- $P(X) = 1/(1+X) = 1 - X + X^2 -X^3 + \dots$.

$1/4 = 1/(1+p) = 1 - p + p^2 - p^3 + p^4 - p^5 + p^6 + \cdots$

However, we don't really know how to interpret $(-1 \cdot p)$, since we assumed
the coefficients are always non-negative. What we can do is to rewrite $p^2 = 3p$,
and then use this to make the coefficient positive. Performing this transformation
for every negative coefficient, we arrive at:
$\begin{aligned}
1/4 &= 1/(1+p) = 1 - p + p^2 - p^3 + p^4 + \cdots \\
&= 1 + (- p + 3p) + (- p^3 + 3p^3) + \cdots \\
&= 1 + 2p + 2p^3 + \cdots
\end{aligned}$

We can verify that this is indeed correct, by multiplying with $4 = (1 + p)$
and checking that the result is $1$:
$\begin{aligned}
&(1 + p)(1 + 2p + 2p^3 + \cdots) \\
&= (1 + p) + (2p + 2p^2) + (2p^3 + 2p^4) + \cdots \\
&= 1 + 3p + 2p^2 + 2p^3 + 2p^4 + \cdots \\
&\text{(Rewrite $3p = p \cdot p = p^2$)} \\
&= 1 + (p^2 + 2p^2) + 2p^3 + 2p^4 + \cdots \\
&= 1 + 3p^2 + 2p^3 + 2p^4 + \cdots \\
&\text{(Rewrite $3p^2 = p^3$ and collect $p^3$)} \\
&= 1 + 3p^3 + 2p^4 + \cdots \\
&= 1 + 3p^4 + \cdots \\
&= 1 + \cdots = 1
\end{aligned}$

What winds up happening is that all the numbers after $1$ end up being cleared
due to the carrying of $(3p^i \mapsto p^{i+1})$.
This little calculation indicates that we can also define take the $p$-adic
expansion of $\begin{aligned}
-1 &= -1 + p - p \qquad \text{(borrow $p$, and subtract to keep equality)} \\
&= (p - 1) - p \qquad \text{(Now we have a problem of $-p$)} \\
&= (p - 1) - p + p^2 - p^2 \\
&= (p - 1) + p(p - 1) - p^2 \\
&= (p - 1) + p(p - 1) - p^2 + p^3 - p^3 \\
&= (p - 1) + p(p - 1) + p^2(p - 1) - p^3 \\
&\text{(Generalizing the above pattern)} \\
-1 &= (p - 1) + p(p - 1) + p^2(p - 1) + p^3(p - 1) + p^4(p - 1) + \cdots \\
\end{aligned}$

This now gives us access to negative numbers, since we can formally multiply
the series of two numbers, to write $-a = -1 \cdot a$.
Notice that this definition of $-1$ also curiously matches the 2s complement
definition, where we have $-1 = 11\dots 1$. In this case, the expansion is
- $X \equiv 3 \mod 3$
- $X \equiv 5 \mod 9$
- $X \equiv 5 \mod 27$
- At this point, the answer remains constant.

- $X \equiv -5 \equiv 1 \mod 3$
- $X \equiv -5 \equiv 4 = 1 + 3 \mod 9$
- $X \equiv -5 \equiv 22 = 1 + 3 + 2 \cdot 9 \mod 27$
- $X \equiv -5 \equiv 76 = 1 + 3 + 2 \cdot 9 + 2 \cdot 27 \mod 81$

- $X = -5 = 1 + 1\cdot 3 + 2\cdot 3^2 + 2\cdot 3^3 + \dots$

- $\alpha_{n+1} = \alpha_n \mod p^n$.

- $2 \rightarrow 5 \rightarrow 5 \rightarrow 5 \rightarrow \dots$
- $1 \rightarrow 4 \rightarrow 22 \rightarrow 76 \rightarrow \dots$

- $(3 + 7k)^2 = 2 \mod 49$, $(4 + 7k)^2 = 2 \mod 49$.

$\begin{aligned}
(3 + 7k)^2 &\equiv 2 \mod 49 \\
9 + 42 k + 49k^2 &\equiv 2 \mod 49 \\
9 + 42 k + 0k^2 &\equiv 2 \mod 49 \\
7 + 42 k &\equiv 0 \mod 49 \\
1 + 6 k &\equiv 0 \mod 49 \\
k &\equiv 1 \mod 49
\end{aligned}$

This gives the solution $X \equiv 10 \mod 49$. The other branch ( $X = 4 + 7k$)
gives us $X \equiv 39 \equiv -10 \mod 49$.
We can continue this process indefinitely ( - $3 \rightarrow 10 \rightarrow 108 \rightarrow 2166 \rightarrow \dots$
- $4 \rightarrow 39 \rightarrow 235 \rightarrow 235 \rightarrow \dots$

- $X_0 = 1$
- $X_1 = 1 + 3$
- $X_2 = 1 + 3 + 3^2$
- $X_n = 1 + 3 + \dots + 3^n$

- $1/(1 - 3) = -1/2$

- $\lvert x \rvert = 0 \iff x = 0$
- $\lvert xy \rvert = \lvert x \rvert \lvert y \rvert$ for all $x, y \in K$
- $\lvert x + y \rvert \leq \lvert x \rvert + \lvert y \rvert$, for all $x, y \in K$.

- $n = p^{v_p(n)} m$, where $p \nmid m$.
- We extend the valuation to the rationals by defining $v_p(a/b) = v_p(a) - v_p(b)$.
- We set $v_p(0) = +\infty$. The intuition is that $0$ can be divided by $p$an infinite number of times.

- the norm of $0$ is $p^{-v_p(0)} = p^{-\infty} = 0$.
- If $p^{-v_p(n)} = 0$, then $-v_p(n) = \log_p 0 = -\infty$, and hence $n = 0$.
- The norm is multiplicative since $v_p$ is additive.
- Since $v_p(x + y) \geq \min (v_p(x), v_p(y)), |x + y|_p \leq max(|x|_p, |y|_p) \leq |x|_p + |y|_p$. Hence, the triangle inequality is also satisfied.

`math.se`

,
and the answer is that this choosing a base $p$ gives us the nice formula
$\forall x \in \mathbb Z, \prod_{\{p : p~\text{is prime}\} \cup \{ \infty \}} |x|_p = 1$

That is, the product of all $p$ norms and the usual norm
(denoted by $\lvert x \rvert_\infty$ )
give us the number 1. The reason is that the $\lvert x\rvert_p$ give us
multiples $p^{-v_p(x)}$,
while the usual norm $\lvert x \rvert_\infty$ contains a multiple
$p^{v_p(x)}$, thereby cancelling each other out.