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Let $p$ be a prime and $b \in \mathbb Q$ some rational number of $p$-adic value $\lvert b \rvert_p \le 1$. Further, let $(a_0, a_1, a_2, a_3,\dotsc)$ be a strictly increasing sequence of natural numbers. I'm studying series of the form

$$S(b, (a_n)_n) := bp^{a_1} + b^2p^{a_2}+b^3p^{a_3}+\dotsc$$

which under the above hypotheses converge in $\mathbb Q_p$.

It's easy to see that this limit is rational (i.e $\in \mathbb Q$) when the series is "periodic" in the sense that there are $j \le k \in \mathbb N$ such that $a_{i+mj} = a_i + mk$ for all $m \in \mathbb N_0$ and $0 \le i \le j-1$, i.e. $$(a_0,a_1,a_2,\dotsc ,a_0+k, a_1+k, \dotsc ,a_0+2k, a_1+ 2k ,\dotsc).$$

It's also rational if the sequence is eventually periodic: that is, periodic after some finite number of terms.

By analogy with $p$-adic expansions generally I feel that the converse must also be true -- that the limit must be irrational when the sequence $(a_n)_n$ is not eventually periodic in the above sense. Am I correct, and if so how can I prove it?

Torsten Schoeneberg
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Joe Slater
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    No such thing as an infinite polynomial, by definition. – Thomas Andrews Apr 14 '21 at 01:51
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    What do you mean by “periodic.” Under the usual definiti9n, this series only converges in the $p$-adics if $|b|_p<1.$ – Thomas Andrews Apr 14 '21 at 01:54
  • E.g., a_i=(0,1,2,3...) is periodic, and the 2-adic \sum (3^i*2^i) = -3/5. – Joe Slater Apr 14 '21 at 03:07
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    Do you have a practical case where this is useful/natural? The same proof as with $u=1$ gives that for any $u\in \Bbb{Z}p^\times \cap \Bbb{Q}$, every $x\in \Bbb{Q}_p\cap \Bbb{Q}$ is of the form $\sum{n\ge -N} c_n u^n p^n$ with $c_n\in 0\ldots p-1$ and $c_n = c_{n+m}$ for $n$ large enough. – reuns Apr 14 '21 at 03:41
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    Well, that’s not what periodic means. $1,2,1,2,1,2,\dots$ is periodic. – Thomas Andrews Apr 14 '21 at 03:47
  • If $a_j+j v_p(b)$ isn't required strictly increasing then the answer to your question is no (with the strictly increasing assumption there is only one way to write a $p$-adic number, without there are uncountably many ways) – reuns Apr 14 '21 at 03:51
  • I assume even with this unusual periodicity definition, you mean eventually periodic? Otherwise there are trivial counterexamples like $b=1$ with $a$'s $(1,2,3,5,7,9,11,13,15,...)$. – Torsten Schoeneberg Apr 14 '21 at 14:43
  • I have now clarified that I class eventually-periodic sequences together with periodic ones. My definition is by analogy to $p$-adic numbers, in which a number like $\dotsc 10101_5$ may be called periodic even though it actually represents $5^0+5^2+5^4\dotsc$. As you note, when $b=0$ a sequence of the sort I describe is simply $p$-adic, and in that case the sum of a wholly aperiodic sequence would certainly be irrational. – Joe Slater Apr 14 '21 at 23:12
  • Er, that should have been "when $b=1", sorry. – Joe Slater Apr 15 '21 at 23:15

1 Answers1

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Every $p$-adic number can be written as a quotient of two $p$-adic integers, so in that sense, every $p$-adic number is “rational”. In fact, for $p > 2$, $e^p \in \Bbb{Z}_p$, even though $e^p$ is not a rational number in $\Bbb{R}$.

Edit: @TorstenSchoeneberg in the comments tried to tell me that "$e^p$" (as defined by the usual convergent series $\sum_{n \geq 0} \frac{p^n}{n!}$) is not a number that exists in both $\Bbb{R}$ and $\Bbb{Q}_p$, because the metrics are different.
But the same logic he cites renders nonsensical his (and other people's) assertion that the OP is "obviously" looking for answers in "$\Bbb{Q} \cap \Bbb{Q}_p$", since the type of infinite series OP is discussing, $$b p^{a_1} + b^2 p^{a_2} + b^3 p^{a_3} + ...$$ with $0 < a_1 < a_2 < a_3 < ...,$ does not even converge in $\Bbb{Q}$. Commenters who think the OP wants an answer in "$\Bbb{Q} \cap \Bbb{Q}_p$" (whatever that is) are wrong, and have failed to appreciate this subtle but essential distinction.

Rivers McForge
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    Clearly not what the OP meant, $\Bbb{Q}_p\cap \Bbb{Q}$ is perfectly defined – reuns Apr 14 '21 at 03:35
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    Yeah, not a helpful answer. – Thomas Andrews Apr 14 '21 at 03:48
  • It's a correct answer, and helpful in understanding how the meaning of "rational" might change from one context to another. In fact, it clarifies the notation $\Bbb{Q}_p$ (as opposed to, say, $\Bbb{R}_p$) for the metric completion of $\Bbb{Q}$ with regards to the $p$-metric. – Rivers McForge Apr 14 '21 at 16:37
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    Actually, it is not correct. It is a $p$-adic rational number, but it is not a rational number. – Thomas Andrews Apr 15 '21 at 00:17
  • @thomasandrews Actually, it is correct. Your comment suggests you didn’t read my answer, since your comment says “It is a $p$-adic rational number...” which was the precise content of my answer. – Rivers McForge Apr 15 '21 at 22:30
  • But the question asked about rationals. Good luck. @RiversMcForge – Thomas Andrews Apr 15 '21 at 23:01
  • And the answer was about rationals. I'd respectfully request that you reverse your downvote, since it was clearly based on a failure to read or correctly assess the relevance of my answer to the question. @ThomasAndrews – Rivers McForge Apr 15 '21 at 23:16
  • Good luck with that. You’re answer is about $p$-adic rationals, not rationals. @RiversMcForge Maybe you should take the number of downvotes as a sign. – Thomas Andrews Apr 15 '21 at 23:51
  • @thomasandrews Respectfully, it’s clear that you didn’t understand my answer, and if all the downvotes are coming from a place of ill-humored (not to say malicious) incomprehension, as yours apparently did, then I take them as a sign of nothing but the gratuitous and unwarranted hostility of the downvoter. – Rivers McForge Apr 16 '21 at 00:24
  • @thomasandrews The original post is about circumstances under which a $p$-adic expression is rational, so it’s quite relevant to point out that every such $p$-adic expression is a $p$-adic rational. You really have to tie yourself in knots to argue that there’s no possible relevance to such an answer—or, more politely, simply fail to read the answer correctly, and irately pick a fight with me based on this misreading, which I have respectfully tried to bring to your attention. – Rivers McForge Apr 16 '21 at 00:35
  • FYI, I downvoted as per reuns' comment, it's clear that OP is interested in "rational" $p$-adics in the sense of $x \in \mathbb Q \cap \mathbb Q_p$. Also "$e^p$" is not a number which exists in both $\mathbb R$ and $\mathbb Z_p$. Surely the series $\sum p^n/n!$ converges in both w.r.t. to respective metrics, but the limits have no reason to be in any way related. Look at https://math.stackexchange.com/a/3061353/96384 and comments to https://math.stackexchange.com/a/3765961/96384. Besides, one can come up with easier $x \in \mathbb Z_p$ which are $\notin \mathbb Q$, if that's what is aimed for. – Torsten Schoeneberg Apr 16 '21 at 23:36
  • @TorstenSchoeneberg There's a lot of mind-reading going on in these comments about what OP "really meant". I stand by my assertions that my answer was a) correct, b) relevant to the question, and c) didn't deserve the downvotes, or hate, that it is getting. If other people want to post answers that highlight different aspects of the situation mentioned in the OP, there was nothing that stopped them from doing that, instead of nitpicking and sniping at my answer for no good reason. – Rivers McForge Apr 17 '21 at 02:03
  • @TorstenSchoeneberg Actually, having read your most recent critique more carefully, I realize that it's self-refuting. reuns and thomasandrews please advise--does this not undermine the entire substance of your carping? – Rivers McForge Apr 18 '21 at 21:37
  • Happy that you followed my links and tried to understand them, sad that you did not quite succeed; and while you must admit the possibility that it's you who misunderstands (not the one who just gave you that nudge in the right direction plus three other downvoters), maybe re-read e.g. reuns' very first comment to this post. Also, https://math.stackexchange.com/a/4007515/96384. The series OP asks about converges in $\mathbb Q_p$, OP rightly asserts he has a sufficient criterion for its limit to be in $\mathbb Q$, and his question is whether that criterion is necessary as well. No nonsense. – Torsten Schoeneberg Apr 19 '21 at 05:27
  • @torstenschoeneberg Your argument is that these p-adic series mean something different in $\Bbb{Q}_p$ and $\Bbb{R}$ and it’s inappropriate to conflate their limits just because they have similar properties. Well, last I checked, $\Bbb{Q} \subset \Bbb{R}$, and if reuns’ comment refers to “$\Bbb{Q} \cap \Bbb{Q}_p$”, the same objection applies. Of course, $\Bbb{Q} \subset \Bbb{Q}_p$ with a different metric, but then that undercuts ThomasAndrews’ assertion that what is being asked for is not “p-adic rationals”. The original question is clearly about p-adic rationals. – Rivers McForge Apr 19 '21 at 12:19
  • No, $\mathbb Q$ is the one field which quite obviously identifies with a uniquely determined subfield of $\mathbb Q_p$ (or any other characteristic $0$ field), and that is what everyone except you is talking about. It's actually a neat beginner's exercise to find the unique $p$-adic expansions of given rationals $\in \mathbb Q$, see e.g. https://math.stackexchange.com/q/3600639/96384, https://math.stackexchange.com/q/3600639/96384 and many duplicates. One must not mix up any of that with convergence w.r.t. real metric / expansions as real numbers, but here, nobody except you has done so. – Torsten Schoeneberg Apr 19 '21 at 20:04
  • @torstenschoeneberg Then the objection that my answer is irrelevant because what is asked for has nothing to do with “p-adic rationals” is invalid, because every element of $\Bbb{Q}$ is a p-adic rational! Face it, these objections are inconsistent and absurd. I don’t know how you can write these comments with a straight face, lol – Rivers McForge Apr 20 '21 at 14:46
  • You're almost there. Every element in $\mathbb Q$ can be seen as an element of $\mathbb Q_p$, but by far not every element of $\mathbb Q_p$ is in $\mathbb Q$. OP defines certain elements of $\mathbb Q_p$; from the way he defines them, it's clear they are in $\mathbb Q_p$ but not clear whether they are in $\mathbb Q$. So he asks for a criterion to decide if they are in $\mathbb Q$. That's what everyone except you has understood from the beginning. -- It's not easy to keep a straight face, but I've dealt with otherwise talented people stubbornly insisting everyone else is wrong before. – Torsten Schoeneberg Apr 20 '21 at 20:41
  • @torsten But that’s not what OP actually asked (even if that’s what they might have “meant to” ask). They asked when this $p$-adic expression is “rational or irrational”, and $\Bbb{Q}_p$ has no “irrationals”, because every element is a quotient of two $p$-adic integers (elements of $\Bbb{Z}_p$), as noted originally. I answered the question as written, without trying to psychically see into the OP’s intentions, as you, reuns, and Thomas Andrews think I should have done. You may think it’s pedantic to answer the actual words of a question, but you have no basis to claim it’s wrong. – Rivers McForge Apr 22 '21 at 01:06