Arithmetic of p-Adic Numbers

Anyone who has not touched upon the p-adic numbers might find the topic rather confusing. Truth be told, it is confusing. The arithmetic part of the process can be a bit challenging due to its abstraction. Assuming that the reader is not familiar with the p-adic numbers, we will consider an example to clarify the matter.

Let’s consider the equation $x+64=0$. Any one individual might tell us that the solution is $x=-64$ when we ask them to solve the equation and of course, they would be right. But we wish to know the 5-adic representation of $-64$. We follow the following procedure:

Solution to $x + 64 = 0$ $mod$ $5$ is $x = 1 = 1 * 5^0$

Solution to $x + 64 = 0$ $mod$ $5^2$ is $x = 11 = 1 * 5^0 + 2 * 5^1$

Solution to $x + 64 = 0$ $mod$ $5^3$ is $x = 61 = 1 * 50^0 + 2 * 5^1 + 2 * 5^2$

Solution to $x + 64 = 0$ $mod$ $5^4$ is $x = 561 = 1 * 50^0 + 2 * 5^1 + 2 * 5^2 + 4 * 5^3$

Solution to $x + 64 = 0$ $mod$ $5^5$ is $x = 3061 = 1 * 50^0 + 2 * 5^1 + 2 * 5^2 + 4 * 5^3 + 4 * 5^4$

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Solution to $x + 64 = 0$ $mod$ $5^n$ is $x = 1 *5^0 + 2 * 5^1 . . . 4 * 5^{n-1}$

For  $x + 64 = 0$ $mod$ $5$, we can instantly see that the solution is $x = 1$ since $1 + 64 = 65$ is evenly divisible by $5$. Notice that $x = 1$ can be written as $x = 1 * 5^0$. Now for $x + 64 = 0$ $mod$ $5^2$, we know that when we add $11$ to $64$ which gives us $75$, it will be divisible by $25$. Once again, notice that we can write $x=11$ as $11 = 1 * 5^0 + 2 * 5^1$. The same procedure applies from $x + 64 = 0$ $mod$ $5^3$ all the way up to $x + 64 = 0$ $mod$ $5^n$. In general, we take our $x$ value for $x + 64 = 0$ $mod$ $5^n$ and write the linear combination for it in the form of $x = 1 *5^0 + 2 * 5^1 . . . 4 * 5^{n-1}$.

Following the same procedure, we represent $-64$ in its 5-adic representation as $(1,2,2,4,4,…)$. Although quite interesting, the arithmetic for p-adic representation can be a tedious process. Now that we have seen an example, next we will consider p-adic representation of rational numbers and absolute value on $Q$.

7 thoughts on “Arithmetic of p-Adic Numbers

  1. Samuel Coskey

    It is very interesting to see that p-adic numbers have a “digit” type of representation, just the same way as ordinary decimal numbers have.

    One big difference seems to be that while ordinary decimal integers have finite decimal representations (like -64), p-adic integers (let alone rational or irrational numbers) may already have infinite digit representations like 1,2,2,4,4,….

  2. grantrosandick

    This is an interesting post. I am having a little trouble following your example however. I’m not sure where you are finding the x value for each individual mod5. For example, when you had x+64=0 mod 5, you then have x=1=1+5^0. I guess I get confused when you state that x=1. Is it because it’s mod 5^0? I’m sure if I were studying this lab I would have a better understanding, but for now it is hard for me to fully understand.

  3. Kenny

    Why may I want to represent the solution to some linear equation in $p$-adics? Or, maybe worded better, what is the benefit or application of such a representation?

    1. Ken Coiteux

      One interesting thing is that there are no negative numbers in the p-adics. By solving $x+64=0$ in the fashion above will give a p-adic representation for the inverse of the solution for $x-64=0$.

      1. Ken Coiteux

        After thinking about what I posted, I need to clarify a bit. This expansion of p-adics is a way to represent additive inverses.

  4. Jordan

    As I was somewhat interested in this topic its great to see more of what it is about. I really liked the example you gave to clear up what a p-adic number is. The example was easy to follow and how you ended the post with what will happen next is helpful to see what else will be talked about.

  5. Pingback: Quick Intro to p-adics | MATH 287

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