TITLE: Prime decomposition over the real numbers?
QUESTION [9 upvotes]: Over the natural numbers, each number can be uniquely decomposed as a product of prime numbers. Can we generalize this idea to the reals? Here is my attempt:
Theorem
Can there exist a subset $P\subseteq \mathbb R$ of the reals such that

For every $0\neq x\in\mathbb R$ exist $F_x := \{(x_1,p_1),\dots,(x_n,p_n)\}\subseteq P\times \mathbb N $ with $x=\prod\limits_{i=1}^n x_i^{p_i}$
For every $x\neq \pm 1$ this decomposition is unique, aside from a representation of $\pm 1$ (i.e. For 2 decompositions $F_x=\{(x_1,p_1),\dots,(x_n,p_n)\}$ and $F_x'=\{(x_1',p_1'),\dots,(x_m',p_m')\}$ there exist $F_1:=\{(z_1,q_1),\dots,(z_a,q_a)\}\subseteq X$ and $F_1':=\{(z_1',q_1'),\dots,(z_b',q_b')\}\subset X'$ with $\prod\limits_{i=1}^p z_i^{q_i}=\pm 1$ and $\prod\limits_{i=1}^q z_i'^{q_i'}=\pm 1$ such that $F_x\setminus F_1 = F_x'\setminus F_1'$

This is essentially Integer factorization generalized on the reals with an additional condition to allow uniqueness. Obviously this set is ambiguous, but the decomposition for each number, given such a set $P$, is unique.
My attemps/results until now
First of all, I am pretty sure this theorem is true. I am aware of the fact that it is very likely to involve the Axiom of Choice.

If $P$ exists, then it is uncountable.
Why? If $P$ were countable, the set of finite subsets of $P$ is countable. Per definition of $P$, that is equinumerous to $\mathbb R$ via $\sigma:\{(x_1,p_1),\dots,(x_n,p_n) \} \mapsto \prod\limits_{i=1}^nx_i^{p_i}$.
$\sigma:\mathbb (Q\times N)^*\to \mathbb R\setminus \{0\}$ ($(Q\times \mathbb N)^*$ is the set of all finite subsets on $Q\times \mathbb N$, $Q\subseteq\mathbb R$) is surjective and multiplicative, i.e. $A\cap B=\varnothing \Rightarrow \sigma(A\dot\cup B)=\sigma(A)\sigma(B)$
$X \sim Y :\Leftrightarrow X$ only differs by a representation of 1 from $Y$ is a equivalence relation.
Why? reflexivity and symmetry are trivial. For transitivity, $X\sim Y\Leftrightarrow X\setminus F_1 = Y\setminus F_1' \Leftrightarrow \sigma (X) = \sigma(X\setminus F_1\; \dot\cup\; F_1)=\sigma(X\setminus F_1)\cdot 1 = \sigma(Y\setminus F_1')\cdot 1=\sigma(Y)$
Over the rationals, such a (countable) set exists: $P_{\mathbb Q}\bigcup\limits_{p\text{ prime}}\{p,\frac{1}{p}\}$
Iteratively adding numbers, which arent expressible by the numbers in the set, to $P_\mathbb Q$ works fine, yet will never yield the set we are searching for.

I am having trouble with finishing the last argument, i.e. doing the step from countably to uncountably infinite elements.
So how can you prove this theorem, or is there a counterexample?

REPLY [2 votes]: The prime factorization of an integer represents a pairing function. If we consider the prime factorization of an integer $n = p_1^{k_1} *  p_2^{k_2} *  p_3^{k_3} ..... *  p_m^{k_m} $
Then we can form a pairing function where $P(n)=\left< k_1,k_2,....k_m \right>$
Some values of $P(N)$:
\begin{align}
P(1) &= \left<0,0\right>\\
P(2) &= \left<1,0\right>\\
P(3) &= \left<0,1\right>\\
P(4) &= \left<2,0\right>\\
P(5) &= \left<0,0,1\right>\\
\end{align}
Since the values of $PP$ are points in multidimensional space, we can connect these points and then get a unique canonical factorization of all reals.
The convex hull is convenient for this. 
Say $P(n)=\left< k_1,k_2,....k_m \right>$
and $P(n+1)=\left< j_1,j_2,....j_m \right>$
The convex hull is equal to $(x) * P(n) + (1-x) * P(n+1) $ for $ 0 \leq x \leq 1.0 $ and will give us all points in the line from $P(n)$ to $P(n+1)$
Consider the inverse function:
$P^{-1}(\left< k_1,k_2,....k_m \right>) = p_1^{k_1} *  p_2^{k_2} *  p_3^{k_3} ..... *  p_m^{k_m} $
It is clearly continuous, as exponentiation is a continuous function. This means that taking the inverse of the convex hull will give us a unique canonical factorization for all reals between n and n+1. 
Consider all reals between 3 and 4.
Let us look at plot of $P^{-1}((x)*P(3) + (1-x)*P(4))$ for $x \in [0,1]$

Perfect, we capture all the reals like we desired.
Let us finalize our solution. 
Given a positive real number $r$ (including naturals), we must solve this equation in order to get the unique canonical prime factorization that corresponds to it:
$r = P^{-1}((x)*P(\lfloor r \rfloor) + (1-x)*P(\lfloor r \rfloor + 1))$ for $x \in [0,1]$
We can simplify the equation and solve it in the general case.
Let $u = P(\lfloor r \rfloor),\ v = P(\lfloor r \rfloor + 1) $
Then $ \displaystyle r = \prod\limits_{n} p_n^{(1-x) * u_n + (x) * v_n} $
Since $p^k = e^{Log(p) * k}$ , we can rewrite the right hand side as a natural exponentiation so all powers will add.
$
\begin{align}
\displaystyle r &= e^{(\sum\limits_{n} (1-x) * Log(p_n) * u_n + (x) * Log(p_n) * v_n)} \cr
Log(r) &= {\sum\limits_{n} (1-x) * Log(p_n) * u_n + (x) * Log(p_n) * v_n} \cr
Log(r) &= x {\sum\limits_{n} Log(p_n) * v_n - Log(p_n) * u_n} + {\sum\limits_{n} Log(p_n) * u_n} \cr
x &= \dfrac{(Log(r)-{\sum\limits_{n} Log(p_n) * u_n})}{\sum\limits_{n} Log(p_n) * v_n - Log(p_n) * u_n} \cr
\end{align}
$
Here are some neat examples of the factorization:
In[383]:= CanonicalRealFactorization[$\Pi$]
Out[383]= $\left<2\to \frac{2 (\log (\pi )-\log (3))}{2 \log (2)-\log (3)},3\to 1-\frac{\log (\pi )-\log (3)}{2 \log (2)-\log (3)}\right>$
In[384]:= CanonicalRealFactorization[$e$]
Out[384]= $\left<2\to 1-\frac{1-\log (2)}{\log (3)-\log (2)},3\to \frac{1-\log (2)}{\log (3)-\log (2)}\right>$
In[388]:= CanonicalRealFactorization[$e^{10}$]
Out[388]= 
$$
\begin{align}
\displaystyle
&\left< 2\to 1-\frac{10-\log (2)-\log (3)-\log (3671)}{-\log (2)-\log (3)-\log (3671)+\log (22027)}, \right. \cr 
&\left.3\to 1-\frac{10-\log (2)-\log (3)-\log (3671)}{-\log (2)-\log (3)-\log (3671)+\log (22027)}, \right.  \cr 
&\left.3671\to 1-\frac{10-\log (2)-\log (3)-\log (3671)}{-\log (2)-\log (3)-\log (3671)+\log (22027)}, \right. \cr
  &\left.22027 \to \frac{10-\log (2)-\log (3)-\log (3671)}{-\log (2)-\log (3)-\log (3671)+\log (22027)}  \right>
\end{align}
$$
In[385]:= CanonicalRealFactorization[$\phi$]
Out[385]= $\left<1\to 1-\frac{\log (\phi )}{\log (2)},2\to \frac{\log (\phi )}{\log (2)}\right>$
I've included Mathematica code for the plot, as well as the canonical factorization.
CanonicalRealFactorizationPlot[r_] := Block[{
    p1 = FactorInteger@Floor@r,
    p2 = FactorInteger@Floor[r + 1],
    f1, f2
    },
   f1 = Product[p[[1]]^(p[[2]]*(1 - x)), {p, p1}];
   f2 = Product[p[[1]]^(p[[2]]*(x)), {p, p2}];
   (* plot convex hull of two points *)
   Plot[f1*f2, {x, 0, 1}]
   ];

CanonicalRealFactorization[r_] := Block[{
    p1 = <|Rule @@@ FactorInteger@Floor@r|>,
    p2 = <|Rule @@@ FactorInteger@Floor[r + 1]|>,
    c, d, x
    },
   c = Plus @@ KeyValueMap[Log[#1]*#2 &, p1];
   d = Plus @@ KeyValueMap[Log[#1]*#2 &, p2];
   x = (Log[r] - c)/(d - c);
   Merge[{(1 - x)*p1, (x)*p2}, Total]
   ];

Example usage:
In[332]:= CanonicalRealFactorizationPlot[3.4]

Out[332]= (*Plot Graphic*)

In[333]:= CanonicalRealFactorization[3.4]

Out[333]= <|3 -> 0.564925, 2 -> 0.870149|>

In[334]:= 2^0.8701490634247632`*3^0.5649254682876184`

Out[334]= 3.4

Below is a quick visualization of the 2d factorization point space for the interval [1,4] (because the point space becomes 3d at 5.)
pts = Table[
   Lookup[CanonicalRealFactorization[r], {2, 3}, 0]
   , {r, 1, 4, 0.01}];
ListPlot@pts