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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.

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    $\begingroup$ $i$ is algebraic and it is not in your set. $\endgroup$ – user657449 Apr 19 at 5:57
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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 $N,M\in\Bbb N$.

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