217 votes
Accepted

Why does an argument similiar to 0.999...=1 show 999...=-1?

If you want to understand the mathematics behind these things, it is all based upon the notions of 'convergence' and of 'limits'. If you read any first course in analysis textbook you will find the ...
114 votes

Why does an argument similiar to 0.999...=1 show 999...=-1?

In the $10$-adic numbers it is true that $\dots 9999 = -1$. More precisely, the series $\sum_{n=0}^{\infty} 9 \cdot 10^n$ converges in $\mathbb{Q}_{10}$ and its limit there is $-1$.
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46 votes

Why does an argument similiar to 0.999...=1 show 999...=-1?

As other users have noted, it doesn't make sense to have infinitely many nines to the left of the decimal point. This is because the sequence $9, 99, 999, \ldots$ doesn't converge to anything, unlike ...
33 votes
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Units of p-adic integers

This must be covered in almost every text on the $p$-adic numbers; I think the book of Gouvêa is the best of these. And your statement is not quite true: for $p=2$, $\mu_{p-1}$ is trivial all right, ...
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26 votes
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Ring of the integer $p$-adic numbers $\mathbb{Z}_p$

Let's look at a different ring first: $R=k[[X]]$, the ring of formal power series over a field. We can also think of $R$ as the ring of sequences $f_0, f_1, f_2,\ldots$, where each $f_i$ is a ...
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25 votes
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Method of finding a p-adic expansion to a rational number

The short answer is, long division. Say you want to find the $5$-adic expansion of $1/17$. You start by writing $$ \frac{1}{17}=k+5q $$ with $k \in \{0,1,2,3,4\}$ and $q$ a $5$-adic integer (that is, ...
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23 votes

Why does an argument similiar to 0.999...=1 show 999...=-1?

One thing to think about is how this plays with modular arithmetic, if you're familiar. Basically, arithmetic mod $10^n$ is arithmetic where we only care about the last $n$ digits of a number and is ...
  • 59.1k
23 votes

p-adic numbers vs real numbers

Could anyone give a concrete example of a p-adic number that is not a "real number"? 1. If one is very very formal, one could say that no $p$-adic number is a real number. The sets $\mathbb ...
19 votes
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Can $\pi$ be defined in a p-adic context?

In a very flattering note @saulspatz has asked me to weigh in on this issue. But many of the commenters to your question have already evinced rather greater familiarity with deep $p$-adic matters than ...
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18 votes
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Roadmap to Iwasawa Theory

Iwasawa theory has at least two aspects, algebraic and $p$-adic analytic. Initially, they were developped independently until they merged into one single subject dominated by the so-called Main ...
15 votes

Why are $p$-adic numbers and $p$-adic integers only defined for $p$ prime?

You can define the $n$-adic integers, even if $n$ is not prime. For example, every $10$-adic integer has a $10$-adic decimal representation, and you can add them and carry as usual. One thing that I ...
15 votes

Visualizing $\textbf{Q}_p$ vs. $\textbf{F}_p((t))$?

Let's visualize the fragments $\mathbf Z_p$ and $\mathbf F_p[[t]]$ instead, since they're compact, and focus only on their structure as abelian topological groups. Then the left figure is $\mathbf Z_3$...
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14 votes

The group $(1+p\mathbb Z_p)/(1+p^{n}\mathbb Z_p)$

To describe elements of $(1+p\mathbf Z_p)/(1+p^n\mathbf Z_p)$ concretely, observe for $p$-adic units $u$ and $v$ that $|u/v -1|_p = |u-v|_p$, so the multiplicative congruence $u \equiv v \bmod (1+p^n\...
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14 votes
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Is the real number $\sqrt{6}$ in $\mathbb{R}$ equal to the 5-adic number $\sqrt{6}$ in $\mathbb{Q}_5$?

The question of whether two "numbers" are "equal" is a somewhat subtle one. For example, lets work with a simpler number, namely "2". Certainly $2\in\mathbb{Z}$, but also $2\in\mathbb{Q},2\in\mathbb{...
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14 votes

Why are $p$-adic numbers and $p$-adic integers only defined for $p$ prime?

I was going on at too great length in a comment to a discussion between Henning Makholm and Hurkyl, so let me put it all into an answer instead: To show that $\Bbb Z_{10}\cong\Bbb Z_2\times\Bbb Z_5$, ...
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13 votes
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sequence $\{a^{p^{n}}\}$ converges in the p-adic numbers.

Recall that the Euler totient function has values $\phi(p^n)=p^{n-1}(p-1)=p^n-p^{n-1}$ for all $n$. This means that for all $a$ coprime to $p$ we have the congruence $$ a^{p^n}\equiv a^{p^{n-1}}\pmod{...
13 votes

Method of finding a p-adic expansion to a rational number

In the case of $-1/6$, it’s very easy: $$-\frac{1}{6} = \frac{1}{1-7} = \sum_{k=0}^∞ 7^k.$$ What happens: In the $p$-adic numbers, the sequence $p^k$ is a null sequence (as $|p^k|_p = p^{-k} \overset{...
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13 votes
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Totally ramified extensions of $\mathbb{Q}_p$

This is an exercise I’ve never done, but it should be a lot of fun. What is the general Eisenstein polynomial in this case? it’ll be $$ X^3 + 2aX^2+2bX+2(1+2c)\,, $$ where $a$, $b$, and $c$ can be any ...
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13 votes
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A puzzle involving $10$-adic numbers

I saw this observation in a math book once when I was 16 or so and was totally baffled at the time. It's nice to know I understand it now! As you say, the starting point is to use CRT, which allows us ...
12 votes
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The maximal unramified extension of a local field may not be complete

This is a natural question, because it’s really easy to get overwhelmed by the situation. In the case of the completion of the maximal unramified of a local field $k$, here’s the way that I look at ...
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12 votes

If $B$ is an abelian group, then is $B{\otimes}_{\mathbb Z}{\mathbb Z}_p$ isomorphic to ${\varprojlim}B/p^{n}B$?

This is false in general: consider $B = \mathbb{Q}$. Then $\mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Z}_p \cong \mathbb{Q}_p$, but $\mathbb{Q} / p^n \mathbb{Q} = 0$ for every $n$, so $\varprojlim_n \...
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12 votes
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Various $p$-adic integrals

Not really an full answer, but some comments (that hopefully answer some of your queries). There seems to be a big confusion here : what do we want to integrate, i.e. to define $\int_{\mathbb{Z}_p} f(...
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12 votes

Why are $p$-adic numbers and $p$-adic integers only defined for $p$ prime?

You can speak about $10$-adic numbers just fine, but they don't behave as nicely as $p$-adic numbers. For example, the $10$-adic integers have zero divisors, so no matter how you complete or massage ...
11 votes
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examples of unramified extensions of $\mathbb{Q}_p$

You get unramified extensions of $\Bbb Q_p$ by adjoining roots of unity of order prime to $p$; alternatively, by adjoining $(p^n-1)$-th roots of unity. The finite unramified extensions of $\Bbb Q_p$ ...
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11 votes

Why introduce the $p$-adic numbers?

On one hand, the $p$-adic numbers are extremely natural objects of study: by Ostrowski's theorem every nontrivial absolute value on $\mathbf Q$ is equivalent to either the usual absolute value or the $...
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11 votes
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Classical number theoretic applications of the $p$-adic numbers

One of my favourite classical results using $p$-adic methods in elementary number theory is the theorem of Skolem-Mahler-Lech: This is a theorem about linear recurrence sequences, which are sequences ...
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10 votes

Which $p$-adic fields contain these numbers?

There are several ways of attacking questions like this. The most general is to transform your defining equation (such as for $\sqrt{-7}$, which I’ll use as the type example) into something to which ...
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10 votes
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Is $123456788910111121314\cdots$ a $p$-adic integer?

This is not an $n$-adic integer or a $p$-adic integer. A pivotal idea of a $p$-adic integer is that it can be well-represented in a sort-of-base-$p$, $$ x = \sum_{k \geq \ell}a_kp^k,$$ and in ...
  • 87.6k
10 votes

Why does an argument similiar to 0.999...=1 show 999...=-1?

This argument is similar to the one $$ \sum_{n=1}^\infty n = -1/12 $$ which went viral a few years ago. You are actually using methods which were originally designed for manipulating absolutly ...
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10 votes
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Is there concept of continuous curve and surfaces in p-adic field?

The $p$-adic numbers are a metric space, and thus a topological space. Therefore continuity is well defined, either by the $\varepsilon,\delta$ definition that you know from Calculus, or by the ...
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