Fundamental theorem of arithmetic^{1}: every whole number can be split into a product of prime numbers in a *unique* way.

Examples:

\[6 = 2 \cdot 3 = (2,3)\] \[8 = 2 \cdot 2 \cdot 2 =2^3 = (2, 2, 2)\]What is prime factorization of one? An empty sequence, because it is the multiplicative identity and an empty product is the identity.

Now given this theorem, let’s think of 6 from a different POV:

\[6 = (2,3)\]Then dividing by a prime p is equivalent to dropping a term of value p from the sequence on the right hand side.

Now let’s take a bold step and declare 0 has a factorization. Name it \(F\)(actorization).

\[0 = F\]But \(\frac{0}{2} = 0\), so dropping a term from \(F\) should give us the factorization of zero. Which is \(F\) itself!

This dropping property holds for any prime \(p\), and by extension any nonzero number at all^{2}.

So \(F\):

- Has to be infinite length, else dropping a term from it would strictly shorten it.
- Contains every prime infinitely many times since 0 can be divided arbitrarily many times without changing.

So the prime factorization of 0 is *all* the prime numbers, repeated a infinite number of times. Unlike every other standard natural number, not a finite sequence.

So if you take every prime and multiply them all by each other infinitely many times, you get zero. Trippy.