3 ms·
> If you allow one to be a prime number, then you can no longer say that each natural number has a unique prime factorization. I don't dispute that, by definit
by Igelau 5y ago
> If you allow one to be a prime number, then you can no longer say that each natural number has a unique prime factorization.
I don't dispute that, by definition, 1 is not prime, but I don't see how this statement would follow if we considered it prime.
Edit: it seems more like it would be that every factorization would implicitly have 1^n tacked onto it, and while that isn't exactly useful, it doesn't break the game.
- drdec 5y agoThat breaks the uniqueness of the prime factorization. Mathematicians love uniqueness, almost as much as existence.
- Igelau 5y agoCould you please demonstrate an example of such a break? I think if we really look at it, we might see that it's really just convention and semantics. I don't want to seem like I'm cherry picking by providing my own example.
- drdec 5y agoThe unique prime factorization of 15 is 3⋅5. Let's call a number an Igelau prime if it is the number 1 or a prime number. Then an Igelau prime factorization of 15 is 3⋅5. Another Igelau prime factorization of 15 is 1⋅3⋅5. Another Igelau prime factorization of 15 is 1²⋅3⋅5. And so on. There are infinitely many Igelau prime factorizations of 15, thus there is no uniqueness. Edit: Clarifying what Igelau primes are.