What subset of the reals have unique prime factorizations if you allow rational exponents? By the fundamental theorem of arithmetic we know that all positive integers have a unique prime factorization. So if $n$ is some positive integer, then
$$n = \prod_{i\in\mathbb{N}}{p_i^{e_i}}$$
where $p_i$ is the $i$th prime number and $\{e_i\}$ is a sequence of natural (I'm including $0$) numbers, all but finitely many of which are $0$.
If we allow the elements of $\{e_i\}$ to be integers and not just naturals, then the subset of $\mathbb{R}$ with "prime factorizations" is $\mathbb{Q}$, but what happens when we allow the exponents to be rationals? What reals have a unique "prime factorization" then? It's some subset of the algebraic numbers, but I don't know if there's any other characterization of this set.
Also, in the case of integer or natural number exponents, it was necessary for all but finitely many exponents to be 0 for the product to converge, but given that rationals are dense in the reals, it seems like you could have products of the form
$$\prod_{i\in\mathbb{N}}{p_i^{e_i}}$$
that converge without needing only finitely many non-zero exponents so long as they converged to $0$ quickly enough. What subset of the reals have unique "factorizations" in this sense? It seems plausible to me that these sorts of factorizations might not be unique too.
 A: For the case where the factorizations are required to contain only finitely many nonzero exponents:
Let $\alpha$ be a nonzero element of $\mathbb{R}$ with a "rational exponent prime factorization": $$\alpha = \pm\prod\limits_{i \in \mathbb{N}} p_i^{r_i/s_i}$$
where $p_i$ are the positive primes, $r_i$ and $s_i$ are coprime integers, all the $s_i > 0$ and only finitely many of the $r_i$ are nonzero.
Then we may choose $d$ as the LCM of the finitely-many $s_i$ corresponding to these nonzero $r_i$.
We then find that $\alpha^d$ has a factorization where each exponent is an integer, so $\alpha^d$ must be rational.
Therefore, the subset of real numbers with such "rational exponent prime factorizations" consists of the real numbers that are $d$-th roots of rational numbers, for some $d \in \mathbb{N}$. Conversely, any $d$-th root of a rational number has an obvious factorization of the desired form.
The factorization in the form given above is unique. The result for positive and negative integer exponents is well-known (Fundamental Theorem of Arithmetic and properties of multiplication in $\mathbb{Q}$), and the case of rational exponents can be reduced to the previous case after raising to the $d$-th power.

To flesh the argument out in more detail... Consider two candidate factorizations: $$\alpha = \pm\prod\limits_{i\in\mathbb{N}}p_i^{r_i/s_i} = \pm\prod\limits_{i\in\mathbb{N}}p_i^{r_i'/s_i'}$$
The signs are uniquely defined ($+$ if $\alpha$ is positive, $-$ if $\alpha$ is negative), so we can take absolute values.
We can again define $d$ as the LCM of the $s_i$ associated with nonzero $r_i$, and similarly $d'$ as the LCM of the $s_i'$ associated with nonzero $r_i'$. We then raise to the $dd'$-th power:
$$|\alpha|^{dd'} = \prod\limits_{i\in\mathbb{N}}p_i^{r_id'\frac{d}{s_i}} = \prod\limits_{i\in\mathbb{N}}p_i^{r_i'd\frac{d'}{s_i'}}$$
By the choice of $d$ and $d'$, both products only contain integer exponents. By uniqueness of the factorization for integer exponents, we see that $r_id'\frac{d}{s_i} = r_i'd\frac{d'}{s_i'}$ for all $i \in \mathbb{N}$, from which it follows that $r_i/s_i = r_i'/s_i'$ for all $i \in \mathbb{N}$ (note that $d, d' \gt 0$).

Not all real algebraic numbers are of this form, however. The set of real numbers with "rational exponent prime factorization" can be described as the subset $S \subset \bar{\mathbb{Q}} \cap \mathbb{R}$: $$S = \{\alpha \in \mathbb{R}: \text{there exists some }d \in \mathbb{N}\text{ such that } \alpha^d \in \mathbb{Q}\}$$
So in particular this only includes elements in radical extensions of $\mathbb{Q}$, and it is well-known from Galois theory that not all real algebraic extensions of $\mathbb{Q}$ are radical (this is probably overkill for the current purposes, but it's enough to show that $S$ is a proper subset of the real algebraic numbers).

The case of infinitely many nonzero exponents in the factorization looks more difficult.
One issue you can run into is that lots of products converge to $0$, simply by choosing lots of negative exponents, for example $\prod\limits_{i\in\mathbb{N}}p_i^{-1} = 0$.
Besides this issue, it might also be possible that the factorizations of nonzero real numbers are not unique, but it requires finding sequences $a_i$ of rational numbers, not all zero, with the following property:
$$\prod\limits_{i\in\mathbb{N}}p_i^{a_i} = 1$$
This can also be rewritten as an infinite series:
$$\sum\limits_{i\in\mathbb{N}} a_i\log p_i = 0$$
So you would need to pick the sequence of rational exponents in such a way that the "weighted average" of the $\log p_i$ tends to $0$. I'm not sure at the moment whether this is possible or not. I think it is possible, the idea would be that you alternate positive and negative values for $a_i$, and you tweak their magnitudes in such a way that the partial sums always oscillate around $0$ with diminishing amplitude. If this idea works, it should also prove that all real numbers have an infinite prime factorization with rational exponents. But I haven't checked the details yet.

Here's a fully fleshed-out argument showing that all real numbers have a rational exponent prime factorization if you allow infinite products. Moreover, there are infinitely many distinct factorizations. This follows from the first claim by using nontrivial factorizations of $1$.
Given a real number $\alpha \in \mathbb{R}$, we wish to find a sequence of rational numbers $a_i \in \mathbb{Q}$ such that $\alpha = \pm\prod\limits_{i \in \mathbb{N}} p_i^{a_i}$. The case where $\alpha = 0$ was already discussed, and the case of negative $\alpha$ is trivially reducible to the remaining case of positive $\alpha$ by choosing the sign, so we assume $\alpha \gt 0$.
We may then take logarithms and define $L = \log \alpha$. Now the problem becomes, as mentioned before, that of finding a sequence of $a_i \in \mathbb{Q}$ such that $$\sum\limits_{i \in \mathbb{N}}a_i\log p_i = L$$
The sequence will be chosen in such a way that for every $i \in \mathbb{N}$, we satisfy $$|s_i - L| < \frac{1}{i+1}$$ from which it follows that the sum converges to $L$. Here $s_i$ is the sequence of partial sums, so $s_1 = a_1\log p_1$ and $s_i = s_{i-1} + a_i\log p_i$ for all $i \geq 2$.
The sequence is defined inductively. First choose some rational number $a_1$ such that $$\frac{L - \frac{1}{2}}{\log p_1} < a_1 < \frac{L + \frac{1}{2}}{\log p_1}$$
This makes sure that $|s_1 - L| < \frac{1}{2}$.
Now assume that for some $i \geq 2$, the values $a_1, \ldots, a_{i-1}$ were chosen so that $|s_{i-1} - L| < \frac{1}{i}$. There are two cases to consider for the choice of $a_i$.

*

*$L - \frac{1}{i} < s_{i-1} \leq L$:

Then we choose $a_i$ to be a rational number satisfying $$\frac{1}{i(i+1)\log p_i} < a_i < \frac{1}{(i+1)\log p_i}$$
For $i \geq 2$, this defines a nonempty open interval, and since $\mathbb{Q}$ is dense in $\mathbb{R}$, this means that such a choice is always possible.


*$L < s_{i-1} < L + \frac{1}{i}$:

Then we choose $a_i$ to be a rational number satisfying $$-\frac{1}{(i+1)\log p_i} < a_i < -\frac{1}{i(i+1)\log p_i}$$
And the choice is again possible for $i \geq 2$.
By induction, the sequence of rational numbers $a_i$ is constructed in such a way that $|s_i - L| < \frac{1}{i+1}$ for all $i \in \mathbb{N}$, from which the result follows.
