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 Wed Jan 21 2004 - 07:35:50 PST
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