Aaron Toponce aaron.toponce at
Thu Jun 2 09:05:49 MDT 2011

On Wed, Jun 01, 2011 at 08:51:14PM -0600, Shane Hathaway wrote:
> On 06/01/2011 02:27 PM, Aaron Toponce wrote:
> > Similar proofs can be constructed for any countable set:
> Related to this, I've been wondering why irrationals are not considered
> countable.  Is it not true that for any irrational number, a computer
> program can be written that converges to that number as the number of
> iterations reaches infinity?  Any computer program can be represented as
> a large integer, so computer programs are countable, and by extension,
> any number that a computer program can represent (but not necessarily
> produce) ought to be considered countable.

Careful. _Some_ irrational numbers are countable. For example, sqrt(2) is
irrational, and the set conntaining sqrt(2)/n for n=0 to infinity is also
countable. The roots to some polynomials might be irrational, but also
algebraic, thus the set containing this solutions is countable.

In other words, don't mix up your terminology. Sets can be countable or
uncountable, based on what the set contains. If the set contains an
interval, finite or infinite, then the set is uncountable.

So, taking the set that contains the interval [0,1] is uncountable. Reason
being, you cannot make a 1:1 correlation of every element in that set to
every element in the set of natural numbers. Cantor gave a proof of this.
It's known as "Cantor's diagonal argument":'s_diagonal_argument

The proof is easy to read, and usually a homework exercise for ungeraduate
analysis students.

While it may be true that you can converge on any point in the interval,
you need infinity to pull it off, and not even the best quantum computer
has that much time nor processing power. So, we must accept some degree of
error, an epsilon distance, to which we'll rationalize the irrational, and
call it good.

. o .   o . o   . . o   o . .   . o .
. . o   . o o   o . o   . o o   . . o
o o o   . o .   . o o   o o .   o o o
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