Explanation of the statement of the fundamental theorem of arithmetic.

Question

It's from the book Mathematics Made Difficult.

If $L^+(P,N_0)$ is the set of functions $f:P\rightarrow N_0$ with a property such that $$\exists\; n_0 \in N_0 \; \forall \; p \in P \;$$

$$ p\ge n_0 \implies f(p) = 0 $$ then there exists a bijection $N_1 \rightarrow L^+(P,N_0) $ such that if $n \rightarrow f$ then $$n = \prod_{p\in P} p^{f(p)}$$ Here $P$ is the prime numbers and $N_1 = N_0 - \{0\} $

I know this is related to the fundamental theorem of arithmetic and the book mentioned above made this statement a bit complicated, intentionally. Out of curiosity, I am interested in what this statement is actually saying?

Answer 1

Here $L^+$ is the set of functions which takes a prime number $p$ and outputs some natural number $n$ with the bonus property that $f(p) = 0$ for all but finitely many primes.

For instance, we may consider the function

$$ \begin{cases} g(2) = 3 \\ g(3) = 1 \\ g(5) = 0 \\ g(7) = 3 \\ g(p) = 0 & \text{for all other primes} \end{cases}. $$

The theorem is saying that the set of all such functions canonically corresponds to the natural numbers. How? Perhaps a definition by examples is the best option:

To $g$ as defined above, we can associate this function with the number

$$ 8232 = 2^3 \cdot 3^1 \cdot 5^0 \cdot 7^3 = 2^{g(2)} \cdot 3^{g(3)} \cdot 5^{g(5)} \cdot 7^{g(7)} $$

Here we think of $g(p)$ as telling us what the exponent on $p$ should be. The fact that all but finitely many $g(p)$ are $0$ means that we get a finite number when we multiply them all together (do you see why?).

Conversely, say we're given a number like $80864$. Then we can factor it as $2^5 \cdot 7 \cdot 19^2$, and so we associate it to the function

$$ \begin{cases} f(2) = 5 \\ f(7) = 1 \\ f(19) = 2 \\ f(p) = 0 & \text{for all other primes} \end{cases} $$

It's easy to see that these two maneuvers undo each other, and so they define a bijection.

This is a fancy way of saying that we can identify a natural number with the exponents of its prime factors. But this is a fancy way to say that every natural number admits a unique prime factorization. I.e. the fundamental theorem of arithmetic.

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I hope this helps ^_^

Answer 2

The statement essentially says that each natural number is a unique product of prime numbers.

Answer 3

Here's one way to "translate" it.

If $L^+(P,N_0)$ is the set of functions $f:P\rightarrow N_0$ with a property such that $$\exists\; n_0 \in N_0 \; \forall \; p \in P \;$$

$$ p\ge n_0 \implies f(p) = 0 $$

Define $L^+(P,N_0)$ as the set of all expressions of the form

$$2^{x_2} \ 3^{x_3} \ 5^{x_5} \ \cdots,$$

where the bases are the prime numbers, and $x_2, x_3, x_5, \ldots$ are non-negative integers, only finitely many of which are nonzero.

In other words, $L^+(P,N_0)$ is the set of all possible prime factorizations.

then there exists a bijection $N_1 \rightarrow L^+(P,N_0) $ such that if $n \rightarrow f$ then $$n = \prod_{p\in P} p^{f(p)}$$

Then there is a function which maps positive integers to prime factorizations, such that the value of the prime factorization is (as we would hope) the number which produced it. Furthermore, this function is a bijection.

In other words, each positive integer has exactly one prime factorization, and furthermore, each possible prime factorization is in fact the prime factorization of exactly one positive integer.