RE: Is the universe computable

From: David Barrett-Lennard <dbl.domain.name.hidden>
Date: Thu, 22 Jan 2004 09:41:57 +0800

Yes, I agree that my definition (although well defined) doesn't have a
useful interpretation given your example of perfect squares interleaved
with the non perfect-squares.

- David

> -----Original Message-----
> From: Kory Heath [mailto:kory.heath.domain.name.hidden]
> Sent: Wednesday, 21 January 2004 8:30 PM
> To: everything-list.domain.name.hidden
> Subject: RE: Is the universe computable
>
> At 1/21/04, David Barrett-Lennard wrote:
> >Saying that the probability that a given integer is even is 0.5 seems
> >intuitively to me and can be made precise (see my last post).
>
> We can say with precision that a certain sequence of rational numbers
> (generated by looking at larger and larger finite sets of integers
from 0
> -
> n) converges to 0.5. What we can't say with precision is that this
result
> means that "the probability that a given integer is even is 0.5". I
don't
> think it's even coherent to talk about "the probability of a given
> integer". What could that mean? "Pick a random integer between 0 and
> infinity"? As Jesse recently pointed out, it's not clear that this
idea is
> even coherent.
>
> >For me, there *is* an intuitive reason why the probability that an
> >integer is a perfect square is zero. It simply relates to the fact
that
> >the squares become ever more sparse, and in the limit they become so
> >sparse that the chance of finding a perfect square approaches zero.
>
> Once again, I fully agree that, given the natural ordering of the
> integers,
> the perfect squares become ever more sparse. What isn't clear to me is
> that
> this sparseness has any affect on "the probability that a given
integer is
> a perfect square". Your conclusion implies: "Pick a random integer
between
> 0 and infinity. The probability that it's a perfect square is zero."
That
> seems flatly paradoxical to me. If the probability of choosing "25" is
> zero, then surely the probability of choosing "24", or any other
specified
> integer, is also zero. A more intuitive answer would be that the
> probability of choosing any pre-specified integer is "infinitesimal"
(also
> a notoriously knotty concept), but that's not the result your method
is
> providing. Your method is saying that the chances of choosing *any*
> perfect
> square is exactly zero. Maybe there are other possible diagnoses for
this
> problem, but my diagnosis is that there's something wrong with the
idea of
> picking a random integer from the set of all possible integers.
>
> Here's another angle on it. Consider the following sequence of
integers:
>
> 0, 1, 2, 4, 3, 9, 5, 16, 6, 25 ...
>
> Here we have the perfect squares interleaved with the non
perfect-squares.
> In the limit, this represents the exact same set of integers that
we've
> been talking about all along - every integer appears once and only
once in
> this sequence. Yet, following your logic, we can prove that the
> probability
> that a given integer from this set is a perfect square is 0.5. Can't
we?
>
> -- Kory
Received on Thu Jan 22 2004 - 06:46:40 PST