Date: 2010-09-29 11:01 pm (UTC)
Actually, here's an interestingly similar sort of argument.

Let probability on natural numbers be defined in the way they say (that is, for any statement about numbers, choose a cutoff, see what the probability is that the statement holds for a uniformly chosen set of numbers below that cutoff, and take the limit as the cutoff tends to infinity).

Then, it's a standard result that the probability that two randomly chosen natural numbers have no factor in common is 6/π^2. (It's a nice exercise to show this, using the fact that 1+1/4+1/9+1/16+...=π^2/6, and rewriting that sum as the product of (1+1/p^2+1/p^4+...) as p ranges over all primes.)

However, in an argument parallel to theirs, it looks like we can say that the probability that a natural number is greater than every power of 2 is 1/2, since as n goes to infinity, the probability that a random number below n is greater than every power of 2 below n is close to 1/2. (I probably have to be a bit more careful in setting that up.)

Of course, in that case, it's obvious how I'm messing things up by taking two nested quantifiers and making them depend on the same n, rather than making the second depend on the first. And their argument is making the same mistake, I think.
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Domino Valdano

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