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Are p-adic numbers real numbers?Why or why not?I came across the idea that it is not a real number from-Is 0.9999... equal to -1? (last comment)

Community
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Harmonic
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  • Soham, you see that the question has been put on hold as unclear. While I managed to think of an IMHO useful interpretation for the question, my level of confidence on that particular interpretation is not very high. My answer is probably meaningless to you unless you know: A) what a field is? and B) what a $p$-adic number is? Therefore it would help immensely, if you told a bit about your background. For example I cannot rule out the possibility that you were just curious about the term "$p$-adic number" in that other thread. – Jyrki Lahtonen Apr 23 '15 at 06:03
  • The context there is that in the usual metric of the set of rational numbers, the sequence $0.9$, $0.99$, $0.999$, $0.9999$ et cetera converges, and has the limit $0.999\ldots=1$. The "usual metric" means that the distance between two numbers, $x$ and $y$, is $|x-y|$. When we turn to $p$-adics the metric is vastly different. The distance between two integers $x$ and $y$ is $1/p^t$, where $t$ is the largest exponent with the property that $p^t\mid x-y$. The sequence $9$, $99$, $999$, $9999$, $\ldots$ converges to $-1$ in $2$-adics as well as $5$-adics, because of this. Do you see why? – Jyrki Lahtonen Apr 23 '15 at 06:13
  • Anyway, if you want to understand this better, then you could either study a book introducing you to the concept, or clarify the question pinpointing a concept or a line of thought from that other thread that you have trouble understanding. No pressure - you can just let the matter rest, when the question will probably be deleted in due course. Because it is unclear in this sense. – Jyrki Lahtonen Apr 23 '15 at 06:17
  • Related: https://math.stackexchange.com/q/4006006/96384 – Torsten Schoeneberg Mar 07 '25 at 15:12

2 Answers2

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No, the field of $p$-adic numbers, $\Bbb{Q}_p$, is not isomorphic to a subfield of $\Bbb{R}$, and thus I don't see a way of calling $p$-adic numbers "real". A way of seeing this is that, by Hensel lifting, the negative integer $-4p+1$ has a square root that is a $p$-adic integer. It is well known that $-4p+1$ has no square roots in $\Bbb{R}$.

As others pointed out the $p$-adics really are something more or less independent from the reals. A construction called completion takes us from $\Bbb{Q}$ to $\Bbb{R}$ as well as to $\Bbb{Q}_p$, but the related metrics are all independent (in a sense that can be made precise), and the completions look very different.

The only reason I used $-4p+1$ above instead of $-p+1$ is to make the argument also work for $p=2$.

Jyrki Lahtonen
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There are two possible types of completion of $\mathbb Q$. With one you get $\mathbb R$ and with the other you get the p-adics. I don't have much experience in the area, but I don't think that it makes sense to say $\mathbb Q_p \subset \mathbb R$ or $\mathbb R \subset \mathbb Q_p.$

James
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  • It is possible to embed the algebraic closure of $\mathbb Q_p$ into $\mathbb C$, and I'm sure there is such an embedding, under which $\mathbb Q_p$ is mapped to $\mathbb R$. But I don't think this is particularly helpful to know. – Thomas Poguntke Apr 22 '15 at 15:38
  • @Engloutie: I agree with you in that such an embedding would not be helpful. But it cannot even exist! See my answer. – Jyrki Lahtonen Apr 22 '15 at 15:56
  • Right, that was quite silly of me! I didn't really think about it because I only wanted to make the other point. Thank you! – Thomas Poguntke Apr 22 '15 at 15:58