Re: JSH: Trivially easy math

In sci.math, jstevh@xxxxxxx
<jstevh@xxxxxxx>
wrote
on 27 Jan 2006 19:01:03 -0800
<1138417263.680967.122920@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>:
> The clever thing you can say about the mathematical ideas I use is that
> I figured out a way to put algebraic functions that give roots to a
> cubic opposite constants in such a way that the functions are forced to
> have a factor based on those constants.
>
> The gist of how it works is already known to you as polynomial factors
> contain simple functions:
>
> P(x) = x^2 + 3x + 2 = (x+1)(x+2)
>
> has two polynomial factors that have linear functions plus the constant
> terms as you have x, which is a function of x, and next to it 1 or 2.

OK thus far. All C[x] polynomials have n factors of the type
(x-z), z in C [*].

If you want to restrict yourself to Z[x] one has the problem of
dealing with x^2 + 1, which cannot be factored over Z, or even
3*x+2, which cannot be factored over Z either (though it resolves
into 3*(x+2/3) if one works in Q).

>
> Multiply by 7, and you can do it an INFINITE NUMBER OF WAYS but you are
> compelled to pick ONE WAY to show the factorization, like
>
> 7P(x) = 7(x^2 + 3x + 2) = (7x+7)(x+2)

Do *what* an infinite number of ways? If you're referring to breakup
of factorization of 7*P(x) over Q, R, or C, sure:

7*P(x) = 7 * (x+1) * (x+2)
= (7x+7) * (x+2)
= (x+1) * (7x+14)
= (3.5x+3.5) * (2x + 4)
= (pi*x+pi) * (7/pi*x + 14/pi)
= 4 * (7x + 7) * (x/4 + 1/2)

You're going to have to be a wee bit more specific. Even within Z[x]
there are three factorizations -- the first three equalities above.
However, the factorization of an arbitrary n-degree polynomial

P(x) = sum(i=1,n) (a_n * x^n)

is unique in C[x] (within ordering of the factors) if one starts
with a_n as the first factor and requires that each factor be
irreducible in C[x].

The factorization is unique in Q[x] as well, though one might
get factors of degree greater than 1 (e.g., x^2+1, x^2 - 2,
x^3 - 4/5).

Z[x] introduces some problems; the polynomial

K(x) = 14*x^2+15*x+4 = (2*x+1)*(7*x+4) = 14*(x+1/2)*(x+4/7).

Of course the last factorization is over Q[x], not Z[x].

I think this can be fixed by requiring that any common factors
of all [integer] coefficients be divided out as the first factor.
It turns out no integers divides into all three of 14, 15, and 4,
though 14 and 4 have a common factor 2.

In your case 7*x^2 + 21*x + 14, the common 7 can be
divided out first:

7*x^2 + 21*x + 14 = 7*(x^2 + 3*x + 2) = 7*(x+1)*(x+2)

This factorization is unique in Z[x], apart from ordering.

>
> as that is just one way where I multiplied the first polynomial factor
> by 7.
>
> So yeah, the mathematical ideas I use are basic, and notice that the
> distributive property is how you know how to multiply an expression
> like that by 7.
>
> If I started with
>
> 7P(x) = 7(x^2 + 3x + 2) = (7x+7)(x+2)
>
> notice that it is FORCED that setting x=0 to clear out functions of x,
> will reveal that only the first factor was multiplied by 7 and that is
> NOT a special case!!!
>
> It doesn't change at x=1, or x=6, or x=8347837. You know that many
> ways, where one is to see it, but it's also possible to prove it.

If one sets x to 5 one gets the identity

294=2*3*7^2 = 7*(6)*(7) = (42)*(7)

If one sets x to 6 one gets the identity

392=2^3*7^2 = 7*(7)*(8) = (49)*(8)

If one sets x to 12 one gets the identity

1274=2*7^2*13 = 7*(13)*(14) = (91)*(14)

If one sets x to 8347837=127*65731, one gets

(58434866) * (8347839) = (2*7*4173919) * (3 * 2782613)

OK thus far, but if one uses x=8641974=2*3*11*23*5693 instead,
one gets

(60493825) * (8641976) = (5^2*7*345679) * (2^3*7*154321)

Seven, seven, who has the seven? (Hint: how do 8641974 and 7 relate?)

>
> Simplest thing is to note that the constants are, constant, and
> independent of x.

That is a consequence of problem setup.

>
> Some of you, with my equations, seemed to fall into the trap of
> thinking that functions can force constants, when you should know it's
> the other way around, as constants are immutable.
>
> They are constant. The number 2 is a bit more powerful in ways you can
> say than the symbol x, as the 2 is just 2, while the x can change.
>
> Given
>
> x+2 = 3
>
> it is the constants that force x, and NOT THE OTHER WAY AROUND.

This is interesting philosophical ground, to be sure. Of course,
all you've really done here is write an assertion (equation). The
assertion

x+2 = 3

is that we can find a number x such that when 2 is added to it,
yields the value 3.

Depending on the ring this may or may not be possible,
though in most arithmetic contexts the answer of course
is x = 1. However, if one defines "+" as "string
concatenation over the alphabet [0-9]", there is an
identity (the empty string) but no inverse, and no
solution to that particular equation.

Of course one focus of math is to solve the assertion, and
most people don't go for string concatenation in math anyway. :-)

A slightly more relevant problem was to find the number i
such that i^2 = -1. Since all reals have nonnegative
squares i was dubbed imaginary, and yielded complex numbers.

>
> Functions don't force constants. Constants force functions.
>
> The real story here is not difficulty in understanding the mathematics.
>
> What really bugged me was not so much the refusal of this newsgroup to
> accept the truth, as hey, you're a group, and you display group
> behavior, but the mathematicians I contacted, like Barry Mazur, and
> especially Andrew Granville and Ralph McKenzie didn't react right.
>
> I still wonder what were they thinking?

[rest snipped]

[*] Z = set of integers
Q = set of rational numbers
R = set of reals
C = set of complex numbers
S[x] = set of all possible polynomials of x with coefficients in S

--
#191, ewill3@xxxxxxxxxxxxx
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