Previously, we discussed the absolute values on $Q$, the real numbers, Cauchy sequence, and basic process of computing the p-adic representation of a number. Now, we move ahead and discuss how to construct these p-adic numbers. We do this by using the absolute value in the usual sense as to how it is used with the reals.

$Definition$. The field of p-adic numbers $Q_p$ is defined to be the set of all equivalence classes of p-adic Cauchy sequence.

As mentioned earlier, we can represent p-adic numbers using the Cauchy sequence. With the given definition, now we know that for any p-adic Cauchy sequence, the sequence converges to a p-adic number. Natural question to arise would be to ask *how*. To address this *how* and to convince ourselves, we will consider a quick example before generalizing it.

Let’s take the 3-adic representation of $241$. We surly know that $241$ is a Cauchy sequence that converges to $241$ for {$241$, $241$, $241$, $…$} . The 3-adic representation of $241$ is $1 + 2*3^1 + 2*3^2 + 2*3^3 + 2*3^4$ or $(1, 2, 2, 2)$. This expansion represents the class of all Cauchy sequence equivalent to {$1$, $1+2*3^1$, ${1+2*3^1 + 2*3^3}$, $…$}. Looking closely, we notice the sequence as follows {$3$, $3^2$, $3^3$, $…$}. Taking the limit, we see that $lim_{n\to\infty}$ $3^n = 0$ for 3-adics. How? As we recall from absolute values of $Q$ for any prime $p$, we write $|p^0*k|=1$, $|p*k|=p^{-1}$, $|p^2*k|=p^{-2}$ and $|p^n*k|=p^{-n}$ in general for some constant $k$ as $n\to \infty$. It would make intuitive sense for $|3^n|=3^{-n}$ to approach zero as $n\to \infty$ because of the inverse. Let’s note the existence of the inverse relation we see here with the p-adic numbers. As $3^n$ grows without bounds, its 3-adic representation converges to zero.

Looking at the general case of p-adic Cauchy sequence, we see that any number in its p-adic representation will converge to zero and can be written as

{$a_n$} = {$a_0 * p^0, a_0 * p^0 + a_1*p^1, … , a_0 * p^0 + a_1*p^1 + a_2*p^2 + … + a_n*p^n$}

where $a_0, a_1, …, a_n$ represent the coefficients.  This p-adic expansion for the Cauchy sequence is abbreviated in the form of $a_0.a_1a_2a_{3,…,p}$. So, the 3-adic representation for $241$ can be abbreviated as $1.22\bar{2}_3$ where $0 \leq a_i \leq p-1$.

Now that we have a general knowledge of how the p-adic numbers work and how it’s constructed, we will consider another interesting representation of negative integers with p-adics. Let’s consider the p-adic expansion of $-1$ for any $p$. To start with, we begin by considering the 5-adic expansion for it.

Solution to $x + 1 = 0$ $mod$ $5$ $is$ $x = 4 = 4*5^0$

Solution to $x + 1 = 0$ $mod$ $5^2$ is $x = 24 = 4*5^0 + 4*5^1$

Solution to $x + 1 = 0$ $mod$ $5^3$ is $x = 124 = 4*5^0 + 4*5^1 + 4*5^2$

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Solution to $x + 1 = 0$ $mod$ $5^n$ is $x = 4*5^0 + 4*5^1 + 4*5^2 + … + 4*5^{(n-1)}$

Next, we consider the *7-adic* expansion of $-1$.

Solution to $x + 1 = 0$ $mod$ $7$ is $x = 6 = 6*7^0$

Solution to $x + 1 = 0$ $mod$ $7^2$ is $x = 48 = 6*7^0 + 6*7^1$

Solution to $x + 1 = 0$ $mod$ $7^3$ is $x = 342 = 6*7^0 + 6*7^1 + 6*7^2$

.

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Solution to $x + 1 = 0$ $mod$ $7^n$ is $x = 6*7^0 + 6*7^1 + 6*7^2 + … + 6*7^{(n-1)}$

$Conjecture$. The *p-adic* expansion of $-1$ for any $p$ is represented as $(p-1, p-1, p-1, …)$.

# The Real Numbers and the Cauchy Sequence

To further continue with our path into studying the p-acid numbers, we must be able to construct it to have a deeper understanding of it. In order to do so, we break from learning the properties and how the p-adic numbers work. Instead we will take time to observe what a Cauchy sequence is and it’s relation to the p-adics. Continuing on, we will first define the limits of the Cauchy sequence of the rational numbers.

$Definition$. For a given sequence of rational numbers ${a_n} = ({a_1, a_2,…})$, it is known as being real Cauchy sequence if for some $ε > 0$ , there exists a positive integer $N$ such that for all $i,j>N, |a_i-a_j|_{\infty} < ε$.

Let’s consider the following real number $1.4142135…$. We can represent it by the limit of the sequence as

$a_1 = 1 = 1*10^0$

$a_2 = 1.4 = 1*10^0 + 4*10^{-1}$

$a_3 = 1.41 = 1*10^0 + 4*10^{-1} + 1*10^{-2}$

$a_4 = 1.414 = 1*10^0 + 4*10^{-1} + 1*10^{-2} + 4*10^{-3}$

$…$

Can the reader recognize the sequence? The expansion of $1.4142135…$ is the converging Cauchy sequence representation of the constant $\sqrt{2}$. This sequence can also be written as $({1, 1.4, 1.41, 1.414, 1.4142, …})$. To be completely sure that $\sqrt{2}$ is a Cauchy sequence, next we set $ε = 0.05$. We see that $N = 2$ and verify this by computing $|1.41-1.414|_{\infty} = 0.004 < 0.05 = ε$ which is true.

If we are given two real Cauchy sequences ${a_n}$ and ${b_n}$, we know that the both of the sequences represent the same real number if ${|a_n – b_n|}$ converge to 0. This is fairly easy to demonstrate and see. Taking ${a_n} = ({1, 1, 1, 1, 1, …})$ and ${b_n} = ({0.9, 0.99, 0.999, 0.9999, 0.99999, …})$ as an example, we see that ${|a_n – b_n|} = ({0.1, 0.01, 0.001, 0.0001, 0.00001, …})$ converges to 0 as $n\rightarrow{\infty}$ . The Cauchy sequence here for ${a_n}$ can seem counter intuitive at first if the reader is countering such case for the first time. After all, how can $0.999\bar{9} = 1$ be true? To clearly understand $how$ this works, we observe the sum of the infinite series for $0.999\bar{9}$.

$$0.999\bar{9} = lim_{n\rightarrow\infty} 9 \frac{1}{10} + 9\frac{1}{100} + 9\frac{1}{1000} + … + 9\frac{1}{10^n}$$

Now we apply the geometric sum given by

$$1 + r + r^2 + … + r^{n-1} = \frac{1 – r^n}{1 – r}$$

to our sequence to get

$$= \frac{9 *(1/10)}{1- \frac{1}{10}} = \frac{9*10}{10*9} = 1$$

Our $r$ value here is $1/10$ because that is our common ratio. As we can see, it is true that $0.999\bar{9}=1$ for the sequence of $0.999\bar{9}$. Note that this is true because we consider the *infinite* sequence of $0.999\bar{9}$. If we did have a cutoff, then we could say that $0.9999$ is approximately equal to $1$ with a slight error. But that’s clearly not the case here.

Now that we have observed the Cauchy sequences of rational numbers, we will next study the p-adic numbers and how we construct it with our thus far accumulated knowledge regarding absolute value and the Cauchy sequence.

# Absolute values on Q (Part II)

Continuing on from absolute values on Q part I/ , I am going to show that the definition of p-adic absolute value satisfies the three properties of absolute values.

Property 1

Let $x=0$ and p be a fixed prime.  By definition then, $|x|_p = 0$.  This satisfies the first property ($|x|_p=0$ if and only if $x=0$).

Suppose now $x \ne 0$.  Then $|x|_p = p^{-ord_p(x)}$.  Using Conjecture 1, $|x|_p=|p^n \cdot a|_p =p^{-n}$.  Since $p^{-n}$ cannot be 0, $p^{-n} > 0$.

Property 2

Let $x=0$ and $y \ne 0$ and p be a fixed prime.  Then $|x \cdot y|_p = |0|_p = 0 = |x|_p \cdot |y|_p$.

Suppose $|x|_p = |p^n \cdot a|$ and $|y|_p = |p^n \cdot b|$.  In other words, x and y have the same exponent on p. Then, by definition,

$|x \cdot y|_p = |p^n \cdot a \cdot p^n \cdot b|_p = |p^{2n} \cdot ab|_p = p^{-2n}$.

By the second property,

$|x \cdot y|_p = |x|_p \cdot |y|_p = |p^n c\dot a|_p \cdot |p^n \cdot b|_p = p^{-n}p{-n}=p{-2n}$

Suppose x and y have different exponents on the p.  $|x|_p=|p^n \cdot a|_p and |y|_p=|p^m cdot b|$.  Multiply x by y….

$|xy|_p=|p^n \cdot a \cdot p^m \cdot b|_p=|p^{n+m} \cdot ab|_p=p^{-n-m}$

By the second property,

$|xy|_p=|x|_p \cdot |y|_p = p^{-n} \cdot p^{-m}=p^{-n-m}$.

Property 3

Let $x=y=0$ and p be a fixed prime.  Then by definition $|x+y|_p = 0$.  This partly satisfies the property for $0 \le 0+0$  Now consider $x \ne 0$.  Then $|x+y|_p = |x|_p = |p^n \cdot a|_p = p^{-n}$.    This contributes to satisfying the property for $p^{-n} \le p^{-n} + 0$.

Now let x and y have the same exponent on p.

$|x+y|_p = |p^n \cdot a + p^n \cdot b|_p = |p^n(a+b)|_p=p^{-n}$

If the definition satisfies the property, then $p^{-n}$ should be less than $|x|_p+|y|_p$.  In fact this is true because $|x|_p+|y|_p = p^{-n}+p^{-n}$ which is larger than $p^{-n}$.

Last case is when x and y have different exponents on p.

$|x+y|_p=|p^n \cdot a + p^m \cdot b|_p=p^{-n}$ where $n<m$

Now consider property 3.

$|x|_p+|y|_p=p^{-n}+p^{-m}$

Now that p-adic absolute values do satisfy the properties of absolute values, there is one more thing to look at:  Archimedean absolute values.  A little warning for you, the following definition is counter-intuitive.

Definition 2:  An absolute value on Q is said to be non-Archimedean if the properties of absolute values are satisfied along with an additional property:

$|x+y| \le max(|x|,|y|)$ for all x,y \in Q.

Absolute values that satisfy the three properties but not the fourth are said to be Archimedean.

Theorem 1:  p-adic absolute values are non-Archimedean.

It has already been shown that p-adic absolute values satisfy the first three conditions.  Now all that needs to be shown is that p-adic absolute values satisfy the fourth property.  Assume p is a fixed prime

Case 1: $x=y=0$

$|x+y|_p = |0|_p = 0 \le max(|x|_p,|y|_p) = 0$

Case 2: $x \ne 0, y=0$

$|x+y|_p = |x|_p = p^{-n} \le max(|x|_p,|y|_p) = max(p^{-n}, 0) = p^{-n}$

Case 3:  x and y have the same exponent on p

$|x+y|_p = p^{-n} \le max(|x|_p,|y|_p) = max(p^{-n}, p^{-n}) = p^{-n}$

Case 4:  x and y have different exponents on p (n<m)

$|x+y|_p = p^{-n} \le max(|x|_p,|y|_p) = max(p^{-n}, p^{-m}) = p^{-m}$

QED

# Absolute Values on Q (Part I)

Real numbers, Q, can be constructed from the rational numbers Q={a/b where a,b ϵ the integers Z}.  Construction of p-adic numbers, Qp, is done exactly the same way.

Let’s now consider absolute values.  An absolute value is a map from Q to [0,∞) that has the following properties for any x,y ϵ Q:

1. |x| ≥ 0, and |x| = 0 if and only if x = 0,
2. |x∙y| = |x|∙|y|,
3. |x+y| ≤ |x|+|y| (The triangle inequality).

Now that we have some properties, we can look at types of absolute values that can be put on the rationals.  Alexander Ostrowski showed in 1935 that there are only three types.

Definition 1:  Let x be a rational number.

The trivial absolute value of x, denoted |x|0, is defined by

$|x|_0 = \left\{ \begin{array}{l l} 1 & \quad \text{if x\ne0 }\\ 0 & \quad \text{if x=0} \end{array} \right.$

The usual absolute value of x, denoted |x|, is defined by

$|x|_{\infty} = \left\{ \begin{array}{l l} x & \quad \text{if x\ge0 }\\ -x & \quad \text{if x\le0} \end{array} \right.$

The p-adic absolute value of x, denoted |x|p, is defined for a given prime p by

$|x|_p = \left\{ \begin{array}{l l} 1/p^{ord_p(x)} & \quad \text{if x\ne0 }\\ 0 & \quad \text{if x=0} \end{array} \right.$

The quantity ordp(x) is called the order of x.  It is the highest power of p that divides x.  For example, let p=5 and let x=75.  Then ord5(75)=2.  Since 25=52 and 25 is the highest power of 5 that divides 75, then the order of x is 2.  Confused yet?  Here are some examples of p-adic absolute values:

|75|5 = |52 ∙ 3|5 = 5-2

|10|5 = |51 ∙ 2|5 = 5-1

|13|5 = |50 ∙ 13|5 = 1

$|\frac{2}{75}|5 = |5-2 ∙ \frac{2}{3} |5 = 52$

|-375|5 = |53 ∙ -3|5 = 5-3

In these examples, we were looking at 5-adic absolute values.  We broke the number up into the form 5n ∙ a.  The 5-adic absolute value of 5n ∙ a is 5-n.

Conjecture 1:  Given a rational number x, the p-adic absolute value of x is p-n, where pn divides x.

|x|p = |pn ∙ a|p = p-n

In the introduction post (http://math.boisestate.edu/m287/quick-intro-to-p-adics/), it was mentioned that large p-adic numbers are closer to 0 than smaller p-adic numbers.  If one was to think of absolute value as a measure of distance, notice in the above examples that the larger the exponent on p, the resulting absolute value (or distance from 0) gets smaller.

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 $Q$.
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