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The wikipedia article on p-adic numbers warns about $b$-adic expansions where $b$ is not a prime:

Although for p-adic numbers p should be a prime, base 10 was chosen to highlight the analogy with decimals. The decadic numbers are generally not used in mathematics: since 10 is not prime, the decadics are not a field.
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As noted above, 10-adic numbers have a major drawback. It is possible to find pairs of non-zero 10-adic numbers (having an infinite number of digits, and thus not rational) whose product is 0.

Hence I was quite surprised to learn that quote notation, which "follows the approach of Kurt Hensel's p-adic numbers", also works if $b$ is not a prime. So I wonder whether anything will go wrong if also non-terminating expansions which are not periodic will be allowed.

Let $b=p_1^{\alpha_1}\dots p_k^{\alpha_k}$ with prime numbers $p_i$ and positive integers $\alpha_i$. My guess is that any number $x \in\mathbb Q_{p_1} \cap\dots\cap \mathbb Q_{p_k}$ can be represented by a potentially non-terminating $b$-adic expansions, and that every non-terminating $b$-adic expansions is equivalent to such a number. You may object that this "formal intersection" is meaningless, but because $\mathbb Q_p$ can be represented by equivalence classes of certain sequences of rational numbers, it should be possible to compute this intersection in the space of sequences of rational numbers.

How could such a $b$-adic expansions be defined? My guess is that $x=p_1^{\beta_1}\dots p_k^{\beta_k}\sum_{i=0}^\infty d_i b^i$ with (potentially negative) integers $\beta_i$, integers $d_i$ with $0\leq d_i < b$ and $\operatorname{gcd}(d_0,b)=1$. Note that using $p_1^{\beta_1}\dots p_k^{\beta_k}$ instead of $b^\beta$ (which allows for $\operatorname{gcd}(d_0,b)=1$) avoids the counterexample given in the wikipedia article. One obvious question is whether addition, subtraction, multiplication and division will be well defined for this representation. This seems to be easy to check, at least for addition, subtraction and multiplication. But is there anything else which needs to be checked, before one can conclude that this is a valid number representation for the field $\mathbb Q_{p_1} \cap\dots\cap \mathbb Q_{p_k}$?

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    $\begingroup$ I see a potential problem now, if I only check whether the arithmetic operations are well defined: After the addition of two numbers in the representation, one of the $\beta_i$ might go towards infinity. Hence the result must be set to zero. Which is fine in a certain sense, but associativity (and distributivity) of addition might be lost (not sure whether this can happen). So in addition to being well defined, I also have to check that addition is associative and distributive. $\endgroup$ Oct 30, 2014 at 10:54

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$\mathbb Q_p$ is the completion of $\mathbb Q$ under the $p$-adic metric. With several norms $|\cdot|_{p-1}, \ldots, |\cdot|_{p_k}$, you can combine these to a single norms e.g. by adding: $|x|:=|x|_{p_1}+\ldots+|x|_{p_k}$, and consider the completion with respect to this norm.

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    $\begingroup$ I also thought about this. My only worry was that the $p$-adic metric is induced by an absolute value (i.e we have $|xy|_p=|x|_p|y|_p$), but the combined norm only satisfies $|xy|\leq|x||y|$. So I wonder whether this is enough to ensure that the completion is a field. $\endgroup$ Oct 30, 2014 at 11:03
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Can the b-adic representation of rational numbers (by quote notation) be extended to non-terminating expansions?

In a certain way, the answer to this question is yes (as indicated by Hagen von Eitzen). For $b=p_1^{\alpha_1}\dots p_k^{\alpha_k}$, define $\mathbb Q_b$ to be the completion of $\mathbb Q$ under the norm $|x|:=|x|_{p_1}+\dots+|x|_{p_k}$. Maybe it would be more honest use the norm $|x|:=\max(|x|_{p_1}^{-\alpha_1\ln p_1},\dots,|x|_{p_k}^{-\alpha_k\ln p_k})^{\ln b}$, but these norms should be equivalent anyway. Then $x\in \mathbb Q_b$ for $x=b^{\beta}\sum_{i=0}^\infty d_i b^i$ and any $x\in \mathbb Q_b$ can be represented that way.

However, the ignored comment about $|xy|_p=|x|_p|y|_p$ vs. $|xy|\leq|x||y|$ is a serious issue here. It boils down to the fact that there are Cauchy sequences $x_i$ and $y_i$ such that both $|x_i|$ and $|y_i|$ converge against non-zero numbers, but $|x_iy_i|$ converges against zero. So $\mathbb Q_b$ contains zero-divisors, and hence is not a field.


But the question didn't ask about $x=b^{\beta}\sum_{i=0}^\infty d_i b^i$, because the wikipedia article already warned that these numbers don't form a field. The question was about $x=p_1^{\beta_1}\dots p_k^{\beta_k}\sum_{i=0}^\infty d_i b^i$ with (potentially negative) integers $\beta_i$, integers $d_i$ with $0\leq d_i < b$ and $\operatorname{gcd}(d_0,b)=1$. Denote the group of units of a ring $R$ as $R^*$. Then $x\in \mathbb Q_b^*$ and any $x\in \mathbb Q_b^*$ can be represented that way. Because $\mathbb Q \subset \mathbb Q_b$ we also have $\mathbb Q^* \subset \mathbb Q_b^*$. This is sufficient to understand why quote notation can allow that $b$ is not a prime. And this also makes it clear that addition is the problematic operation which can fail in this notation for non-terminating expansions which are not periodic.


What about $\mathbb Q_{p_1} \cap\dots\cap \mathbb Q_{p_k}$? If this "formal intersection" would be a real intersection, then this would be a field with $\mathbb Q \subset\mathbb Q_{p_1} \cap\dots\cap \mathbb Q_{p_k}$. The problem of the proposed construction of this intersection is that two Cauchy sequences of rational numbers (with respect to all $|\cdot|_{p_i}$) might be equivalent with respect to one of the norms, but not with respect to the others. If the two sequences are treated as non-equivalent, then we get a subset of the Cartesian product of the $\mathbb Q_{p_i}$ and not a subset of each $\mathbb Q_{p_i}$. And if the two sequences are treated as equivalent, then we get a subset of quotients of the $\mathbb Q_{p_i}$ and not a subset of each $\mathbb Q_{p_i}$. In both cases $\mathbb Q_{p_1} \cap\dots\cap \mathbb Q_{p_k}$ won't be a subset of each $\mathbb Q_{p_i}$, hence its better to declare this "formal intersection" as meaningless.

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