mm-427
Subject: Re: JSH: Distributive property, math argumentthis is
cause for a party.> F(x) = x^2 + x + 3.> P(x) = 2> Q(x) =
(1/2)(x^2+x+3).> Then F(x) = P(x)*Q(x) in the ring of all
polynomials over Q. IN> ADDITION, Q(x) is a function from Z to
Z, and P(x) is a function from> Z to Z, with the properties
that, AS FUNCTIONS, F(x) = P(x)*Q(x).> That is, for each value
of x, Q(x) divides F(x) in Z, which is a> different statement
from the polynomial Q(x) divides the polynomial> F(x) in
Z[x].> Amazing, that after only about two years he finally
realizes what he--ils duces
d'Enron!http://www.movisol.org/http://members.tripod.com/~
american_almanac/===Subject: Re: JSH: Distributive property,
math argument>...> The start is simple enough, where the x
shown is in the ring of> algebraic integers.> Let P(x) =
14706125 x^3 - 900375 x^2 - 17640 x + 1078>Why does it not
work with my polynomial?>It's not your polynomial, of course,
it's the *factorization* that you>use.>At the start of the
post I point out that if f(x) = x + 3, and f(x)>has 3 as a
factor, then x must have it as a factor as well.>That's
*basic* Dik Winter, and if you accept that result, then
at>least some things should be clear to you.>Notice I didn't
mention a ring for x, as the statement is true,>without
regard to the commutative ring.>yes, you did.>You said
Consider f(x) = x+3, where x is an algebraic integer.>Yup,
you're right. I'd forgotten that quickly and didn't go back
to>check before making the reply.>In any event, maybe I
should say that it doesn't matter what the ring>is.>Consider
that if I have f(x) = x + 3, where 3 has itself as a
factor>and f(x) has 3 as a factor, then x *must* have 3 as a
factor.>Now let's say x=7. Then, 7/3 must be in the
ring.>See?>No, I don't understand . You said it was the ring
of algebraic integers>and that f(x), given as f(x) = x+3, has
3 as a factor. Then instead of>concluding that this places a
*restricition* on the values that x can take>(i.e. x must be
an algebraic integer that is divisible by 3) you choose x
to>be 7 and conclude that therefore 7/3 =2.33333... must be in
the ring of>algebraic integers.!!!!> Nope. I'm saying that it
doesn't matter what you might *say* the ring> is, given f(x) =
x+3, if you also have that f(x) has 3 as a factor and> 3 has
itself as a factor, as then necessarily x has 3 as a factor,>
which means that if x=7, which IS an algebraic integer, it
*still*> must have 3 as a factor for the other statements to
be true.> Look at it as a logical argument:> 1. f(x) = x + 3>
2. 3 has itself as a factor> 3. f(x) has 3 as a factorThis is
false for x=7 in the algebraic integers.> 4. Therefore, x has
3 as a factor.This is false for x=7 in the algebraic
integers.Therefore, since the other statements are *not* all
true, you are correct.===Subject: Re: JSH: Distributive
property, math argument>...> The start is simple enough, where
the x shown is in the ring of> algebraic integers.Let P(x) =
14706125 x^3 - 900375 x^2 - 17640 x + 1078>Why does it not
work with my polynomial?>It's not your polynomial, of course,
it's the *factorization* that you>use.>At the start of the
post I point out that if f(x) = x + 3, and f(x)>has 3 as a
factor, then x must have it as a factor as well.>That's
*basic* Dik Winter, and if you accept that result, then
at>least some things should be clear to you.>Notice I didn't
mention a ring for x, as the statement is true,>without regard
to the commutative ring.>yes, you did.>You said Consider f(x) =
x+3, where x is an algebraic integer.> Yup, you're right. I'd
forgotten that quickly and didn't go back to> check before
making the reply.> In any event, maybe I should say that it
doesn't matter what the ring> is.> Consider that if I have
f(x) = x + 3, where 3 has itself as a factor> and f(x) has 3
as a factor, then x *must* have 3 as a factor.> Now let's say
x=7. Then, 7/3 must be in the ring.You missed a step. You
first have to assert that f(7) is divisible by 3. *Then* you
can can state that 7 is divisible by 3.> See?> It's rather
simple and depends on accepting the distributive property:>
a(b+c) = ab + ac> Which is why those who are arguing with me
are amusing in one sense,> as inevitably they are attacking
VERY basic mathematics to try and> keep playing silly games.>
It's funny to see how much math society actually cares about
its> foundations: like the distributive property.This is an
example of the problem you have. We aren't arguing the
distributive property, but something else entirely. You tend
to make sloppy mistakes, like not checking all the conditions.