A bit of everything

# Prime decomposition

Every integer (superior to $0$) can be expressed as a product of prime numbers. In other words, for any $n \in \mathbb{N} \setminus \{ 0 \}$,

$$n = \prod_{i = 0}^{\inf} p_i^{c(i)}$$

where $p_i$ is the $i$-th prime number and $c_i$ is an integer-valued function of $i$. Furthermore, this decomposition is unique, i.e., there is a bijection between sequences $\{ c(i) \}_i$ and $\mathbb{N}$. There is no specific need for the product to be infinite, but it will simplify some aspects of it somewhat (the decomposition is unique without referring to a specific number of factors, avoids situations like $1 = 2^0 = 3^0$).

## Basic proof

First, let's show that such a decomposition does indeed exist. A prime number is defined as any integer with exactly two different divisors, $1$ and itself (this definition is here specifically to prevent $1$ from being considered as a prime number, which invalidates the theorem). We define our infinite product in the usual way :

There exists an integer $n$ such that

$$n = \prod_{i = 0}^{\inf} a_i$$

if, for every $N \in \mathbb{N}$, there exists an integer

$$n_N = \prod_{i = 0}^{N} a_i$$

such that the sequence $\{ n_N \}$ converges to some value.

Let's consider three cases here : $n = 1$, $n$ is a prime number, and $n$ is neither $1$ nor a prime number.

### Decomposition of $1$

The most obvious case for $1$ is simply the sequence $\{ 0 \}$. This can be shown by recursion. Consider the sequence

$$n_N = \prod_{i = 0}^{N} p_i^{0}$$

For $N = 1$, we have

$$n_1 = p_1^{0} = 1$$

and assuming that $n_N = 1$, we also have

$$n_{N + 1} = 1 \times p_{N+1}^0 = 1 \times 1 = 1$$

Therefore $n_N = 1$ for every value of $N$. For any other sequence of integers, we have that the associated number is not $1$, since for any sequence $\{ a_i \}$ with any $a_i \neq 0$, we have

$$\prod_{i = 0}^{\infty} p_i^{a_i} = 1 \cdot \prod_{a_i \neq 0} p_i^{a_i}$$