p-adic distance

The Euclidian distance represents the most intuitive concept of distance on the real line.However, we can define other distances between rational numbers, which do not correspond to any intuitive concept.

Although we will only present the corresponding definitions, these new distances are a key factor to obtain arithmetical results.

Now we are going to define the p-adic norm:

We fix a prime number $$p$$. To define the norm of a rational number $$\dfrac{a}{b}\neq 0$$ we first have to factorize both $$a$$ and $$b$$. In fact it it is enough to see how many times they are divisible by $$p$$.

Let's then suppose that $$a=m\cdot p^r$$ so that $$p$$ does not divide $$m$$, and $$b=n\cdot p^s$$ with $$p$$ not dividing $$n$$.

According to these definitions, we define the norm of $$\dfrac{a}{b}$$ as $$$p^{s-r}$$$ We will denote the norm of $$\dfrac{a}{b}$$ as $$\Big| \dfrac{a}{b}\Big|_p$$ and we will call it p-adic norm.

If $$\dfrac{a}{b}=0$$ then we set $$\Big| \dfrac{a}{b}\Big|_p=0$$.

We must not confuse the notation between $$\Big| \dfrac{a}{b}\Big|$$ and $$\Big| \dfrac{a}{b}\Big|_p$$. When we do not write subscript we will always be referring to the euclidean norm defined previously.

Also, we must bear in mind that when talking about p-adic norms, $$p$$ must be a prime. Therefore, it does not make sense if we say, for example 4-adic or 6-adic norm since $$4$$ and $$6$$ are not prime.

Also we can see that to calculate $$\Big| \dfrac{a}{b}\Big|_p$$ it is not necessary that $$a$$ and $$b$$ have common factors.

As a last comment we must emphasize that the p-adic distance corresponds to the rational numbers and does not make sense for irrational numbers. For example, we cannot write $$|\sqrt{2}|_p$$.

Let's consider the rational number $$\dfrac{10}{12}$$. We calculate the 2-adic norm.

The factorization of $$10$$ is $$5 \cdot 2$$ and of $$12$$ is $$3 \cdot 2^2$$. According to the previous notations we have $$a=10$$ and $$b=12$$ with $$m=5$$ and $$n=3$$ and also $$r=1$$ and $$s=2$$. Then, we have the 2-adic norm of $$\dfrac{10}{12}$$ is

$$$\Big| \dfrac{10}{12}\Big|_2=2^{s-r}=2^{2-1}=2$$$

We make use of the previous factorization to calculate the 5-adic and 7-adic norms.

For the 5-adic norm we have, according to the presented definitions, $$m=2$$ and $$n=12$$ and also $$r=1$$ and $$s=0$$. And then

$$$\Big| \dfrac{10}{12}\Big|_5=5^{s-r}=2^{0-1}=\dfrac{1}{5}$$$

For the 7-adic norm we obtain, according to the presented definitions, $$m=10$$ and $$n=12$$ and also $$r=0$$ and $$s=0$$. And then

$$$\Big| \dfrac{10}{12}\Big|_7=7^{s-r}=7^{0-0}=1$$$

The p-adic norm has the same properties as the euclidean norm and allows us to define a distance, which we will call p-adic distance. To define this distance we do an analogy with the euclidean distance defining the p-adic distance as: $$$d_p(a,b)=|b-a|_p$$$

This distance has the properties already commented on for the euclidean distance. Let's remember them:

  • $$d_p(a,b)>0$$; and $$d_p(a,b)=0$$ if and only if $$a=b$$.
  • $$d_p(a,b)=d_p(b,a)$$.
  • $$d_p(a,b) \leq d_p(a,c) + d_p(c,b)$$

    We must not confuse the notation $$d(a,b)$$ and $$d_p(a,b)$$, since the first one always corresponds to the euclidean distance and the second one to the p-adic distance, in which $$p$$ is a prime number.

Let's see an example in which it is observed that this distance does not keep the intuitive sense of proximity that the euclidean distance has.

We consider the 3-adic distance. We choose, for example $$b=1$$. We are going to see that we can choose a rational number, $$a$$ , so that $$d(a,b)$$ is big but $$d_3(a,b)$$ is small.

If we choose $$a=82$$ we have $$$d(a,b)=d(82,1)=|82-1|=81$$$

Let's see, on the other hand, that the 3-adic distance is small. We have: $$$d_3(a,b)=d_3(82,1)=|81|_3$$$

Factorizing we obtain that $$81=3^4$$ and consequently we have $$$|81|_3=3^{-4}=\dfrac{1}{81}$$$

Therefore, the numbers $$1$$ and $$82$$ are far from one another according to the euclidean distance but close according to the 3-adic norm.

In the same way, if we choose $$a=1+3^m$$ we obtain $$$d(a,b)=d(1+3^m,1)=|1+3^m-1|=3^m$$$

And, on the other hand we, have $$$d_3(a,b)=d_3(1+3^m,1)=|1+3^m-1|_3=|3^m|_3=3^{-m}$$$

Then, the euclidean distance is becoming bigger on increasing $$m$$ and, on the other hand, the 3-adic distance is becoming smaller every time.