Re: Rational numbers, irrational numbers: each dense in real numbers
- From: "Timothy Golden BandTechnology.com" <tttpppggg@xxxxxxxxx>
- Date: Wed, 24 Oct 2007 02:28:17 -0700
Hi Ross. I have to admit that I have barely even read your post yet I
think I understand it.
When we consider the concept of continuum then we do not need real
analysis or its portions which develop as the rational and irrational
numbers. If we simply consider the differential from calculus and
accept there there is always something smaller then we can have
continuum.
To what degree is continuum at the heart of your argument?
In some ways the extension to this argument is that the division
operator is not a necessary part of numerical representation and so as
a complication it should be eliminated.
The division operator is only semiexistent. Division is merely the
inverse of a product operator. So the construction of numerical
representations such as 1/3 are not fundamental.
-Tim
On Sep 18, 3:21 pm, "Ross A. Finlayson" <r...@xxxxxxxxxxxxxxx> wrote:
In ZFC, with standard definitions of the real, rational, and
irrational numbers, let p_i be an irrational number between zero and
one for i from a suitably large well-ordered index set X. With the
well-ordering of the index set, let the i'th element p_{i+1} be an
irrational number between zero and p_i, where i+1 is the least element
of the well-ordering X_i setminus i, that is defined to equal X_{i
+1}. There are uncountably many irrational numbers less than each p_i
and greater than zero, else the irrational numbers are countable (or
the real numbers are not standard). Define P to be comprised of the
p_i's. There exists a rational number q_i between p_i and p_{i+1},
else the rational numbers are not dense in the reals thus that between
any two irrational numbers there is a rational number. For each of
the irrational p_i's, there thus exists at least one unique rational
q_i between p_i and p_{i+1}, and infinitely many. Let the ordered
pair (p_i, q_i) be an element of a function, as a set, from P to Q.
If there is an uncountable set P of irrational numbers in (0,1), then
there is a 1-to-1 function defined by the set {(p_i, q_i), i E X} from
uncountable P to a subset of Q the rational numbers, and there is thus
an injection from an uncountable set of irrational numbers to a subset
of the rational numbers, a subset of a countable set is countable.
Contradiction ensues, from that an uncountable set injects into a
countable set. Which presupposition is false? Perhaps it is so that
the irrationals can not be well-ordered in their normal ordering, but,
via separation, for any given subset of the irrational numbers in (0,
p_i) there exists (0, p_{i+1}) for any p_i such that 0 < p_{i+1} <
p_i, or the irrationals are not dense in the reals. Perhaps it is so
that there are no uncountable subsets of the irrationals in (0,1), but
then the irrationals wouldn't be uncountable. The rationals are dense
in the reals so between any distinct p_i and p_{i+1} there exists a
q_i, or p_i = p_{i+1} and p_i =/= p_{i+1}. In ZFC there exists a
suitably large well-ordered index set.
Quantify over sets, ZF(C) and/or the standard definitions of the real,
rational, and irrational numbers are thus inconsistent.
Ross
--
Finlayson Consulting
.
- Prev by Date: Re: smallest positive integer that has exactly k divisors
- Next by Date: Re: number of topologies on all non-empty subsets of R
- Previous by thread: Re: Rational numbers, irrational numbers: each dense in real numbers
- Next by thread: solution manual
- Index(es):
Relevant Pages
|
