Talk:P-adic number/Archive 1

More material moved here from article
[ilan]: There is no need to limit the base to a prime number. In fact, base 10 will do just fine, and 10-adic numbers are just ordinary integers, except with possibly infinitely many digits to the left, otherwise, all rules of addition and multiplication as usual. A good exercise is to understand why the 10-adic number ...111 = -1/9. A good application of this result is the non Archimedean Zeno paradox: http://www.lix.polytechnique.fr/Labo/Ilan.Vardi/zeno.html P.S. To the people who feel the need to remove what I write: Do some research on my person, and decide whether you are more qualified than I am to write about such subjects and therefore whether you are qualified to delete what I have written (I guess you didn't find the p-adic Zeno paradox very interesting). Otherwise, am I mistaken, or is anyone allowed to contribute here? Rx StrangeLove 22:52, 22 May 2005 (UTC)
 * I've removed this same material again, mostly commentary but it looks like there's some math content as well, I'm not sure what to make of it. Maybe someone can look and see if it should be returned to the article page. Thanks! Rx StrangeLove 23:17, 22 May 2005 (UTC)

Hmmm - do you know this area of mathematics well? I only ask since you don't generally seem to edit mathematical articles. Ilan is a mathematician. Charles Matthews 19:19, 28 July 2005 (UTC)

Infinite expansions
Article says "Real numbers are obtained by allowing for infinite expansions to the right; p-adic numbers are obtained by allowing for infinite expansions to the left." What if you allow infinite expansions in both directions? -- SJK


 * Doesn't work. You get numbers that cannot be calculated with. Any repeating p-imal in both directions is equal to zero; if you add e (which is in no p-adic field) and anti-e (the sum of all factorials, which is in all p-adic fields) you get a number whose square cannot be taken. -phma


 * But AxelBoldt says below: "...the p-adic numbers (of which you can think as infinite p-adic expansions to the left which also have finitely many digits to the right of the "decimal" point)." Isn't that numbers with infinite expansion in both directions? -ReiVaX 21:46, 5 November 2005 (UTC)


 * (I've moved this discussion to the bottom of the talk page, which is where new topics are usually added) Doesn't work. The p-adic metric is defined so that sequences which expand to the left (such as 1, 11, 111, 1111 etc.) will converge in the p-adic metric, whereas in the normal metric they would diverge. But the price you pay is that sequences which expand to the right (such as 0.1, 0.11, 0.111, 0.111 etc.), which would converge in the normal metric, actually diverge in the p-adic metric. So a p-adic expansion can only have a finite number of digits to the left of the "decimal" point. Gandalf61 11:31, 6 November 2005 (UTC)

Extension vs Completion
It seems to me to not be helpful to call Qp and extension of Q, since that basically just says it has characteristic zero. As an extension, it has uncountable transcendence degree. What would make sense is to say completion, so I think I'll try to figure out how to word this. Gene Ward Smith 08:06, 6 May 2006 (UTC)

p-adic numbers notation
Gene - nice re-write on P-adic number - I liked most of your updates. However, I am not sure about your new "backwards arithmetic" section. I have always seen p-adic numbers written down as strings extended to the left. I believe that writing them as strings extended to the right is non-standard - is this your own notation ? And you have your "decimal" point in a strange place. Take -1/5 as a 5-adic number, for example - I would write it ...444.45 - how would you write it ? Personally, I thought the first part of the old "Motivation" section was a clearer introduction - introducing 10-adic expansions first at least has the advantage that newcomers are working in a base that they are familiar with.Gandalf61 09:56, 7 May 2006 (UTC)

Writing strings to the left is what Koblitz did. I find it hard to read, and don't like it. I don't know of anyone who does it to the right, but there's nothing wrong with it and the section is on trying to make the p-adics intuitive. Writing to the right makes p-adic numbers look like real numbers, and to my mind is far more digestible.

Anyway I am quite sure they are intutitive this way, because I discovered the 10-adics when I was 16 and sitting bored in an algebra class. When the teacher said carry to the left, I asked myself why not to the right? I proved to my satisfaction that this was a ring, and spent more than a week trying to show it was a field, until I discovered that in fact it wasn't, and that it depended on the base. So, I find this a very intuitive approach. I recommend it.

As for the 10-adics, I think it isn't such a good idea to fixate on them, since they aren't important. Gene Ward Smith 22:18, 7 May 2006 (UTC)


 * It sounds as if writing p-adic numbers to the right is your own convention. Nothing wrong with that in itself, but I think the article should stick to the standard notation. Putting your own personal notation into Wikipedia would seem to breach the No original research policy. Gandalf61 09:42, 8 May 2006 (UTC)


 * I don't think there is a standard notation. It's usual in computer algebra packages to write things to the right, not the left, giving p-adics as power series in a prime number. However, would "Backwards arithmetic" be a suitable topic for Wikipedia? Gene Ward Smith 01:41, 10 May 2006 (UTC)


 * I took a look at "no original research" and I don't think notation qualifies. Gene Ward Smith 06:17, 10 May 2006 (UTC)


 * As far as I can tell, the majority of authors use the "writing to the left" notation - examples are Cassels in Local Fields; Alain M. Robert in A Course in p-adic Analysis; and various sets of on-line notes such as these. The only place I can find that uses the "writing to the right" notation is here, and that does not use your unusual placement of the "decimal" point. Perhaps we should invite comments from the wider Wiki mathematics community at WikiProject Mathematics to see if there is a concensus on whether the article should use one or the other notation, or both ? Gandalf61 11:28, 12 May 2006 (UTC)


 * It seems to me that it's his decimal point which is non-standard, not mine. Standard in base b is that we have a0 b0 &middot; a1 b-1 ..., and in p-adic numbers 1/p corresponds to b, so that we should get a0 p0 &middot; a1 p1 ... .As I remarked, to-the-right is actually more common that to-the-left if you count the power series notation, where you write out the exponents and additions explicitly. Decimal notation nothing more tha a compact power series notation, and hence most logically, it seems to me, follows the same proceedure. I'll take a look at the WikiProject and see if it's a good place to stick a note. Gene Ward Smith 21:07, 13 May 2006 (UTC)


 * I admit I haven't worked with p-adic numbers for some time, but I've never seen the "writing to the right" notation before. (Even if you count writing out the power series as an example of "writing to right", I've seen that done backwards, as well, as in ... + 0*34 + 1*33 + 0*32 + 2*3 + 2.)
 * I also think that a new, non-standard, notation may be a violation of WP:NOR, if there is a standard notation. Why is a new notation not a neologism?  00:26, 16 May 2006 (UTC)
 * And I'm not wikistalking you, Gene, although it looks as if it would be a good use of my time and abilities. Just because I disagree with you on Real numbers, Zeration, and here, doesn't mean that I've followed you. &mdash; Arthur Rubin |  (talk) 00:30, 16 May 2006 (UTC)
 * The "standard" notation has the feature (advantage, or disadvantage) that 234.35 has the "same" meaning as a member of Q5 or of Q. &mdash; Arthur Rubin | (talk) 00:36, 16 May 2006 (UTC)


 * I have mentioned this debate on the WikiProject Mathematics take page, to see whether anyone else has views on the notation used in this article - and I have re-named this talk page section to make it easier to locate. Gandalf61 08:25, 16 May 2006 (UTC)


 * Notation is meant to communicate. Agreement (convention) is good, and so is logic. Mathematical notation sometimes freezes its form before the underlying theory is well understood, and sometimes evolves differently at different times and places. So here we are again. We write in service of the reader. If a notation is logically far superior to its alternatives, we may consider it for purposes of explanation, but must alert the reader to what is standard and what is not. If a notation differs from overwhelmingly common usage, then it becomes that much harder to justify its use.
 * The notation proposed by Gene Ward Smith is not one that readers are likely to encounter elsewhere, and does not offer significant advantages over common notation. Therefore we should mention it, but not adopt it for the bulk of the article. If facts refuting my premises are brought into evidence (major advantages, frequent usage), I'll reconsider. --KSmrqT 13:12, 16 May 2006 (UTC)


 * I propose that we revert to the "right to left" notation in the body of the article, and mention the two variants of the "left to right" notation (with the units digit to the right or to the left of the "decimal" point) in a footnote. Gandalf61 14:32, 18 May 2006 (UTC)
 * I concur. &mdash; Arthur Rubin | (talk) 18:06, 18 May 2006 (UTC)


 * Please do not make a claim in the article that "right to left" is a standard notation. It's a notation used to introduce p-adic numbers, it's not something number theorists are in love with. In an actual math paper, if you needed a specific p-adic number, the two usual ways of representing it would be to give the number mod pn for some suitable n, or to give the number as a power series expansion (and so of course, "right to left".) The power series idea goes back to Hensel, incidentially.


 * If you want to see an example of how you might use and display actual p-adic computations, I recently did a lot of work on the algebraic number field article. In the section an example, I work out an example in the 23-adics. This is how you'd probably want to do it in an actual math article. Gene Ward Smith 03:42, 19 May 2006 (UTC)


 * Isn't that what we're doing and should be doing &mdash; introducing p-adic numbers? You're convincing me that we should use easily understandable notation.  &mdash; Arthur Rubin |  (talk) 05:32, 19 May 2006 (UTC)


 * What number theorists do is in a context where people know what p-adic numbers are; the question here is how to explain them to people who only know the real and complex numbers. From that point of view, saying p-adics carry to the right, reals carry to the left, and function fields over finite fields don't carry at all is one way of approaching it, and does bring home the idea that there is an underlying similarity. What is most likely to get people to see the analogy? Gene Ward Smith 09:01, 20 May 2006 (UTC)

Teichmüller expansions
By the way, it's by no means true that number theorists are wedded to the {0, ..., p-1} digit set for the p-adics either. The prime p has all p-1 of the p-1 roots of unity in Zp, and these, together with 0, can be used for the digits of a representation of the p-adics. Rounding off gives the Teichmüller character.Gene Ward Smith 03:56, 19 May 2006 (UTC)


 * Yes, there are many choices for the digit set. Yes, Teichmüller digits have advantages in some circumstances. Yes, number theorists seldom need to write out an explicit representation of a specific p-adic number. All of these issues could be discussed in a new section on notation. But I still maintain that for the purposes of an introductory section written for newcomers to the subject, the digit set {0,1,...,p-1} written right to left is the best notation, because the p-adic representation of a positive rational integer is then the same as its base p representation. If I re-write the first main section of the article with right-to-left notation, but without saying that this is a standard notation, and add a new section describing alternative notations, would you be happy with that ? Gandalf61 11:19, 20 May 2006 (UTC)


 * That would be fine. I'm not insisting on left-to-right, which is why I put the topic up for discussion in the first place. I think some place or other it would be nice to mention that you can carry to the left, carry to the right, or not carry at all, and all three choices produce things like numbers. Gene Ward Smith 23:20, 20 May 2006 (UTC)


 * Okay, I have re-written the "backwards arithmetic" section and re-named it as Introduction; added a new Notation section to describe various different notations (but without claiming that any one of them is more standard than the others); and taken out references to backwards arithmetic in other places in the article to avoid confusion. I next plan to add references for the various properties of Zp, Qp and Cp listed in the Properties section. Gandalf61 09:28, 26 May 2006 (UTC)


 * Looks good--at least to the extent right-to-left can ever look good. Gene Ward Smith 06:50, 27 May 2006 (UTC)

Topological approach
I like the idea. Unfortunately, it seems to be covered under "#Analytic approach", as the completion with respect to the specified metric. &mdash; Arthur Rubin | (talk) 16:12, 16 July 2006 (UTC)

What's the meaning or derivation of the suffix "-adic"?
The first paragraph of the current article explains that 'p' stands for 'prime number', but doesn't tell what an '-adic' means, signifies, or anything about its derivation. I'll qualify that, not having examined the whole article -- the suffix '-adic' is not explained in or near the beginning of the article, the most appropriate location. A hop over to p-adic analysis didn't help. Presumably this system's founder Kurt Hensel would have recorded his reason for choosing this term, unless it was too obvious in German? --AC 06:54, 31 August 2007 (UTC)

...99999 = -1?
Hi, I hope this comes out right because I usually just stick to fixing spelling mistakes!

The very first section, about how the 10-adic number ...99999 = -1 does not make things clear to me. I have seen on another occasion, elsewhere, the following idea used:

$$9+1 = 10$$

$$99+1 = 100$$

$$...9 + 1 = ...0$$

where in the last one we keep passing the "carry" to the left up to infinity, and so since $$x+1=0$$, we have $$x=-1$$, where $$x=...9999$$

That seems clearer to me than the mentioning of "being close together if they differ by a large power of ten." --MarkHudson 15:51, 19 July 2006 (UTC)

I think that this type of discussion is directly attributable to Archimedes Plutonium, otherwise known as Ludwig Plutonium. Maybe others discussed in it this way, but this was the first time I saw it. He is also the first to write ellipses in front of a number.Likebox 16:21, 20 October 2007 (UTC)


 * In his usenet posts, Archimedes Plutonium appropriated and to some extent distorted the mainstream mathematical concept of p-adic numbers, which goes back over 100 years to German mathematician Kurt Hensel. Writing an ellipsis to the left of a p-adic number is a widespread and natural mathematical notation - see here for example. Gandalf61 17:01, 20 October 2007 (UTC)

Motivations
Should there perhaps be a clear indication early in the article about why one would want to have p-adic numbers? Maybe a simple one or two-line example from p-adic analysis? SmartPatrol (talk) 06:12, 15 March 2008 (UTC)

p-adic NORM???
Umm, the p-adic function defines a METRIC, but not a NORM: For a Norm, ||k*x||=|k|*||x||

But taking, say, the 2-adic Norm, ||6||=1/2, ||18|| is still 1/2!

Anyone disagree?

So that section needs rewriting... any volunteers? —Preceding unsigned comment added by 131.111.200.200 (talk) 00:36, 26 November 2008 (UTC)


 * I imagine the term norm here comes from field norm rather than vector norm. However, it may remove a possible source of confusion if we describe the p-adic valuation as an absolute value instead. I have changed the relevant section of the article. Gandalf61 (talk) 08:58, 26 November 2008 (UTC)

too complex?
In my opinion the introduction needs a re-write. It is too complex for everyday Joes to undestand, and acts more like a handy reference/reminder for the likes of mathemetical geniuses a la Einstein. —Preceding unsigned comment added by 122.107.166.42 (talk) 03:32, 2 January 2009 (UTC)


 * Well, p-adic numbers are a somewhat complex subject, and for a non-mathematician it is difficult to give a concise explanation that says any more than "p-adic numbers are a number system used in number theory" - which is what the first couple of sentences of the article say now. When you refer to "the introduction", do you mean the opening paragraphs of the article, or the longer section headed "Introduction" that follows ? Which specific parts of the introduction do you find hard to understand ? Gandalf61 (talk) 13:55, 2 January 2009 (UTC)

I find the present introduction very good and intuitive! Marozols (talk) 18:59, 26 February 2009 (UTC)

p-adic expansion
"If p is a fixed prime number, then ...."

p does not need to be a prime number, it can be any positive integer not less than two. —Preceding unsigned comment added by 129.13.186.1 (talk)


 * Yes, the base p expansion works for any integer p>1, but if p is not a prime number then the p-adic numbers are not a field, which limits their usefulness. Gandalf61 19:16, 26 November 2006 (UTC)

maybe to you gandalf61 -- is that prime? -- but some of us are not so blinded by our numerical prejudices as to discarded a technically infinite collection of numbers off hand. there are only a limited number of primes, i suppose that works to the advantage of the monopoly on number systems -- lee smolin


 * Funny question. 61 is prime, and the string gandalf61 is prime when interpreted as numbers in base 25 & 28, the respective primes are 2508187837651 and 6191075089753 in base 10.  I checked from base 24 (the lower limit for a string containing 'n', the 14th letter) to base 36, (26 letters + 10 digits), which is as high as the case-insensitive qalculate program goes.  --AC 08:01, 31 August 2007 (UTC)


 * Well, he may be discarding an infinite number of bases, but there are as many primes as there are integers, so it's all good. Eebster the Great (talk) 06:01, 20 November 2009 (UTC)

Lead must be generalized to include non-prime p-adic numbers
Presently, the lead acts as if only primes can serve at the "p" in p-adic. Then the information induction jumps to using 10-adic numbers. It is later mentioned that non-prime p's lead to zero divisors but that this is rather out-of-order logically speaking. The article should introduce p-adic number allowing for all valid p's. It should then quickly mention why only prime p's as usually used. Jason Quinn (talk) 15:25, 18 June 2011 (UTC)
 * I agree that there's a problem. However, most people working in number theory or algebra define p-adics only for prime values of p (or, more generally, for prime ideals of a ring).  Changing the lead to say otherwise would be misleading (apologies for the pun).  I've inserted some explanation at the top of the introductory/10-adic section.  Does that help? Jowa fan (talk) 01:53, 19 June 2011 (UTC)

Etymology
I've just removed the following text from the beginning of the article: "The modern term p-adic is formed in a similar way to words of Greek origin like monadic, dyadic, triadic, where the suffix -adic relates to the number of terms in an expression. See Adicity on Wiktionary. -adic corresponds with the -ary in words with Latin roots like unary, binary, ternary. See Arity." I think it would be nice to have an "etymology" section somewhere (although maybe not right at the top of the page). But I find the aove paragraph confusing: following the links, it seems that arity/adicity refers specifically to arguments of a function, and it's not clear how it directly relates to p-adic numbers. Also, I think a discussion of etymology in this case needs to be supported by a reference to a suitable source. Jowa fan (talk) 12:37, 31 October 2011 (UTC)


 * Hi, I was trying to incorporate an answer to AC's question in the above section What's the meaning or derivation of the suffix "-adic"?. None of the wiki -adic articles I have found seem to give a quick definition/derivation of the suffix for the interested layman reader like myself. What sort of source for the etymology do you think would be appropriate? A standard dictionary like Chambers, Webster, OED etc.? For a modern secondary source on first usage of p-adic, this page says that it was Hensel who coined the phrase, quoting Katz, Victor J. A History of Mathematics: An Introduction. Harper Collins, 1993, or Addison Wesley, 1998, p. 824


 * BTW, I found what may be an earlier use of the term p-adic than the date referenced in the intro to the main article:"First described by Kurt Hensel in 1897, the p-adic numbers..."


 * (NB above quote includes template). In the paper referenced above (the article text just says he *describes* them) Hensel doesn't appear to use the actual word p-adic (Ger. p-adisch) although he uses p-fach (p-times, like dreifach=three times, thrice). However, the phrase "p-adic number systems" (Ger. p-adischen Zahlensysteme) appears on p. 356 in Crelle's Journal 1896 (Google Books dreaded snippet view). This volume may contain an apparently earlier paper by Hensel (Ǜber den grössten gemeinsamen Theiler aller Zahlen, welche durch eine ganze Function von Veränderlichen darstellbar sind) according to this page but it's not free to view and I can't confirm the link - if it matters at all, that is.

Python code
I boldly removed the Python code from the article. It's long and adds little value to the exposition; there seems no reason to give it in Python rather than a hundred other languages; there's no particular reason to believe it's correct as it is unsourced and hence presumably original research. Deltahedron (talk) 06:20, 23 August 2012 (UTC)

Referenced text missing
The line:

"This sequence can be derived in a similar way to the 10-adic expansion of 1/7 shown above"

is in the published page, but the referenced text is no longer included in the published version. — Preceding unsigned comment added by 76.254.42.229 (talk) 00:37, 6 September 2012 (UTC)

Placeholder variable dispute
was not an improvement. For a mathematician (user:Dark Charles or whoever) the absence of any sense in the letter $p$ might be an obvious consequence of the intro. For a linguist or a geneticist it is not. Do not degrade the article, please. Incnis Mrsi (talk) 10:20, 22 July 2013 (UTC)

Just not very clear
The introduction is not very clear.

When you are presenting a topic that is new to people - non-professional mathematicians

you have to make everything absolutely clear. So give clear and consise/distinct examples.

There is no paper limit from the publisher so over-explain at the beginning so ever intelligent reader can grasp at

least the fundamentals of what is being said. — Preceding unsigned comment added by 50.201.98.122 (talk) 21:35, 28 February 2014 (UTC)


 * This is interesting feedback. It would be easier to act upon if it were more specific. Was there a particular paragraph, sentence, or phrase that stood out as hard to follow? Of the examples given, was one of them particularly unclear? There's no paper limit to the talk page either, so feel free to quote passages and go into as much detail as you can! Melchoir (talk) 04:17, 1 March 2014 (UTC)

Elucidation: limit of norm goes negative?
The "Elucidation" block states that "limit of 5-adic norm of the sequence 4, 44, 444, ... is -1. However, the article on p-adic norm says that norm is, indeed, non-negative. This looks like contradiction for me: limit of a nonnegative sequence is nonnegative too. Mistake? --194.226.199.61 (talk) 10:04, 10 June 2014 (UTC)


 * You are right. The limit in the 5-adic metric of the sequence 4, 44, 444, 4444, ... of integers written in base 5 is indeed -1, but the "Elucidation" box makes a hash of explaining why that is. The norm of the differences between successive terms in the sequence approaches 0 in the 5-adic metric, which is why we can be sure that the sequence converges to a limit in the 5-adic metric (whereas in the "ordinary" metric it diverges). If this limit is x then by adding 1 to each term in the sequence we can see that x + 1 = 0, and hence x = -1. I thought about fixing the "Elucidation" box, but in the end I decided it was so muddled that this was not worthwhile, so I just removed it. Gandalf61 (talk) 10:32, 10 June 2014 (UTC)

Topology
Thanks to user:Gandalf61 for answering my question about the topology (on his talk page). I still don't understand why this would be true:
 * The topology of $Q_{p}$ is that of a Cantor set minus a point (which would naturally be called infinity).

This sentence was put in the very first version of this article, but there's no reference. User:Arthur Rubin, I tried to send you an e-mail about this (did you get it?).

Eric Kvaalen (talk) 10:47, 16 July 2014 (UTC)


 * No, I didn't get the E-mail, but I got notified. The topology of $Z_{p}$ is that of a Cantor set. This can be seen by directly mapping
 * $$ \cdots d_2 d_1 d_0 $$
 * in $Z_{2}$ to
 * $$ 0. e_0 e_1 e_2 \cdots_3$$
 * in $C$, where
 * $$ e_n = 2 d_n. $$
 * (This is reported in some of the unreliable references obtained by doing a Google search on "p-adic" + "Cantor set". One of the references seems to be a Springer magazine article, but the relevant text is not accessible without subscription.)
 * I think you can map $Q_{2}$ to $C-{1}$ by mapping
 * $$\cdots d_2 d_1 d_0 . d_{-1} d_{-2} \cdots d_{-m}(=1) $$
 * to
 * $$0. \underbrace {2 \cdots 2}_m 0 e_{-m} \cdots e_{-2} e_{-1} e_0 e_1 e_2 \cdots _3$$
 * but I don't have and have not seen a proof. I don't see the appropriate analog for $Q_{3}$.  — Arthur Rubin  (talk) 15:52, 16 July 2014 (UTC)
 * Try  Deltahedron (talk) 17:32, 16 July 2014 (UTC)

Thanks, Arthur. I don't understand why you wrote (=1) though.

One could use the same approach for Q3. Consider just the triadic integers first. If the units digit is 0, then start your base-3 number in the Cantor set with 0.00. If it's 1, start with 0.020, and if it's 2, start with 0.022. Then if the next digit to the left (d1) is 0, add (concatenate) 0, if it's 1 add 20, and if it's 2 add 22. Keep doing that ad infinitum. That way we map the triadic integers to [0, 1/3]. Then you can similarly map the numbers that have one digit after the "decimal point" to the interval [2/3, 7/9], and so on.

I think you should check what e-mail address Wikipedia has for you. Maybe it's out of date. I'd like it if you could e-mail me (go to my user page) because there's something else I'd like to discuss with you.

Eric Kvaalen (talk) 18:59, 16 July 2014 (UTC)
 * In $Q_{2}$, the last non-zero digit (in fact, all non-zero digits) must be "1".... This means, of course, that
 * $$\cdots d_2 d_1 d_0 . d_{-1} d_{-2} \cdots d_{-m}(=1) $$
 * should be mapped to
 * $$0. \underbrace {2 \cdots 2}_m 0 e_{-m^*} \cdots e_{-2} e_{-1} e_0 e_1 e_2 \cdots _3$$
 * where $m^{*}$ is 0 if m = 0 (by definition m cannot be less than 0), and $m &minus; 1$ if m > 0. A similar modification can (and must) be done for your Q3 construction, but it should work.  — Arthur Rubin  (talk) 20:58, 16 July 2014 (UTC)
 * To be more precise, let
 * $$s_k = \begin{cases}

\underbrace { 0 \cdots 0}_k 2 & 0 \le k < p-1 \\ \underbrace { 0 \cdots 0}_k  & k = p-1 \\ \end{cases} $$
 * $$\cdots d_2 d_1 d_0 . d_{-1} d_{-2} \cdots d_{-m} $$
 * in $Q_{p}$ maps to
 * $$ \begin{cases}

0.0 s_{d_0} s_{d_1} s_{d_2} \cdots _3 & m=0 \\ 0. \underbrace { 2 \cdots 2}_m s_{d_{-m}} \cdots s_{d_{-2}} s_{d_{-1}} s_{d_0} s_{d_1} s_{d_2} \cdots _3 & m > 0 \\ \end{cases} $$
 * Should be a topological isomorphism from $Q_{p}$ to $C&minus;{1}$.
 * (We choose m so that if m > 0, then $$d_{-m} > 0$$, and $$s_{d_{-m}}$$ starts with 0.)  — Arthur Rubin  (talk) 21:21, 16 July 2014 (UTC)
 * E-mail sent — Arthur Rubin (talk) 21:42, 16 July 2014 (UTC)

Use of this talk page as a source...
Although I am a mathematician, I have no published papers in the fields of topology or of p-adic numbers, so my comments above are not a reliable source for the maps between Zp and C and between Qp and C'. — Arthur Rubin (talk) 17:41, 22 July 2014 (UTC)

p-adic order of zero
As I write this, the article text states in the "Constructions" section that "We also define $&#124;0&#124;_{p} = 0$." This is incorrect. The convention is that the order of 0 in a p-adic system is &infin;. E.g., see A Course in p-adic Analysis, top of page 5. I will now correct this error.

I had norm confused with order. The norm is zero, the order is &infin;. The article was correct and I have backed out the mistake.

John (talk) 19:01, 2 November 2014 (UTC)

n-adic numbers where n is composite
The ring of n-adic numbers is isomorphic to the product of the rings of p-adic numbers where p ranges over the distant prime factors of n. In particular, it depends only on the radical rad(n) of n. GeoffreyT2000 (talk) 22:02, 12 March 2015 (UTC)

Capitalization of p-adic
Rules for the proper capitalization of prefixed words like p-adic are not spelled out in any copy-editing authority that I was able to find, including the Chicago Manual of Style and APA style, so the "correct" guideline will have to result from a compromise between usage, similar rules from other disciplines, and common sense. Of course this is a subject on which reasonable editors can and will differ.

For whatever it is worth (not much, I realize), here is my reasoning:


 * Publishers of mathematical books on p-adic numbers are not consistent. A search in Amazon.com for book titles containing "p-adic" yields all of the following examples on the very first page of the search results:
 * 1) p-adic Numbers: An Introduction
 * 2) P-adic Analysis Compared With Real
 * 3) P-Adic Numbers:
 * 4) p-Adic Lie Groups
 * The p in "p-adic" is a mathematical symbol, so its case should never be changed. Thus examples (2) and (3) above are clearly wrong. Example (1), no capitalization at all, is a travesty of capitalization. That leaves example (4), p-Adic, for the "correct" capitalized version of the word.
 * At least one scientific discipline does have a rule: The IUPAC has standardized the rules for capitalizing chemical names: Naming conventions (chemistry). The relevant recommendation is this (key part in boldface):


 * Another chemical example from Wikipedia: trans-3-Methyl-2-hexonoic acid. Note the capital M.
 * Here are three examples of capitalization after a prefix in mathematics: pre-Hilbert, non-Abelian, post-Cantor. See the discussion in Stack Exchange.

The rules for the capitalization of chemical names seem very reasonable to me, and they yield easily-understood chemical names. I suggest that a similar rule should be invoked when words like "p-adic" need to be capitalized. Thus: p-Adic, n-Adic, 3-Adic, etc.

I suggest this guideline:

I attempted to use this form of capitalization in this article, using p-Adic at the beginning of a sentence, but my edit was immediately reverted. Perhaps a discussion might be in order.

Thanks for your time. —Loren Cobb Aetheling (talk) 17:35, 5 April 2015 (UTC)


 * I understand your argument. I don't really like it, though.  One could (IMO) equally well remove the second sentence of your proposed guideline. This differs from the IUPAC convention.  In your other examples of mid"word" capitalization, the capitalized component is a proper name.  Perhaps this should be discussed on the WP:MOSCAP talk page?  — Arthur Rubin  (talk) 16:01, 7 April 2015 (UTC)
 * To me, this rule feels like "when the first letter of the first word of a sentence is not capitalizable, search for the next character that is capitalizable and capitalize it instead". Logically, for instance, this would mean that I should write "bell hooks Likes her name to be written in lower case" because the L in Likes is the first capitalizable letter. I don't think that is correct and I don't want to write like that in mathematics or in other subjects. I might tolerate it in chemistry only because you have found a style guide promoting that style, but even there I think it's wrong. —David Eppstein (talk) 16:25, 7 April 2015 (UTC)

"The reason for this property turns out to be that 10 is a composite number which is not a power of a prime."
OK, so what happens if we take p to be a prime power, say 8 (would those be the octadic numbers)? Double sharp (talk) 14:42, 31 October 2015 (UTC)
 * It's the same as for the prime. The "octadic" numbers would be the same as the dyadic numbers. Eric Kvaalen (talk) 12:06, 24 December 2015 (UTC)

$C$ used in 2 different meanings?
Section P-adic number uses $C$ in the close neighbourhood of the Cantor set, quite similar to Talk:P-adic number. But the $C$ in section P-adic number definitely means the complex numbers. This could be kept less ambiguous. --Nomen4Omen (talk) 19:44, 22 January 2016 (UTC)


 * Inserted $$\mathcal{C}$$ for Cantor set, kept $C$ for complex numbers. --Nomen4Omen (talk) 13:09, 26 January 2016 (UTC)

cut a Square into triangles
There is a nice applictation of 2-adic numbers, Theorem states that it is impossible to cut a square into odd number of equal triangles, I thought to write about that, but I do not remeber the names of authors, seach on web gave me nothing (if you know it put it here: User talk:Tosha) Tosha 14:22, 14 Jun 2004 (UTC)

On Dividing a Square Into Triangles. Author(s): Paul Monsky. Source: The American Mathematical Monthly, Vol. 77, No. 2 (Feb., 1970), pp. 161-164 — Preceding unsigned comment added by 129.104.247.2 (talk) 09:48, 27 March 2016 (UTC)

In section "Algebraic approach" I have changed (1, 3, 3, 3, 35, 35, ...) to (1, 3, 3, 3, 3, 35, 35, ...). I think this is the right version.

I am not sure whether the following sentence (in section Properties) is correct: "Thus e is a member of all algebraic extensions of p-adic numbers." Would you give it a thought? Mikolt (How do you put in your name and the date automatically?)


 * Since ep is a p-adic number, let's call it k, then e is a solution of xp = k. Since an algebraic closure (which I think is what is meant here by an "algebraic extension") of Qp must contain all roots of any polynomial whose coefficients are in Qp, then it must contain all roots of xp = k, and hence it must contain e. (and you get your name and date added automatically by typing " ~ " at the end of your message) Gandalf61 18:42, Jun 25, 2004 (UTC)


 * Yes, this is exactly what I thought. However, an extension is by no means a closure! In an extension you can just add the roots of one polynomial, plus the things you get by requiring that the field remain closed under the field-operations. And one more thing: there is the algebraic closure, not an, because up to isomorphism it is unique. I try to fix it in the article. (And thank you for the other (actually both) answer.) Mikolt 11:01, 28 Jun 2004 (UTC)


 * The number e, defined as the sum of the reciprocals of the factorials, is not even an element of the algebraic closure of the p-adic numbers, since the series doesn't even converge in Qp (which is already complete). As ep has p different roots in the algebraic closure, you have no canonical choice which to call e. - Holger

Assessment comment
Substituted at 02:26, 5 May 2016 (UTC)

10-adic example
The page states "Then the number |x-y| can always be expressed uniquely as 10^e·p/q, where p and q are positive integers relatively prime to each other and to 10." This is false, e.g. even when x-y=2, because there is no value of e such that 2 * 10^-e = p/q has both p,q relatively prime to 10 (e=0 makes p=2, e=-1 makes q=5, and all other e clearly don't work). This arises from the fact that, quite generally, there does not exist a p-adic absolute value when p is not prime, and it is why mathematicians do not talk about these other completions of the integers in these terms. The all the computations of 10-adic absolute values that follow are thus mathematically complete nonsense, even if evocative, and have no place in the article, so this section should be rewritten much more carefully. As a mathematician, I'd even remove all examples of nonprime p, and briefly explain why this is a good idea. Sharlaon (talk) 04:00, 2 April 2012 (UTC)


 * There is a discussion of g-adic values for composite g in Mahler's Tract:  This could be used to source a whole article on the subject.  At p.6 he shows that any rational can be written as g^{-f}.R/S where g does not divide R, (R,S) are coprime and (g,S) are coprime.  Then the g-adic value is g^f.  This is of course not what is written here.  However, given that this article is about p-adic numbers for prime p, I'll boldly remove this section as unsourced, undue and inaccurate.  Deltahedron (talk) 04:12, 24 August 2012 (UTC)


 * I'm not sure that 10-adic (and n-adic) numbers should be removed completely, but it does need a credible source. Could you move the deleted section to this talk page, so we can discuss it further?  — Arthur Rubin  (talk) 07:55, 24 August 2012 (UTC)

Section under discussion
''This section is an informal introduction to p-adic numbers, using examples from the ring of 10-adic numbers. (Base 10 was chosen to highlight the analogy with decimals. The 10-adic numbers are generally not used in mathematics: since 10 is not prime, the 10-adics are not a field.)  More formal constructions and properties are given below.''

In the standard decimal representation, almost all real numbers do not have a terminating decimal representation. For example, 1/3 is represented as a non-terminating decimal as follows


 * $$\frac{1}{3}=0.333333\ldots.$$

A real number can be approximated to any required degree of precision by a terminating decimal. If two decimal expansions differ only after the 10th decimal place, they are quite close to one another; and if they differ only after the 20th decimal place, they are even closer.

10-adic numbers use a similar non-terminating expansion, but with a different concept of "closeness". Whereas two decimal expansions are close to one another if their difference is a large negative power of 10, two 10-adic expansions are close if their difference is a large positive power of 10. Thus 3333 and 4333, which differ by 103, are close in the 10-adic world, and 33333333 and 43333333 are even closer, differing by 107.

More precisely, a rational number $r$ can be expressed as $10^{e}·p/q$, where $p$ and $q$ are positive integers and $q$ is relatively prime to $p$ and to 10. For each $r ≠ 0$ there exists the maximal $e$ such that this representation is possible. Let the 10-adic norm of $r$ to be
 * $$|r|_{10} =  \frac {1} {10^e}$$
 * |0|10 = 0.

Closeness in any number system is defined by a metric. Using the 10-adic metric the distance between numbers $x$ and $y$ is given by $&#124;x − y&#124;_{10}$. An interesting consequence of the $p$-adic metric is that there is no longer a need for the negative sign. As an example, by examining the following sequence we can see how unsigned 10-adics can get progressively closer and closer to the number −1:


 * $$9=-1+10 \, $$       so  $$ |9-(-1)|_{10} = \frac {1} {10}$$.
 * $$99=-1+10^2 \, $$      so  $$ |99-(-1)|_{10} = \frac {1} {100}$$.
 * $$999=-1+10^3 \, $$      so  $$ |999-(-1)|_{10} = \frac {1} {1000}$$.
 * $$9999=-1+10^4 \, $$      so  $$ |9999-(-1)|_{10} = \frac {1} {10000}$$.

and taking this sequence to its limit, we can say that the 10-adic expansion of −1 is


 * $$\dots 9999=-1.\,$$

In this notation, 10-adic expansions can be extended indefinitely to the left, in contrast to decimal expansions, which can be extended indefinitely to the right. Note that this is not the only way to write $p$-adic numbers – for alternatives see the Notation section below.

More formally, a 10-adic number can be defined as


 * $$\sum_{i=n}^\infty a_i 10^i$$

where each of the ai is a digit taken from the set {0, 1, … , 9} and the initial index $n$ may be positive, negative or 0, but must be finite. From this definition, it is clear that positive integers and positive rational numbers with terminating decimal expansions will have terminating 10-adic expansions that are identical to their decimal expansions. Other numbers may have non-terminating 10-adic expansions.

It is possible to define addition, subtraction, and multiplication on 10-adic numbers in a consistent way, so that the 10-adic numbers form a commutative ring.

We can create 10-adic expansions for negative numbers as follows


 * $$-100 = -1 \times 100 = \dots 9999 \times 100 = \dots 9900 \, $$
 * $$\Rightarrow -35 = -100+65 = \dots 9900 + 65 = \dots 9965 \, $$
 * $$\Rightarrow -\left(3+\dfrac{1}{2}\right)=\dfrac{-35}{10}= \dfrac{\dots 9965}{10}=\dots 9996.5$$

and fractions which have non-terminating decimal expansions also have non-terminating 10-adic expansions. For example


 * $$\dfrac{10^6-1}{7}=142857;

\dfrac{10^{12}-1}{7}=142857142857; \dfrac{10^{18}-1}{7}=142857142857142857$$
 * $$\Rightarrow-\dfrac{1}{7}=\dots 142857142857142857$$
 * $$\Rightarrow-\dfrac{6}{7}=\dots 142857142857142857 \times 6 = \dots 857142857142857142$$
 * $$\Rightarrow\dfrac{1}{7} = -\dfrac{6}{7}+1 = \dots 857142857142857143.$$

I think it would be beneficial to either show some expansions with numbers to the right of the decimal point or to state that this is not allowed, because is it stands this seems to imply that all rational numbers have a p-adic expansion with no numbers to the right of the decimal point yet it does not explicity state that as a fact. 80.47.68.7 (talk) 09:04, 21 June 2016 (UTC)

Generalizing the last example, we can find a 10-adic expansion for any rational number $p⁄q$ such that $q$ is co-prime to 10; Euler's theorem guarantees that if $q$ is co-prime to 10, then there is an $n$ such that $10^{n} − 1$ is a multiple of $q$.

As noted above, 10-adic numbers have a major drawback. It is possible to find pairs of non-zero 10-adic numbers whose product is 0 (though finding such pairs is not obvious ). This means that 10-adic numbers do not always have multiplicative inverses i.e. valid reciprocals, which in turn implies that though 10-adic numbers form a ring they do not form a field, a deficiency that makes them much less useful as an analytical tool. Another way of saying this is that the ring of 10-adic numbers is not an integral domain because they contain zero divisors. The reason for this property turns out to be that 10 is a composite number which is not a power of a prime. This problem is simply avoided by using a prime number $p$ as the base of the number system instead of 10 and indeed for this reason $p$ in $p$-adic is usually taken to be prime.

Discussion continued
I would just like to point out that it is not at all difficult to find pairs of non-zero decadic numbers whose product is 0, contrary to what was written in the removed section. Let's find a pair of decadic integers u and v such that uv=0. We want the last n digits of u to be equivalent to 0 mod 2n, and the last n digits of v to be equivalent to 0 mod 5n, but the last digit of either should be non-zero. This is easy. We can choose the last digit of u, say 2. We then choose the next digit to the left so that the last two digits are 0 mod 4, such as 12. There are five possible choices. We then choose the next digit to the left to give something 0 mod 8, such as 112. There are always five possibilities. For v, we do something similar. We have to choose the last diget as 5. The previous digit can be 2 or 7, let's take 2, so we have 25. The previous digit has to give us something 0 mod 125, so we may as well say 125 itself (we could have said 625). There are always two choices, separated by 5. The resulting u and v have the property that uv mod 10n is always 0. In other words, uv=0. Here's an example:


 * ...10112 × ...03125 = 0

Obviously we can then multiply either by any decadic number and the product will still be 0. Eric Kvaalen (talk) 15:16, 17 August 2014 (UTC)


 * The book by Mahler, pages 60-61, contins a discussion about zero divisiors in g-adic rings. Deltahedron (talk) 15:48, 17 August 2014 (UTC)

You know what, I'm gonna put that section back in. I found it quite useful when I first redd this article years ago. User:Sharlaon pointed out an error, but that was quickly fixed by User:Incnis Mrsi. User:Arthur Rubin has expressed doubts about whether the section should be removed completely. There is a reference in the section to a "credible source". I have modified or added a couple sentences. Eric Kvaalen (talk) 08:32, 19 August 2014 (UTC)

In correspondence between me and Louis de Branges he has defined a "modulus" that can be used with g-adic numbers when g is not necessarily prime. It's equal to a product of the prime divisors of g, raised to integer powers (which can be negative), such that the modulus times the number gives a g-adic integer having an inverse that is also a g-adic integer. To put it more simply for a rational number, you decompose the number as a quotient of primes, throw out all the primes that don't divide g, and then take the inverse. So in the decadic numbers the modulus of 10 will be 1/10 as above, but the modulus of 2 will be 1/2 instead of 1. The modulus of 4 will be 1/4, but of 6 will be 1/2. The modulus of a zero divisor is zero (it can be considered to have in it an infinite power of one of the prime divisors of g). De Branges points out that this "modulus" is not an absolute value when g is not prime. (An absolute value can only be zero for 0, whereas this modulus is zero for zero divisors as well.) I don't think this is worth putting into our article, but has anyone seen this elsewhere? Eric Kvaalen (talk) 16:51, 27 August 2014 (UTC)

Analytic Approach Chart: Ls and Js
Is there any distinct purpose to the use of black and white, 'L's and 'J's in the chart next to the Analytic Approach section? I see the pattern, but no real meaning is attributed to the letter choices nor the alternation between black and white. I presume there is one, at least for the letters. The color choice seems likely to have been for visibility against each column's color. Can anyone explain? LordQwert (talk) 21:00, 22 November 2016 (UTC)
 * I assume that the use of black and white is to attempt to comply with MOS:CONTRAST, but it doesn't succeed (e.g. the rightmost of the four green columns, with foreground #FFFFFF and background #789966, fails the test at https://snook.ca/technical/colour_contrast/colour.html#fg=FFFFFF,bg=789966 But I have no idea what the J's and L's are supposed to mean. I don't think this is a very good visualization. —David Eppstein (talk) 21:23, 22 November 2016 (UTC)

almost all
Doesn't "almost all" mean: up to a finite number of exceptions? Its usage here seems unconventional. — Preceding unsigned comment added by 193.83.106.80 (talk) 08:09, 7 March 2018 (UTC)
 * If you click on the almost all link you will find that it "has a number of specialised uses" one of which is 'Referring to a subset of an uncountable set, "almost all" is sometimes used to mean "all the elements except for a countable set"'. —David Eppstein (talk) 08:24, 7 March 2018 (UTC)

Backwards arithmetic
Copied from my talk page: — Preceding unsigned comment added by Gene Ward Smith (talk • contribs) 15:22, 7 May 2006 (UTC)

this article is bad, impossible to understand
there is major drawback with this article : we don't understand how to construct the p-adic expansion of a rational a/b.

you have to understand that 99% of the people coming here just want to know that. — Preceding unsigned comment added by 78.227.78.135 (talk) 18:28, 23 March 2016 (UTC)

2-adic binary string representation
Binary numbers where 00 is not equal to 0 can be expressed by a kind of encoding system, a string of 0s and 1s in the form  $$z_i$$, with i varying from 0 to n-1, where n is the bit-length of the string... But it is not an arbitrary  string sequence, it is also a kind of number... It is a  2-adic number, where each positional digit have the value $$a_i 2^i$$... So, something in the article is wrong at

How to calculate $Q_{p}$ or to express binary encodings with ak = 0?

Krauss (talk) 11:10, 15 April 2019 (UTC)


 * OK, this is not in the text. But obviously it corresponds to the empty string or the empty sum. Maybe you take $$k = \infty$$. --Nomen4Omen (talk) 11:25, 15 April 2019 (UTC)

Are there p-adic integers that represent irrational numbers
For example -- is there a p-adic integer whose square is 2?

We know there are p-adic integers that represent the natural numbers, the integers, positive and negative and the rational numbers whose denominators do not have p as a factor. But what about the square root of 2?

Certainly this depends on the prime. If p = 5 there is no such p-adic integer Let us denote q by …d3d2d1. Since d1* d1 must equal 2 we see that only a value of p for which i*i = 2 for some digit,i,  will do. That leaves out 2,3 and 5. But p = 7 is a possibility since 3*3 = 2 mod 7 And d1 for the aquare root of 2 mey be 3 (or 4). Some crude calculations with p = 7 leads me to suspect that (...421216213) is indeed close to a square root of 2.

ABS ((...421216213)*(421216213)-2) <= (1/7)* (1/7) (1/7) *(1/7) *(1/7) *(1/7) *(1/7)

and I stopped when I got bored. I believe I can get a lot closer than that. But I cannot prove that (...421216213) actually converges to 2.

Starting with d1 = 4 does not lead to anything that looks like an approximation to the square root of 2.

Since I do not know how to navigate on wiki pages I would appreciate hearing from you directly if you have a comment. Thanks.

Gus Rabson gusrabson@comcast.net

—Preceding unsigned comment added by 76.19.57.205 (talk) 21:33, 7 January 2011 (UTC)


 * "Is there a p-adic integer whose square is 2?" - yes, for certain values of p. Hensel's lemma, which is an extension of Newton's method, tells us that if we have an integer r such that
 * $$r^2=2 \mod p^{k}$$
 * then there is a unique integer s between 0 and p-1 such that
 * $$(sp^k+r)^2=2 \mod p^{k+1}$$
 * In other words, we can "lift" the solution modulo pk to a solution pk+1 by adding spk, and s is the k+1-th digit of the square root of 2 when expressed as a p-adic integer. If we start with an initial digit a such that
 * $$a^2=2 \mod p$$
 * (as, in your example, 3^2 = 2 mod 7) then Hensel's lemma lets us generate the infinite sequence of digits of the square root of 2 expressed as a p-adic integer. There is also a 7-adic integer whose first digit is 4 and whose square is 2 - it is the "negative" (additive inverse) of the one you have found that starts with 3, so its first few digits are ...0454 . Gandalf61 (talk) 06:37, 8 January 2011 (UTC)

Thanks, Gandalf for your speedy and informative response. You can probably guess my next question. By a countability argument we know that there must be real numbers that are not p-adic integers for any p. Can you name one? What about pi, or e? — Preceding unsigned comment added by Gusrabson (talk • contribs) 18:57, 8 January 2011 (UTC)

How to relate irrational real numbers with irrational p-adic numbers
e is not a p-adic integer for any p, as the series defining it doesn't converge in the p-adic metric for any p. Double sharp (talk) 11:49, 28 June 2019 (UTC)

$$\exp(x) := \sum_{n=0}^\infty \frac{x^n}{n!}$$ converges for $ Thus $$\exp(p)\in\Q_p$$ for $p − 1$. But whether one of the roots $$\sqrt[p]{\exp(p)}$$ can be “identified” with Euler's number $$e = \exp(1) = 2.71828\ldots \in\R$$ and what the meaning of this should be is quite questionable. Furthermore, it is not obvious how to relate (or even identify) irrational real numbers with irrational p-adic numbers. E.g. $$\sqrt{41} \not\in \Q_3$$, but $$\Z_2 \ni \sqrt{41} \in \Z_5$$ and the 2-adic is  $$\sqrt{41} = \sqrt{101001_2} \; \underset{\Z_2}= \; \ldots 11001101_2 = (1, 5, 13, 77, 205, \ldots)_{10}$$, whereas the 5-adic is  $$\sqrt{41} = \sqrt{131_5}\; \; \; \; \; \; \underset{\Z_5}= \; \ldots 32434204_5 = (4, 54, 554, 2429, \ldots)_{10}$$. These 2 expansions can be united by means of the Chinese remainder theorem: $$\sqrt{41} \; \underset{\Z_{10}}= \; \ldots 26452429_{10}$$. Is there more than what is written here what is common between $$\sqrt{41} \; \underset{\R}= \; 6.40312423 \ldots_{10}$$ and these 3 series? --Nomen4Omen (talk) 20:26, 30 June 2019 (UTC)
 * The exponential series


 * Of course, having both a 2-adic and a 5-adic number, we can “cross” them using the Chinese remainder theorem. But what’s the merit in such string of decimal digits? $Q_{2}$ is no field. It is also far from obvious, to me, how to solve quadratic equations in $p > 2$; can only suppose that the number of roots can be determined with quadratic residues. Incnis Mrsi (talk) 20:33, 5 August 2019 (UTC)

Old discussion
A question - I'm confused by the intention of the original article writer to compare the infinite sum &sum; aipi with the algebraic definition of p-adic integer, which appears to distinguish integers not by their sum in any real sense, but by their distinctiveness as an infinite sequence. Is the "sum" analogy a good one? How is it related to the convergence under the p-adic metric? It does seem to make sense that the partial sums of the p-adic metric, where we just extend the final digit to the right, should converge; but it's not clear to me.Chas zzz brown (talk) 09:09, 21 December 2002 (UTC)


 * Yup, the connection between sequences and sums was missing. I'll add it. Basically, if you have the sequence (1, 3, 3, 11, 11, 43, ...) in the 2-adics, you write it as the series 1 + 1*2 + 0*4 + 1*8 + 0*16 + 1*32 + ... The partial sums of this series form the original sequence. AxelBoldt (talk) 00:32, 4 January 2003 (UTC)

Furthermore, it's hard to see how we get a field using the algebraic description as given here; since not every p-adic integer has a multiplicative inverse (any m with p | m canjnot have an inverse). Is the given definition correct? Chas zzz brown (talk) 22:24, 21 December 2002 (UTC)


 * The p-adic integers are an integral domain and therefore have a field of quotients. m doesn't have an inverse in the p-adic integers, but it does have an inverse in the p-adic numbers (of which you can think as infinite p-adic expansions to the left which also have finitely many digits to the right of the "decimal" point). For instance, the inverse of 12 in the 2-adics is 1*2-2 + 1*2-1 + 0 + 1*2 + 0*4 + 1*8 + ... which you can check by multiplying the latter with 12. AxelBoldt (talk) 00:32, 4 January 2003 (UTC)


 * Muchly appreciated Axel. The fact that the indexing can include negative numbers was a missing piece of the original explanation. Chas zzz brown (talk) 08:59, 6 January 2003 (UTC)

Kudos to whoever wrote this article. I had heard of p-adic numbers for a while and was looking for an accesible definition. The clear exposition here encouraged me to look farther into the Wikipedia. (BTW, this comes up 6th on a Google search for p-adic numbers.) JPB (talk) 03:37, 11 July 2003 (UTC)

I changed the function f(x) = (1/|x|p)2 into (|x|p)2 since the former is not defined at 0. (May have been a typo.) -- SirJective (talk) 12:41, 12 August 2003 (UTC)


 * Well, that was a mistake of mine: The given function was correct, I didn't work out the derivative:
 * $$\left| (1/|x|_p^2) / x \right|_p = |x|_p$$.
 * The function I gave is not even continuous at 0. I now corrected the article. --SirJective (talk) 07:27, 26 July 2004 (UTC)

Spherical completeness
Here an end is reached, as &#937;p is algebraically closed.
 * Actually, this may not be the end. The completion of the algebraic closure of Qp is algebraically closed and topologically complete, but not spherically complete. (Meaning: every decreasing sequence of closed balls has a nonempty intersection.) Turns out spherical completeness is something worth having. (See "A Course in p-adic Analysis", Alain Robert) Also, there appears to be a conflict in notation. Some authors use &Omega;p for the spherical completion and Cp for the topological completion of the algebraic closure of Qp. This seems to make much more sense to me, because the "C" part matches with what we expect from the complex numbers, except that of course "C&infin;" = "&Omega;&infin;" since the complex numbers are already spherically complete. Revolver (talk) 13:18, 7 November 2004 (UTC)

p-adic number system
Am I completely wrong in that I believe that one also refers to 'p-adic' number system as synonym of "base-p positional notation" ?

I think we should be not too categorical about the use of "system", which some, feeling very "rigourous", categorically use as synonym of "set", while it really means "(finite(?)) family", and definetly not "semiring" (i.e. numbers themselves (be they real, complex, natural or whatsoever) do not form a system).

Everyday people (and without doubt most dictionaries) use "system" for a collection of conventions, in that sense the "numbering (I mean: number writing) system in base n" seems to me quite well defined.

In any way, a big effort is to be made in interconnecting all that is written about positional notation (decimal etc. etc.) and in making it clear what one speaks about, not in being too bourbakist about definition of a the only true one terminology, but in adding well-explained cross-references on top of each article on the field (or at least one "disambiguation" page). MFH (talk) 14:58, 8 April 2005 (UTC)


 * I do not recall ever seeing p-adic number used to mean base-p notation. They are quite different concepts - one is a number system, the other is a numeral system - the difference is well explained in those two articles. Having said that, I can see no harm in adding a clarification note with a pointer to number system at the beginning of the p-adic number article, to avoid confusion. Gandalf61 (talk) 15:34, 9 April 2005 (UTC)

material moved from the article:
[ilan]: Hello! You don't need any of these technicalities as motivation (I usually think of motivation as being non-technical, but that's me). First off, your p doesn't have to be a prime number, the construction works for any base, including the usual base 10. So, a 10-adic integer is simply an integer with possibly an infinite number of digits to the left. Since most people are used to doing ordinary arithmetic with an infinite number of digits to the right of the decimal point, they should be able to adjust to an infinite number of digits to the left, just arithmetic as usual. A p-adic number is the same thing, but now p is a prime number, that is, you are doing your digit expansions and arithmetic in base p with possibly an infinite number of digits to the left. I don't see why you have to say anything more complicated that that!


 * Indeed the construction works for any composite number, except in some sense the p-adic numbers describe all such completions of the integers. For example, the 10-adic integers you describe are isomorphic to the 2-adic integers cross the 5-adic integers.  (In general, the n-adic integers are isomorphic to the product of the p-adic integers, where p ranges over the distinct prime factors of n.)  However, I do see your point that those not acquainted with college math would find the idea of infinite decimals to the left easier to digest than numbers expanding infinitely to the left in other bases. Stuwanker (talk) 15:08, 22 June 2005 (UTC)

Are there p-adic integers that represent irrational numbers
For example -- is there a p-adic integer whose square is 2?

We know there are p-adic integers that represent the natural numbers, the integers, positive and negative and the rational numbers whose denominators do not have p as a factor. But what about the square root of 2?

Certainly this depends on the prime. If p = 5 there is no such p-adic integer Let us denote q by …d3d2d1. Since d1* d1 must equal 2 we see that only a value of p for which i*i = 2 for some digit,i,  will do. That leaves out 2,3 and 5. But p = 7 is a possibility since 3*3 = 2 mod 7 And d1 for the aquare root of 2 mey be 3 (or 4). Some crude calculations with p = 7 leads me to suspect that (...421216213) is indeed close to a square root of 2.

ABS ((...421216213)*(421216213)-2) <= (1/7)* (1/7) (1/7) *(1/7) *(1/7) *(1/7) *(1/7)

and I stopped when I got bored. I believe I can get a lot closer than that. But I cannot prove that (...421216213) actually converges to 2.

Starting with d1 = 4 does not lead to anything that looks like an approximation to the square root of 2.

Since I do not know how to navigate on wiki pages I would appreciate hearing from you directly if you have a comment. Thanks.

Gus Rabson gusrabson@comcast.net

—Preceding unsigned comment added by 76.19.57.205 (talk) 21:33, 7 January 2011 (UTC)


 * "Is there a p-adic integer whose square is 2?" - yes, for certain values of p. Hensel's lemma, which is an extension of Newton's method, tells us that if we have an integer r such that
 * $$r^2=2 \mod p^{k}$$
 * then there is a unique integer s between 0 and p-1 such that
 * $$(sp^k+r)^2=2 \mod p^{k+1}$$
 * In other words, we can "lift" the solution modulo pk to a solution pk+1 by adding spk, and s is the k+1-th digit of the square root of 2 when expressed as a p-adic integer. If we start with an initial digit a such that
 * $$a^2=2 \mod p$$
 * (as, in your example, 3^2 = 2 mod 7) then Hensel's lemma lets us generate the infinite sequence of digits of the square root of 2 expressed as a p-adic integer. There is also a 7-adic integer whose first digit is 4 and whose square is 2 - it is the "negative" (additive inverse) of the one you have found that starts with 3, so its first few digits are ...0454 . Gandalf61 (talk) 06:37, 8 January 2011 (UTC)

Thanks, Gandalf for your speedy and informative response. You can probably guess my next question. By a countability argument we know that there must be real numbers that are not p-adic integers for any p. Can you name one? What about pi, or e? — Preceding unsigned comment added by Gusrabson (talk • contribs) 18:57, 8 January 2011 (UTC)

How to relate irrational real numbers with irrational p-adic numbers
e is not a p-adic integer for any p, as the series defining it doesn't converge in the p-adic metric for any p. Double sharp (talk) 11:49, 28 June 2019 (UTC)

$$\exp(x) := \sum_{n=0}^\infty \frac{x^n}{n!}$$ converges for $ Thus $$\exp(p)\in\Q_p$$ for $Q_{10}$. But whether one of the roots $$\sqrt[p]{\exp(p)}$$ can be “identified” with Euler's number $$e = \exp(1) = 2.71828\ldots \in\R$$ and what the meaning of this should be is quite questionable. Furthermore, it is not obvious how to relate (or even identify) irrational real numbers with irrational p-adic numbers. E.g. $$\sqrt{41} \not\in \Q_3$$, but $$\Z_2 \ni \sqrt{41} \in \Z_5$$ and the 2-adic is  $$\sqrt{41} = \sqrt{101001_2} \; \underset{\Z_2}= \; \ldots 11001101_2 = (1, 5, 13, 77, 205, \ldots)_{10}$$, whereas the 5-adic is  $$\sqrt{41} = \sqrt{131_5}\; \; \; \; \; \; \underset{\Z_5}= \; \ldots 32434204_5 = (4, 54, 554, 2429, \ldots)_{10}$$. These 2 expansions can be united by means of the Chinese remainder theorem: $$\sqrt{41} \; \underset{\Z_{10}}= \; \ldots 26452429_{10}$$. Is there more than what is written here what is common between $$\sqrt{41} \; \underset{\R}= \; 6.40312423 \ldots_{10}$$ and these 3 series? --Nomen4Omen (talk) 20:26, 30 June 2019 (UTC)
 * The exponential series


 * Of course, having both a 2-adic and a 5-adic number, we can “cross” them using the Chinese remainder theorem. But what’s the merit in such string of decimal digits? $Q_{p}$ is no field. It is also far from obvious, to me, how to solve quadratic equations in $p > 2$; can only suppose that the number of roots can be determined with quadratic residues. Incnis Mrsi (talk) 20:33, 5 August 2019 (UTC)

"and, ultimately, to the reals"
In this talk, the section How to relate irrational real numbers with irrational p-adic numbers contains some examples of non-rational algebraic numbers which are solutions of the same polynomial over Q: in R and several Qp. Their p-adic expansions are apparently uncorrelated. I do not know of any literature which relates them. In a sense beyond the defining equation. Does somebody else know anything? --Nomen4Omen (talk) 19:03, 5 August 2019 (UTC) And then after all, just as you say: most numbers are not algebraic! --Nomen4Omen (talk) 21:03, 5 August 2019 (UTC)
 * am not convinced in the presumed method to solve quadratic equations in $Q_{10}$. Can expand the linked talk section further? Anyway, most numbers are not algebraic (think of cardinality). Incnis Mrsi (talk) 20:27, 5 August 2019 (UTC)
 * You are right with quadratic residue (in the case of quadratic polynomials), then Hensel's lemma. But we already knew that before: that quadratic equations are solvable. And as it appears the square root examples shown above did not help you. And my question is: Does this example somebody tell anything and what? Even in the case of square roots.

How to relate irrational real numbers with irrational p-adic numbers (revisited)
Above are some examples of non-rational algebraic numbers which are solutions of the very same algebraic equation $$X^2-41 = 0 $$ over $Q_{p}$, however, each solution is irrational ($$\notin \Q $$) and developped in $Q_{p}$ and some special $Q$ ($R$).

The existence of the solutions is guaranteed by well-known theorems, e.g. first take quadratic residue then "lift" the residue class using Hensel's lemma. In addition, for $Q_{p}$ the Chinese remainder theorem “crosses” the solution in $p=2,5,10$ with that in $p=10$.

It is easy to prove the correctness of the expansions by squaring them:
 * $$\begin{array}{lrrrrrl}

\Z_2 : & (\ldots 011011001101_2)^2 &=& \ldots 0000101001_2 & \underset{\Z_2}{\longrightarrow} & 101001_2 = & 41_{\mathrm{dec}} \\ \R_2 : & (110._{2}011001110 \ldots)^2 &=& 101000._{2}111111 \ldots &\underset{\R}{\longrightarrow} & 101001_2 = & 41_{\mathrm{dec}} \\ \Z_5 : & (\ldots 32434204_5)^2 &=& \ldots 00000131_5 &\underset{\Z_5}{\longrightarrow}   & 131_5 = & 41_{\mathrm{dec}} \\ \R_5 : & (11._{5}200143402 \ldots)^2 &=& 130._{5}44444444 \ldots &\underset{\R}{\longrightarrow} & 131_5 = & 41_{\mathrm{dec}} \\ \Z_2 \!\!\times\! \Z_5 \,\cong\, \Z_{10} : & (\ldots 26452429_{10})^2 &=& \ldots 0000041_{10} &\underset{\Z_{10}}{\longrightarrow} & 41_{10} = & 41_{\mathrm{dec}} \\ \R : & (6.40312423 \ldots_{\mathrm{dec}})^2 &=& 40.9999999 \ldots_{\mathrm{dec}} &\underset{\R}{\longrightarrow} & & 41_{\mathrm{dec}} \\ \end{array}$$ Thereby $Q_{2}$ shall mean that a real number is written in base (radix) $Q_{5}$ (with the radix as a subscript immediately right adjacent to the radix point – as opposed to the left adjacency when the the expansion is $R_{p}$-adic).

The limits are taken according to the respective metrics, i.e. the metrics in $p$ and $p$. The square root is irrational (is not in $R$); thus no expansion can be finite (nor periodic). All the non-reals are $Q_{p}$-adic and $Q$-adic integers (are in $p$). Digitwise they look completely unrelated.

Question: Is there more relation among them – beyond their being zeroes of $$X^2-41 \in \Z[X] $$? --Nomen4Omen (talk) 18:37, 22 August 2019 (UTC)

Supplementary remark: Even the expansions in the reals according to base $p$ ("$Z_{p}$") have no apparent similarity with the respective expansions in $p$. --2001:1A81:32BB:800:ED6E:807A:8452:99A8 (talk) 11:51, 2 September 2019 (UTC)

(Text even further improved by --Nomen4Omen (talk) 16:42, 3 September 2019 (UTC))

Derivative
The article mentions a function supposed to have "zero derivative everywhere but not locally constant at 0". But how is the "derivative" defined? Obviously the statement does not mean anything without a definition! The phrase links to derivative, but no part of that article is relevant here, it's only about real valued functions. There are several generalisations of the derivative but that article doesn't mention "p-adic". In turn, the article p-adic analysis doesn't mention "derivative" or "differentiation", ... Could anyone give a hint on what is meant? Thanks in advance! &mdash; MFH:Talk 15:59, 12 June 2019 (UTC)

(PS: In the section "old discussion" someone mentioned "I could not work out the derivative of..." but that was silently ignored...)
 * . Of course, we don’t need such a section to say only about total disconnectedness of $R_{p}$ using yet another metaphor. Incnis Mrsi (talk) 08:58, 13 June 2019 (UTC)
 * Moreover, p-adic analysis generally doesn’t use derivatives and sharply distinguishes two scenarios: p-adic-valued functions on $Q_{p}$ vs real (or complex) functions there. For the former, mapping of balls is a much more useful concept than derivatives. For the latter, the analysis relies on the Vladimirov’s operator which isn’t local and resembles the wavelet transform. Incnis Mrsi (talk) 09:10, 13 June 2019 (UTC)
 * The author is using the usual definition of the derivative, the "usual" limit. I don't see any problem, if he was talking about a continuous function, would ask you "what is the definition of a continuous function in the p-adic case"? Probably not. You'd use the good old epislon-delta definition. We are in the same situation, he is using the epsilon delta definition of the derivative. — Preceding unsigned comment added by Alcyon007 (talk • contribs) 19:47, 8 February 2020 (UTC)

Unclear definition of 10-adic "absolute value"
In the introduction, the 10-adic "absolute value" is defined only for powers of 10, but it needs to be defined for any rational number for the definition of the 10-adic metric that follows to make any sense. Unless I'm missing something? — Preceding unsigned comment added by 93.25.93.82 (talk) 10:52, 28 March 2020 (UTC)


 * You are right: $r$ has not been defined. Now it is. - Nomen4Omen (talk) 15:20, 28 March 2020 (UTC)