If
f_1 and f_2 are not polynomials, what are they? If they are general
functions, you need to at least specify the domain and codomain.

Now I say that the distributive property is independent of the value
of x.

Posters disagree

There's no argument against the distributive law. Just make sure your
elements live in some ring. The fact that the P, f_1, f_2 are not
clearly defined makes that problematic, but let's assume that you
could clean that up (although you've had years to do it). So
everything in sight lives in some (as yet unspecified) big ring. Thus,
you can claim a distributive law. Let's see what happens next.

because if you agree with that statement then you can
let x=0, and with examples I have you can find then that f_1(x) or
f_2(x) should have c as a factor, but THEN find that they do not
always in the ring of algebraic integers.

Sorry, the above paragraph is essentially nonsense.

Of course you can set x=0 in your equation.

Thus, the equation

c*P(x)=(f_1(x) + c)*(f_2(x) + c)

becomes

c*P(0)=(f_1(0) + c)*(f_2(0) + c)

and your claim that c divides f_1(x) or f_2(x) is baseless since the
new equation has no terms f_1(x) or f_2(x) to even talk about.

It's pretty clear that you can't even write 10 lines of high school
level math rigorously, or if you can, you don't bother to do it.

If these were the first 10 lines of a paper, and I was the reviewer, I
would stop reading right at this point and send it back.

quasi
.

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