[proofplan]
We verify the three axioms of a non-archimedean absolute value for $|\cdot|_p$: non-degeneracy, multiplicativity, and the strong (ultrametric) triangle inequality. Non-degeneracy follows directly from the definition. Multiplicativity reduces to showing that $v_p(xy) = v_p(x) + v_p(y)$, where $v_p$ is the $p$-adic valuation. The strong triangle inequality is the main content: we write $x + y$ in lowest terms, extract the $p$-adic valuation, and show it is at least $\min(v_p(x), v_p(y))$.
[/proofplan]
[step:Recall the definition of $|\cdot|_p$ and verify non-degeneracy]
The $p$-adic absolute value on $\mathbb{Q}$ is defined by $|0|_p = 0$ and, for $x \in \mathbb{Q}^\times$, by writing $x = p^n \frac{a}{b}$ with $n \in \mathbb{Z}$, $a, b \in \mathbb{Z} \setminus \{0\}$, and $\gcd(a, p) = \gcd(b, p) = 1$, and setting $|x|_p = p^{-n}$. The integer $n = v_p(x)$ is the $p$-adic valuation of $x$.
Non-degeneracy: $|x|_p = 0$ if and only if $x = 0$ (by definition). Positivity: for $x \neq 0$, $|x|_p = p^{-n} > 0$ since $p > 0$.
[/step]
[step:Prove multiplicativity using the additivity of the $p$-adic valuation]
Let $x, y \in \mathbb{Q}^\times$. Write $x = p^n \frac{a}{b}$ and $y = p^m \frac{c}{d}$ with $\gcd(a, p) = \gcd(b, p) = \gcd(c, p) = \gcd(d, p) = 1$. Then
\begin{align*}
xy = p^{n+m} \frac{ac}{bd}.
\end{align*}
Since $\gcd(a, p) = \gcd(c, p) = 1$, we have $\gcd(ac, p) = 1$ (the product of integers coprime to $p$ is coprime to $p$, since $p$ is prime). Similarly $\gcd(bd, p) = 1$. Therefore $v_p(xy) = n + m$, and
\begin{align*}
|xy|_p = p^{-(n+m)} = p^{-n} \cdot p^{-m} = |x|_p \cdot |y|_p.
\end{align*}
The case $x = 0$ or $y = 0$ gives $|xy|_p = |0|_p = 0 = |x|_p |y|_p$.
[guided]
Multiplicativity of $|\cdot|_p$ is equivalent to the statement that $v_p: \mathbb{Q}^\times \to \mathbb{Z}$ is a group homomorphism from $(\mathbb{Q}^\times, \cdot)$ to $(\mathbb{Z}, +)$. The key algebraic fact is that $p$ is prime, so $p \nmid a$ and $p \nmid c$ together imply $p \nmid ac$. This is precisely the statement that $\mathbb{Z}/p\mathbb{Z}$ is an integral domain (equivalently, $p$ is prime), applied to the numerator and denominator separately.
Write $x = p^n \frac{a}{b}$ and $y = p^m \frac{c}{d}$ with $\gcd(a,p) = \gcd(b,p) = \gcd(c,p) = \gcd(d,p) = 1$. Then $xy = p^{n+m} \frac{ac}{bd}$. Since $p$ is prime, $p \nmid ac$ (from $p \nmid a$ and $p \nmid c$) and $p \nmid bd$ (from $p \nmid b$ and $p \nmid d$). So the representation $xy = p^{n+m} \frac{ac}{bd}$ already has numerator and denominator coprime to $p$, giving $v_p(xy) = n + m = v_p(x) + v_p(y)$. Therefore
\begin{align*}
|xy|_p = p^{-(n+m)} = p^{-n} \cdot p^{-m} = |x|_p \cdot |y|_p.
\end{align*}
[/guided]
[/step]
[step:Prove the strong triangle inequality $|x + y|_p \leq \max(|x|_p, |y|_p)$]
We must show that for all $x, y \in \mathbb{Q}$, $|x + y|_p \leq \max(|x|_p, |y|_p)$. If $x = 0$ or $y = 0$, equality holds. If $x + y = 0$, the left side is $0$ and the inequality holds. So assume $x, y, x+y \in \mathbb{Q}^\times$.
Write $x = p^n \frac{a}{b}$ and $y = p^m \frac{c}{d}$ with $\gcd(a, p) = \gcd(b, p) = \gcd(c, p) = \gcd(d, p) = 1$. Assume WLOG that $n \leq m$ (so $|x|_p = p^{-n} \geq p^{-m} = |y|_p$, and $\max(|x|_p, |y|_p) = p^{-n}$). Then
\begin{align*}
x + y = p^n \frac{a}{b} + p^m \frac{c}{d} = p^n \left(\frac{a}{b} + p^{m-n} \frac{c}{d}\right) = p^n \cdot \frac{ad + p^{m-n} bc}{bd}.
\end{align*}
Since $\gcd(b, p) = \gcd(d, p) = 1$, we have $\gcd(bd, p) = 1$. The numerator $ad + p^{m-n} bc$ is an integer. Let $k = v_p(ad + p^{m-n} bc) \geq 0$. Then
\begin{align*}
v_p(x + y) = n + k - 0 = n + k \geq n,
\end{align*}
where the $-0$ reflects that $\gcd(bd, p) = 1$ contributes no factors of $p$ from the denominator. Therefore
\begin{align*}
|x + y|_p = p^{-(n+k)} \leq p^{-n} = \max(|x|_p, |y|_p).
\end{align*}
[guided]
The strong triangle inequality is the defining feature that makes $|\cdot|_p$ non-archimedean. The question reduces to: when we add $x = p^n \frac{a}{b}$ and $y = p^m \frac{c}{d}$, can the $p$-adic valuation of the sum decrease below the minimum of $v_p(x) = n$ and $v_p(y) = m$?
Assume WLOG $n \leq m$. We factor out $p^n$:
\begin{align*}
x + y = p^n \left(\frac{a}{b} + p^{m-n} \frac{c}{d}\right) = p^n \cdot \frac{ad + p^{m-n} bc}{bd}.
\end{align*}
The denominator $bd$ is coprime to $p$ (since $\gcd(b,p) = \gcd(d,p) = 1$ and $p$ is prime). So the entire $p$-content of $x + y$ beyond the factor $p^n$ is determined by the numerator $ad + p^{m-n} bc$.
The numerator $ad + p^{m-n} bc$ is the sum of two integers. The first term $ad$ may or may not be divisible by $p$, but the second term $p^{m-n} bc$ is divisible by $p^{m-n}$ (and $m - n \geq 0$). In any case, the $p$-adic valuation of the numerator is $k \geq 0$. This gives $v_p(x + y) = n + k \geq n = \min(v_p(x), v_p(y))$, and therefore
\begin{align*}
|x + y|_p = p^{-(n+k)} \leq p^{-n} = \max(|x|_p, |y|_p).
\end{align*}
Note the inequality can be strict: if $m = n$ and $ad + bc$ is divisible by $p$, then $k \geq 1$ and $|x+y|_p < \max(|x|_p, |y|_p)$. This cannot happen in the archimedean setting and is a distinctive feature of non-archimedean valuations.
[/guided]
[/step]
[step:Conclude that $|\cdot|_p$ is a non-archimedean absolute value]
We have verified that $|\cdot|_p$ satisfies: (i) $|x|_p = 0 \iff x = 0$ and $|x|_p > 0$ for $x \neq 0$ (non-degeneracy), (ii) $|xy|_p = |x|_p |y|_p$ (multiplicativity), and (iii) $|x + y|_p \leq \max(|x|_p, |y|_p)$ (strong triangle inequality). The strong triangle inequality implies the ordinary triangle inequality $|x + y|_p \leq |x|_p + |y|_p$ (since $\max(a,b) \leq a + b$ for $a, b \geq 0$). Therefore $|\cdot|_p$ is a non-archimedean absolute value on $\mathbb{Q}$.
[/step]
