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Han de Bruijn wrote:
Following the style of reasoning a la Max Planck, we could say:
Please, don't blame Planck for this.
The chance of picking an arbitrary natural from all naturals N is 1/N .
This is an infinitesimal. Integration of these infinitesimals is delayed
until we are asked for a finitary result.
What is the probability that a natural is relative prime to 2, 3 and 5?
Integration starts.
And the outcome is 4/15.
You've been shown over and over that this does not work.
Denote your infinitesimal weight for each natural by e,
then denoted by w(n) the weight of each natural, n,
we have w(n)=e.
Now, you claim that sum_{n \in N} = 1.
Next, let S be any infinite subset of the naturals,
say the set of multiples of naturals relatively
prime to 2, 3 and 5. Then the weight of
S (i.e. the probability of choosing an element of
S) is sum_{n \in S) w(n).
Denote by n_i the ith element of S. Then this
probability is
sum_{i \in N} w(n_i).
But w(n_i) = e = w(i), so
sum_{i \in N} w(n_i) = sum_{i \in N} w(i) = 1
again.
Of course, the complement of S, say T, is also infinite,
so the weight of T is 1 as well.
But then, the weight of S plus the weight of T is 1+1=2,
but SuT is the naturals, which has weight 1, so it's a
consequence of your system that 1+1=2.
Yep, that's a *real* improvement on the standard
approach of using asymptotic densities and accepting
that you can't put a uniform measure on the
naturals.
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