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 $(p-1, p-1, p-1, …)$.