After just having learned about $p$-adic numbers I've now got another question which I can't figure out from the Wikipedia page.

As far as I understand, the $p$-adic numbers are basically completing the rational numbers in the same way the real numbers do, except with a different notion of distance where differences in more-significant digits correspond to small distances, instead of differences in less-significant digits. So if I understand correctly, the $p$-adic numbers contain the rational numbers, but not the irrational numbers, while the non-rational $p$-adic numbers are not in $\mathbb{R}$ (someone please correct me if I'm wrong).

Now the real numbers do not depend on the base you write the numbers in. However the construction of the $p$-adic numbers seems to depend on the $p$ chosen. On the other hand I am sure that the construction of the real numbers can be written in a way that it apparently depends on the base, so the appearance might be misleading.

Therefore my question: Are the $p$-adic numbers the same for each $p$ (that is, are e.g. $2$-adic and $3$-adic numbers the same numbers, only written in different bases), or are they different (except for the rational numbers, of course). For example, take the $2$-adic number $x := ...1000001000010001001011$ (i.e. $\sum_{n=0}^\infty 2^{n(n+1)/2}$), which IIUC isn't rational (because it's not periodic). Can $x$ also be written as $3$-adic number, or is there no $3$-adic number corresponding to this series?

In case they are different, is there some larger field which contains all $p$-adic numbers for arbitrary $p$?

  • 3
    $\begingroup$ The adeles do serve to package up the various completions in some way, but this is a pretty subtle object and it's probably best to not think about it for now. Harry’s answer is excellent. $\endgroup$ Commented Jul 24, 2012 at 22:08
  • $\begingroup$ I took the liberty of editing the title, because at first glance it seemed to be asking whether the set of $p$-adic numbers is a singleton. Anyway, nice question (and nice answer by Harry)! $\endgroup$
    – user856
    Commented Jul 24, 2012 at 22:18

1 Answer 1


No, the different $p$-adic number systems are not in any way compatible with one another.

A $p$-adic number is a not a number that is $p$-adic; it is a $p$-adic number. Similarly, a real number is not a number that is real, it is a real number. There is not some unified notion of "number" that these are all subsets of; they are entirely separate things, though there may be ways of identifying bits of them in some cases (e.g., all of them contain a copy of the rational numbers).

Now, someone here is bound to point out that if we take the algebraic closure of some $\mathbb{Q}_p$, the result will be algebraically isomorphic to $\mathbb{C}$. But when we talk about $p$-adic numbers we are not just talking about their algebra, but also their absolute value, or at least their topology; and once you account for this they are truly different. (And even if you just want algebraic isomorphism, this requires the axiom of choice; you can't actually identify a specific isomorphism, and there's certainly not any natural way to do so.)

How can we see that they are truly different? Well, first let's look at the algebra. The $5$-adics, for instance, contain a square root of $-1$, while the $3$-adics do not. So if you write down a $5$-adic number which squares to $-1$, there cannot be any corresponding $3$-adic number.

But above I claimed something stronger -- that once you account for the topology, there is no way to piece the various $p$-adic number systems together, which the above does not rule out. How can we see this? Well, let's look at the topology when we look at the rational numbers, the various $p$-adic topologies on $\mathbb{Q}$. These topologies are not only distinct -- any finite set of them is independent, meaning that if we let $\mathbb{Q}_i$ be $\mathbb{Q}$ with the $i$'th topology we're considering, then the diagonal is dense in $\mathbb{Q}_1 \times \ldots \times \mathbb{Q}_n$.

Put another way -- since these topologies all come from metrics -- this means that for any $c_1,\ldots,c_n\in\mathbb{Q}$, there exists a sequence of rational numbers $a_1,a_2,\ldots$ such that in topology number 1, this converges to $c_1$, but in topology number two, it converges to $c_2$, and so forth. (In fact, more generally, given any finite set of inequivalent absolute values on a field, the resulting topologies will be independent.)

So even on $\mathbb{Q}$, the different topologies utterly fail to match up, so there is no way they can be pieced together by passing to some larger setting.

  • $\begingroup$ +1 Great answer! But I think you want $c_i\in\mathbb{Q}_i$. $\endgroup$
    – M Turgeon
    Commented Jul 25, 2012 at 2:04
  • 2
    $\begingroup$ Well, the $\mathbb{Q}_i$ are all the same set; it's only the topology that's changing. You can write $c_i\in\mathbb{Q}_i$ if you want, but I thought it was better to emphasize that these are all the same set. $\endgroup$ Commented Jul 25, 2012 at 6:08
  • $\begingroup$ So does it mean, that cardinality of different n-adic numbers can be equal, but all p-adic always will have different cardinalities or is it uncertain? $\endgroup$
    – Sugar
    Commented Jul 28, 2021 at 21:00
  • $\begingroup$ All $\mathbb{Q}_p$ have the same cardinality, namely, $2^{\aleph_0}$. The same is still true if you look at $n$-adics instead of $p$-adics. Cardinality isn't really that relevant to the question. $\endgroup$ Commented Jul 28, 2021 at 23:58
  • $\begingroup$ For the record, recently my essay-like answer math.stackexchange.com/a/4007515/96384 has gotten some attention, but I realize that everything in there is already contained in a nutshell in this excellent answer of yours from nine years ago; and probably I had even read this somewhere along the way before formulating it in my own words. Thank you for that! $\endgroup$ Commented Aug 12, 2021 at 16:46

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