# 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.

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