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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

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

from 0

result

don't

idea is

that

integer is

between

That

specified

(also

is

this

idea of

integers:

perfect-squares.

we've

once in

we?

Received on Thu Jan 22 2004 - 06:46:40 PST

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