# Is it useful to think of the natural numbers as a powerset of the primes?

As a consequence of the fundamental theorem of arithmetic, it seems that the powerset of the prime numbers uniquely identifies each natural number, $\mathbb{N_1}=\mathcal{P}(\mathbb{P})$ (here I'm assuming that the empty set corresponds to $1$). As someone who is still making the connections in mathematics, it would be of interest to know if this is a useful construction of the natural numbers - and if so, to what purpose?

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Square-free supernatural numbers are isomorphic to the power set of the primes: the union and intersection operations on the latter correspond precisely to the least-common-multiple and greater-common-divisor operations on the former. – mjqxxxx Jan 20 '12 at 16:29

This is false for two reasons: (a) infinite sets of primes do not determine a natural number; (b) your proposed representation can not account for natural numbers with a repeated prime factor.

Here are two statements which are true:

The set of square-free natural numbers corresponds naturally to the set of finite sets of primes.

The set of all natural numbers corresponds naturally to the set of finite multi-sets of primes.

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Ah silly me - I realize that the construction forgets about 4 (talk about not checking the small cases!). I don't understand point (a) however, can you elaborate? – Hooked Jan 20 '12 at 14:53
@Hooked, take for example, the set of all primes greater than 10. That's a member of the power set of primes, but which natural number would it correspond to? – Henning Makholm Jan 20 '12 at 14:55
@Henning A supernatural number. – Bill Dubuque Jan 20 '12 at 14:59