Re: JSH: Critique means slow, and thorough



jstevh@xxxxxxx wrote:
Jesse F. Hughes wrote: 

	... stuff deleted ...


You may ask how we know that there *exists* a least ring containing
both Z and 1/2, but that's a different (and easy) matter.  But the
fact is, if Z[1/2] denotes any ring at all, then it denotes a ring
contained in Q.

Honest, it does. 


Well, it'd be nice if it did, but it doesn't.


	AND HERE'S THE PROOF:

Think of coprimeness as being like a filled balloon.

If you prick the balloon by introducing something that contradicts any
coprimeness result in the ring of integers, like sticking in 1/2 so
that 2 is no longer coprime to 1, then you pop the balloon.


Now, *that's* some high-falutin' math, there!

Here's some more:

	Think of your balloon as a flaming sack placed
	on your doorstep just before someone rings the
	doorbell.

	Think of the partial sums of a convergent series
	as your foot, instinctively coming down to put
	out the flames.

	Think of the convergence of this series as your
	foot coming down hard on this flaming sack.

	Think of how much of your own concept of mathematics
	you're going to have to scrape off your shoe.

Why is your model of coprimeness any better than my model of
your model of mathematics?

Now you can *define* a minimal ring all you want, except that the ring
is infinite in size.  Because the ring is infinite in size, what you're
doing is like trying to look at only the foot of an elephant and
*define* that foot to be the entire elephant.


But an you can't fit an elephant, not even an elephant's foot, inside
a filled balloon. The balloon is FULL! Everyone knows that FULL means
that nothing else can get into that thing!

Mathematically the *definition* has no meaning, and you get infinite
sums whether you want them or not.

One way to look at it is that to the mathematics either you have
coprimeness rules or you don't.


Hmm. So, what you're saying, if I get the meaning correctly, is
that there is a single "coprimeness" concept, and it doesn't really
depend on what elements in the ring you're talking about, or even about
the ring itself. It's just "coprimeness".

If 1 and -1 are the only rationals units in the ring, like with
integers, plus my other requirement, then you have a ring where
coprimeness holds.


Your "other requirement" ? What "other requirement" would that be?

If you break the coprimeness rules even a little bit, you break them
completely.

You cannot break them halfway.

The idea that you can just stick in 1/2 and get a partial break is just
wrong.

It's a complete break.

It even sounds bogus when you think about it that arbitrarily you can
remove convergent sums just because they're infinite sums.

Hey, they converge, they're in there.


So, you're saying that the ring of algebraic integers is really the full
field of complex numbers, right? After all, given *any* complex number
z, there is a sequence (well, many sequences, to be sure, but at least
one) of algebraic integers that converges to z. By taking successive
differences of this sequence, we obtain an infinite sum of algebraic
integers that converges to z.

Explicitly: let z be any complex number, and {a_n | n >= 0} a sequence
of algebraic integers that converges to z.

For n >= 1, let b_n = a_n - a_(n-1). Then b_n is an algebraic integer
for each n >= 1.

Consider the infinite sum:

	a_0 + b_1 + b_2 + ... + b_n + ...

It is trivial to verify that for n >= 1 the nth partial sum is equal
to a_n:

	S_n = a_0 + b_1 + ... + b_n

	    = a_0 + (a_1 - a_0) + (a_2 - a_1) + ... + (a_n - a_(n-1))

	    = a_0 + a_1 - a_0 + a_2 - a_1 + ... + a_n - a_(n-1).

	    = a_0 - a_0 + a_1 - a_1 + a_2 - a_2 + ...

			+ a_(n-1) - a_(n-1) + a_n

	    = a_n.

Since the sequence {a_n | n >= 0} converges to the (arbitrarily chosen)
complex number z, by definition

	z = a_0 + sum(b_n, n=1,..., infinity).

According to your reckoning, then, z is in the ring of algebraic
integers.

For instance, 1/2 is an algebraic integer.

So, mathematically, despite what semantics you use, when you break the
integers, you get reals, whether you want to or not.


Here's the statement you want to assert as a theorem:

	CLAIM (JSH): If K is a field, and sum(k_n | n >= 0) is a
	convergent infinite series of elements of K, then the limit

		S = lim_{n --> infinity) (k_0 + ... + k_n)

	is an element of K.

Here's the corollary that you are then obliged to acccept:

	COROLLARY: pi is a rational number.

PROOF:
	1. Q, the set of rational numbers, satisfies the field axioms,
	   and is therefore a field.

	2. The finite set of decimal expansions of pi form the sequence
	   of partial sums for a particular series of rational numbers.

	3. The series of statement 2. converges to pi.

	4. The summation of that series, defined as the limit of the
	   sequence of partial sums, is equal to pi.

	5. pi is therefore an element of Q.

If the above CLAIM(JSH) is assumed, then the above is a correct proof,
modulo details about defining convergence (which I take it you assume).

The justification for each step (each may take a certain amount of work
to verify, but each step is simple to do):

	1. Definition of field, definition of rational numbers.
	2. Take the series formed by adding the sequence of differences.
	3. Convergence in R is guaranteed by the fact that the sequence
	   of finite initial segments of the decimal expansion of any
	   real number is a Cauchy sequence.
	4. Definition of the sum of an infinite series.
	5. CLAIM(JSH).

As I said, given your claim, you must then accept the rationality of pi.
By a similar token, with a different convergent series, you must accept
that pi is an algebraic integer, leading to the following

	ASTONISHING CLAIM (JSH)

		pi is an integer!

	PROOF:

		pi is a rational number
		pi is an algebraic integer

		The intersection of the algebraic integers
		with the rational numbers is equal to the
		set of (ordinary, rational) integers.

		==> pi is an element of Z.

So, not only must you agree that pi is rational, you must stand with
the Great State of Indiana in insisting that pi is an integer!

WOW!


James Harris


This is fun. Just how much trouble can a person get into by holding an
apparently innocent position, and disregarding the constraints imposed
by standard definitions?

Dale
.

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