If $1$ was a prime, could it be possible for the prime factorization of any number to go on forever?

Question

I think this would happen because if you multiply anything by $1$, you get the first factor itself, for example $6 \dot\ 1 = 6$ and $-2 \dot\ 1 = -2$.

That would not be good. What do you think about this? Am I right?

"If $1$ was a prime" is as good a premise as "if pigs could fly". — Hagen von Eitzen, Dec 10 '14 at 18:28
I read somewhere that the ancient Greeks considered $1$ to not have parity, to be neither odd nor even. — , Dec 10 '14 at 19:08
I agree it wouldn't be good. Thank goodness $1$ isn't a prime, right? We really dodged a bullet there. — Qiaochu Yuan, Dec 11 '14 at 06:06

score 0 · Answer 1 · answered Dec 03 '14 at 01:34

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I think I'm right because I just said that anything times 1 is equal to the factor being multiplied by 1. This would make the prime factorization of any number go on forever and ever and ever. This is also why 1 isn't considered as a prime number. So, this might be why (1^$\infty$ x 3 = 3 (this might not be true because 1^$\infty$ is undefined)).

answered Dec 03 '14 at 01:34

VladInTheTaylor

339

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I think bringing up $\infty$ as an exponent only muddles the issue. – Dec 10 '14 at 18:30

score 0 · Answer 2 · answered Dec 10 '14 at 18:24

That's in part why we have the concept of distinct factorization. For example, in $6 = 1^{548} \times 2 \times 3 = (-1)^8 \times 2 \times 3$ these are not considered distinct factorizations because we're multiplying by units. In a unique factorization domain, each number has infinitely many factorizations but only one distinct factorization. And in some non-UFDs like $\mathbb{Z}[\sqrt{-5}]$, a number might have more than one distinct factorization, but the total number of distinct factorizations is still finite.

score 0 · Accepted Answer · answered Dec 11 '14 at 06:04

A factorization has to have finite exponents but there can be infinitely many factorizations. So $3 = 1^{489} \times 3 = 1^{490} \times 3 = 1^{491} \times 3 = \ldots$

As you already know, $(-1)^k = -1$ if $k$ is odd and 1 if $k$ is even. So we have $-3 = (-1)^{489} \times 3 = (-1)^{491} \times 3 = (-1)^{493} \times 3 = \ldots$

And what about if in $\mathbb{Z}[i]$ we regard $i$ as prime? Or if in $\mathbb{Z}[\sqrt{2}]$ we regard all numbers of the form $(1 - \sqrt{2})^k$ as primes? Obviously $(1 - \sqrt{2})^k 3$ grows larger as $k$ grows, but there's something about 3 that doesn't actually change: tell me, what is $(9 - 6\sqrt{2})(9 - 6\sqrt{2})$? What about $(21 - 15\sqrt{2})(21 + 15\sqrt{2})$? $(51 - 36\sqrt{2})(51 + 36\sqrt{2})$?

But we don't have to get into algebraic number theory to understand that 1 differs from the prime numbers in far more important ways than a couple of asterisks on the fundamental theorem of arithmetic.

If $1$ was a prime, could it be possible for the prime factorization of any number to go on forever?

3 Answers3