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The fundamental theorem of arithmetic gives us a unique way of writing any non-zero integer. For any $n \in \mathbb{Z}^*$, we have a unique decomposition : $$n = (-1)^\epsilon \prod\limits_{i \in \mathbb{N}^*} p_i^{\alpha_i}$$ where $\epsilon \in \{0, 1\}$, $p_i$ denotes the $i$-th prime number, and $(\alpha_i)_{i \in \mathbb{N}^*}$ is a sequence of non-negative integers such that only a finite number of terms are different from $0$.

Now, let $S_0$ denote $\mathbb{N}^*$. We construct by induction, for every $k \in \mathbb{N}^*$, the set $S_k$ of numbers that can be written as $$(-1)^\epsilon \prod\limits_{i \in \mathbb{N}^*} p_i^{\alpha_i}$$ where $\epsilon \in \{0, 1\}$, $p_i$ denotes the $i$-th prime number, and $(\alpha_i)_{i \in \mathbb{N}^*}$ is a sequence of elements of $S_{k-1} \cup \{0\}$ such that only a finite number of terms are different from $0$.

  • $S_1 = \mathbb{Z}^*$.
  • $S_2 = \mathbb{Q}^*$. Moreover, the uniqueness of the decomposition is conserved.
  • $S_3 = \{(-1)^\epsilon \sqrt[n]{q} \mid \epsilon \in \{0,1\}, q \in \mathbb{Q}_+^*, n \in \mathbb{N}^* \}$ and the uniqueness of the decomposition is still conserved.
  • With $S_4$ appear the first transcendental numbers, and I am unsure about the uniqueness of the decomposition.

So I have a few questions that maybe some of you could answer : Has this been studied before ? Do we always keep the uniqueness of the decomposition or do we lose it at some step ? Since $(S_k)_{k \in \mathbb{N}}$ is a non-decreasing sequence of sets, what would be its limit ? Would there be any use to this ?

Thanks in advance!

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    $\begingroup$ Using induction, one can show that every $S_k$ is a countable subset of $\mathbb{R}$, hence the union of the $S_k$s is countable. In particular, it is not the whole set of real numbers. $\endgroup$ – Crostul Feb 16 '16 at 16:57
  • $\begingroup$ Thanks for your remark! That was one of my thoughts, too. Another thing to notice is that $S_k$ is a multiplicative group for $k \geq 2$, but it seems that $S_k \cup \{0\}$ is an additive one only for $k = 1$ and $k = 2$. $\endgroup$ – Esperluet Feb 16 '16 at 17:16
  • $\begingroup$ How about continuing $S_k$ with $k$ transfinite ordinal? Instead of choosing from $S_{k-1}$, choose from the union of all $S_j$, $j<k$. $\endgroup$ – GEdgar Feb 16 '16 at 17:55

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