Two way representation of rational numbers with prime factorization. What's the connection?

Question

So whole numbers have a unique representation as the product of primes raised to a certain power. This representation can be extended to rational numbers too if we allow negative powers, since if $$a:=\prod_{i}^{N_a} p_i^{a_i}$$ and $$b:=\prod_{j}^{N_b} p_j^{b_j}$$ are coprimes then their ratio $$x:=\frac{a}{b} = \prod_{i}^{N_a} p_i^{a_i}\times\prod_{j}^{N_b} p_j^{-b_j}$$ is uniqely represented by the set of pairs $(a_i, i)$ and $(b_j, j)$. I denote this representation with $R_1(x)$. Rational numbers have another uniqe representation in the form of finite continued fractions $$x = x_0 + \frac{1}{x_1+\frac{1}{x_2+...}} = [x_0;x_1,x_2...x_N]$$ where each $x_i \in \Bbb{N^+}$. This can be reformulated in a way that $x$ is represented by the set of representations $$R_2(x):=\{(R_1(x_i),i)\,|\,0\le i\le N\}$$ My question basically is how can I go from $R_2$ to $R_1$ without re-calculating prime factorizations of sums arising during the trivial "$R_2$ to $R_1$" process (or is this even possible)? If not, then does $R_2$ contain any "useful" and "easily accessible" information about $R_1$?

Edit: I thought it would be easier to answer the question if you knew my motivation: I'd like to extend the concept of prime factorization further to the set of irrationals. Since we know the $R_2$ of an irrational it would be somehow logical to use this to obtain its $R_1$-like representation, that should fulfill the uniqueness, but can be infinitely long.

Answer 1

I do not believe that continued fractions contain useful information about the prime factorization of the numerator and denominator. Given two large coprime integers $a, b$ of similar magnitude, it is quite fast (using the Euclidean algorithm) to construct the continued fraction of $a/b$, which usually will involve fairly small terms. On the other hand, factoring $a$ and $b$ is notoriously difficult.

Answer 2

When expanded, the $R_2$ form is a ratio of two sums of some products of $x_i$. Given you have the prime factorization of every $x_i$, and a rule that gives you a prime factorization of a sum of two prime factorizations, you could do it. You could perhaps have the first relatively cheaply (so that you are not "reducing" prime factorization to n prime factorizations), but I doubt you will ever have the second.

But think this way. What if you extend the domain of prime exponents in prime factorization to rational, just like you extended them from natural numbers to all integers? You introduce roots, such as $\sqrt2 = 2^{1\over2}$. So you might not have them all, but you already have some irrationals. Such set would probably match the algebraic numbers or some subset of them, i'm not sure, let's call them hyper-rationals. But then you could extend the domain of prime exponents to hyper-rationals and get hyper-hyper-rationals (or maybe hyper-algebraics?). Important thing to notice here is that this way you cannot extend the domain of uniquely prime-factorizable numbers to all reals, since this way you never increase the cardinality of the set. Naturals, integers, rationals, algebraics, they are all countably infinite sets. Hyper-algebraics will be too, as well as any set created by finite number of applications of this method. Maybe applied infinitely many times? Dunno. But that's probably not helpful for you.

You could also straightforward allow real prime exponents, but now the prime factorizations are no longer unique, consider $2 = 3^{log_3 2}$. Many reals are not computable, and having an easily expressible prime factorization probably makes a number computable. But most of the numbers we use in our everyday math applications, even many transcendentals, such as e and pi are computable. Perhaps if you lower your ambition to being able to express only interesting reals uniquely with some kind of prime factorization, you might get somewhere. Maybe you will need to extend the definition of primes so that they include some irrationals.