What is the prime factorization of zero?

Fundamental theorem of arithmetic1: every whole number can be split into a product of prime numbers in a unique way.


\[6 = 2 \cdot 3 = (2,3)\] \[8 = 2 \cdot 2 \cdot 2 =2^3 = (2, 2, 2)\]

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 all2.

So \(F\):

  1. Has to be infinite length, else dropping a term from it would strictly shorten it.
  2. 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.

\[0 = (2^\infty,3^\infty,5^\infty...) = \prod_p p^\infty\]

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

  1. Which deserves its name more than any other fundamental theorem, even the fundamental theorem of calculus. 

  2. not gonna touch \(\frac{0}{0}\) idk with that one 

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