# For two different products of primes with rational powers, are the real number representations always unique?

So from the fundamental theorem of arithmetic we have that every $$\prod_{i=0}^{n}p_{i}^{e_{i}}$$, for some $$e_i \hspace{2px} \epsilon \hspace{2px} \mathbb{N}$$ and prime numbers $$p_i$$, gives a unique number in $$\mathbb{N}$$ and the set of all numbers generated in this way equals $$\mathbb{N}$$. This can be generalised to have a unique representation of every positive element of $$\mathbb{Q}$$ if $$e_i \hspace{2px} \epsilon \hspace{2px} \mathbb{Z}$$.

My question is can we generalise this further to have a unique representation of all elements of the set containing all $$x$$ such that $$x = \prod_{i=0}^{n}s_{i}^{e_i} ,\hspace{3px} e_i \hspace{2px} \epsilon \hspace{2px} \mathbb{Q} ,\hspace{3px} s_i \hspace{2px} \epsilon \hspace{2px} \mathbb{N}$$ using products of prime numbers with rational powers, or is there some case in which $$\prod_{i=0}^{n}p_{i}^{e_{i}} = \prod_{i=0}^{m}q_{i}^{f_{i}}$$ for some $$e_i, f_i \hspace{2px} \epsilon \hspace{2px} \mathbb{Q}$$ and primes $$p_i$$ and $$q_i$$ where not all $$e_k = f_k$$ or not all $$p_k = q_k$$? I suspect that this is the case but I can not think of a way to prove it myself.

On a side note, is the set I defined above the complete set of algebraic numbers or are there elements of the algebraic numbers that do not appear in that set? I doubt the set is the full set of algebraic numbers but it would be useful to know if they are equivalent, or if the set has a special name of some kind.

• $i$ is algebraic and it is not in your set. – user657449 Apr 19 at 5:57

If $$N$$ is the common denominator of all $$e_i$$ and $$f_i$$, we find $$\prod p_i^{Ne_i}=\left(\prod p_i^{e_i}\right)^N=\left(\prod q_i^{f_i}\right)^N=\prod q_i^{Nf_i}$$ and hence (up to permutation) $$q_i=p_i$$, $$e_i=f_i$$. In other words, the representation is still unique if we use rational exponents. At the same time we see that all representable numbers are of the form $$\sqrt[N]M$$ with $$N,M\in\Bbb N$$.