Re: Dial 999 for the real number line

On Tue, 01 May 2007 13:46:22 +0100, Six Letters wrote:

Given one irrational, say pi, there is obviously a kind of spread
to other irrational. Thus if:
.141592.......... (pi expansion)
is a number, then so is:
.131592......... (continuing as pi expansion),
changing the 4 in the 2nd decimal place to a 3.
And generalising, all finite permutations from the integers 0 to 9
length n as initial strings followed by the expansion of pi after the nth
decimal place, these are all good, paid up members of the real club too. We
can make n as large as we like, so one irrational number generates an
infinite number of others.

But there are only countably many real numbers that can be specified by a
formula or an algorithm, leaving uncountably many that cannot be so
specified.

One of those permutations could of course be the initial decimal
expansion of pi itself. More generally and directly for my purpose, we can
say that any irrational number, pi for example, consists of a finitely
long decimal expansion followed by an infinite expansion. Since the finite
part can be as long as we like, we have an infinite sequence of initial
expansions (already a Cauchy sequence) followed by, in each case, an
infinite expansion. Isn't the second infinite part superfluous?

Is 1/3 a real number? It doesn't have a finite decimal representation.

I believe it is superfluous, but it is necessary to understand
infinity correctly. Why can we not just posit some arbitrary irrational,
where an infinite decimal expansion is as it were an infinite collection of
choices for the digits in each place? Because there is not even an actual
infinite expansion for pi; better, what it means for there to be an
infinite expansion here is that there is some method (formula, any device)
for producing successive digits of pi without end.

I can specify the exact value of pi very compactly. Pi is the smallest
positive root of the unique function f satisfying the initial value
problem:

f + f'' = 0
f(0) = 0
f'(0) = 1

It's possible to give precise definitions of some numbers (even
transcendentals) without mentioning decimal digit strings.

The notion of infinity as without bound, limit or end, as radically
sizeless, which is the real lesson of the Galileo paradox, is facile and
cardinal. (Contrast the notion of infinite sets being equinumerous with
proper subsets, which I feel will have future historians of mathematics
chuckling for decades.) That there exists an infinite capacity to extract
the digits of pi, does not mean there literally exists an infinite decimal
expansion of pi. On the contrary, there is literally no end to the
specifying of the digits of pi.

Pi exists, whether we can write down all its digits or not. See the
definition above. A decimal representation is just that, a
representation. It is not the same thing as a number.

By the same token, the systematic listing of the reals in the
following table (in binary for convenience; -> means trailing 0s):

.0 ->
.1 ->
.0 1 ->
.1 1 ->
.0 0 1 ->
.0 1 1 ->
.1 0 1 ->
.1 1 1 ->
etc.

such a table represents all the reals (in the unit interval).

On which line of your table can I find the number 1/3?

To say that the list only produces rational numbers is like saying
the Leibnitz formula (for example) for pi only produces approximations to
pi. Any cut-off produces a rational number in either case, but the Leibnitz
formula is a formala for pi, not for approximations to pi. And the list of
reas is all of them (or if you like, there is no 'all' in such a case).
There are no other reals at the end of or besides this list. Indeed the
decimal system is a tool for constructing rationals. To think it incomplete
is to imagine some reality beyond the model. The model is the reality. So
where are these real ratios or distances, like pi or sq.rt of 2, in this
list? They are there, in the only way they could be, in the endless
successive approximations ocurring there.

Your table contains only binary digit strings, not formulas. On which
line of your table can I find pi? I will accept pi-3, since that is a
number in [0,1].

To believe in some arbitrary real, an infinite number of choices
for the decimal places, is like thinking one can count to infinity. This is
not a matter of ordinary human limitation. What's laughable about counting
to infinity (1,2,3.........) is that there is no end to it, so beginning is
misconceived. Be immortal, and your unceasing years will simply match your
unending task. Nor is it a matter of importing some notion of process
(first you have to say 1, then 2,..). It is simply that there is no end to
the business of making an infinite number of choices, sequentially or
simultaneously, and therefore, again, no real beginning either.

Is your real objection simply that you don't think infinite sets exist?
In that case, you don't accept the natural numbers, let alone the
rationals or the reals. Why don't you just say what your real objection
is?

Cauchy sequences are of no help. Of course, GIVEN some sequence, we
can prove that if it's Cauchy then it converges (and vica versa). But we
are not GIVEN a sequence by some arbitrary real. The Leibnitz formula gives
us a sequence. Any sort of pattern or rule will give us a sequence. But the
sequence that consists of an infinite number of arbitrary choices is a
mirage. Similar considerations apply to Dedekind cuts.
As far as I can tell, the reals so described would still constitute
technically a complete ordered field.

Wrong. The computable reals are an ordered field, but they are not
complete. That is, there is a set of computable reals that is bounded
above, but for which the least upper bound is not computable.


--
Dave Seaman
Oral Arguments in Mumia Abu-Jamal Case to be heard May 17
U.S. Court of Appeals, Third Circuit
<http://www.abu-jamal-news.com/>
.

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