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The concept of a p adic number is something that I haven't quite been able to grasp. Heretofore, I can comprehend the fact that the p adic number system essentially extends the conventional arithmetic of rational numbers to real and complex number systems. P adic numbers have numerous applications in number theory, considered to be "close" when their differences are divisible by a higher power of p. However, I do not quite understand the mathematical reasoning behind a p adic number system, particularly the algebraic approach to the subject.

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  • $\begingroup$ "I can comprehend the fact that the p adic number system essentially extends the conventional arithmetic of rational numbers to real and complex number systems."...what does this mean? $\endgroup$ Commented Apr 16, 2019 at 5:41
  • $\begingroup$ By what do you mean? $\endgroup$ Commented Apr 16, 2019 at 5:42
  • $\begingroup$ Well, you said you understand something, I'm asking you to explain what you understand? $\endgroup$ Commented Apr 16, 2019 at 5:45
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    $\begingroup$ OK well I don't know $p$-adics, so take what I say with caution, but I was under the impression that the $p$-adics(completion of $\mathbb Q$ under a certain metric) have nothing to do with $\mathbb R$ other than they both contain $\mathbb Q$ (but with different metrics) $\endgroup$ Commented Apr 16, 2019 at 5:57
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    $\begingroup$ May I recommend my paper with Alf van der Poorten, Some problems concerning recurrence sequences, The American Mathematical Monthly, Vol. 102, No. 8 (Oct., 1995), pp. 698-705, jstor.org/stable/2974639 which gives a gentle introduction to, and a nice application of, the $p$-adics? $\endgroup$ Commented Apr 16, 2019 at 6:23

2 Answers 2

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When writing a real number in base p you can write an unlimited number of digits to the right of the decimal point. Because the size of the extra amount represented by each new digit decreases exponentially as the number of digits increases, an unlimited sequence of digits will still converge to a specific number.

Informally you can think of p-adic numbers as being numbers written with an unlimited number of digits to the left of the decimal point. Using the p-adic metric, the size of the extra amount represented by each new digit decreases exponentially as the number of digits increases because $|p^{n+1}|=\frac{|p^n|}{p}$. So now an unlimited number of digits to the left of the decimal point will still converge to a specific number.

In $5$-adic numbers (remember we are writing numbers in base $5$) we have

$4+1=10 \\ 44+1 = 100 \\ 444+1 = 1000$

and in the $5$-adic metric the size of the numbers on the right becomes smaller and smaller, so in the limit we have

$\dots 4444+1=0$

So in $5$-adic numbers we have

$\dots 4444=-1 \\ \Rightarrow \dots 1111=-\frac{1}{4} \\ \Rightarrow \dots 1112 = - \frac{1}{4} + 1 = \frac{3}{4}$

Another way to derive the same result is to notice that if $\dots 1112$ converges to a limit $x$ then

$5x = \dots 11120 = \dots 1112 + 3 = x+3 \\ \Rightarrow 4x = 3 \\ \Rightarrow x=\frac{3}{4}$

Similarly

$\dots 2222=-\frac{1}{2} \\ \Rightarrow \dots 2223 = - \frac{1}{2} + 1 = \frac{1}{2}$

See if you can show that:

$\dots 3334 = \frac{1}{4} \\ \dots 131313 = -\frac{1}{3} \\ \dots 131314 = \frac{2}{3} \\ \dots 313132 = \frac{1}{3}$

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  • $\begingroup$ ...3333 = -3/4 ⇒ ...3334 = -3/4 + 1 = 1/4 $\endgroup$ Commented Apr 16, 2019 at 11:04
  • $\begingroup$ However, I am yet to quite understand the other three problems you set me. $\endgroup$ Commented Apr 16, 2019 at 11:08
  • $\begingroup$ Hint: Assume $\dots 131313$ converges to a limit $x$. Then $25x = \dots 131300$. So $25x = x - 8$ ... $\endgroup$
    – gandalf61
    Commented Apr 16, 2019 at 11:47
  • $\begingroup$ 24x = -8, and thus x = -1/3... Could you kindly explain why in your example problem, you allowed 1112 to converge to the limit x, for I am somewhat perplexed. $\endgroup$ Commented Apr 16, 2019 at 11:53
  • $\begingroup$ $x$ is just an arbitrary label for whatever limit we are considering at the time. $x$ does not take the same value in the two problems. Call one of the limits $y$ if you prefer. $\endgroup$
    – gandalf61
    Commented Apr 16, 2019 at 12:31
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There are two definitions :

  • For an integer $n$ let $|n|_p = p^{-k}$ if $n \equiv 0 \bmod p^k, n \not \equiv 0 \bmod p^{k+1}$. Then $|.|_p$ is a metric, a norm and an absolute value on $\Bbb{Z}$, so you can take its completion (limits of Cauchy sequences) just as you did when completing $\Bbb{Q}$ to obtain $\Bbb{R}$.

  • $\Bbb{Z}_p$ is the set of sequences $(a_k)_{k\ge 1}$ such that $a_k \in \Bbb{Z/p^k Z}$ and $a_{k+1} \equiv a_k \bmod p^k$ which becomes a ring with the pointwise addition and multiplication modulo each $p^k$. Then $\Bbb{Z}$ embeds in $\Bbb{Z}_p$ by sending $n$ to $(n \bmod p, n \bmod p^2,n \bmod p^3,\ldots)$.

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