[proofplan]
We classify all non-trivial absolute values on $\mathbb{Q}$ by splitting into two cases based on the behaviour on $\mathbb{Z}$. If $|n| \leq 1$ for all $n \in \mathbb{Z}$ (the bounded case), the absolute value is non-archimedean and we show it is equivalent to some $|\cdot|_p$ by identifying the unique prime at which it is less than $1$. If some integer has absolute value greater than $1$ (the unbounded case), we use base-$n$ expansions and a limiting argument to show $|\cdot|$ is equivalent to the standard archimedean absolute value $|\cdot|_\infty$.
[/proofplan]
[step:Reduce to two cases: bounded versus unbounded on $\mathbb{Z}$]
Let $|\cdot|$ be a non-trivial absolute value on $\mathbb{Q}$. Since $|1| = 1$ (from $|1| = |1 \cdot 1| = |1|^2$ and $|1| \neq 0$), and $|-1|^2 = |(-1)^2| = |1| = 1$ gives $|-1| = 1$, the absolute value is determined by its values on positive integers (since $|a/b| = |a|/|b|$ for $a, b \in \mathbb{Z}$, $b \neq 0$, and $|-a| = |a|$).
Exactly one of the following holds:
**(Case 1):** $|n| \leq 1$ for all $n \in \mathbb{Z}$.
**(Case 2):** There exists $n \in \mathbb{Z}$ with $|n| > 1$.
Since $|\cdot|$ is non-trivial and $|0| = 0$, $|1| = 1$, there exists $n_0 \in \mathbb{Z}_{>0}$ with $|n_0| \neq 1$. In Case 1, $|n_0| < 1$. In Case 2, some integer exceeds $1$.
[/step]
[step:Case 1 (bounded): identify the distinguished prime $p$ with $|p| < 1$]
Assume $|n| \leq 1$ for all $n \in \mathbb{Z}$, and $|\cdot|$ is non-trivial. Then there exists $n_0 \in \mathbb{Z}_{>0}$ with $|n_0| < 1$. Write $n_0 = p_1^{e_1} \cdots p_r^{e_r}$ (prime factorisation). Since $|n_0| = |p_1|^{e_1} \cdots |p_r|^{e_r} < 1$ and each $|p_i| \leq 1$, at least one prime $p := p_j$ satisfies $|p| < 1$.
[claim:For any prime $q \neq p$, $|q| = 1$.]
Since $\gcd(p, q) = 1$, Bezout's identity gives $a, b \in \mathbb{Z}$ with $ap + bq = 1$. Suppose for contradiction that $|q| < 1$. Since $|n| \leq 1$ for all $n \in \mathbb{Z}$, we have $|a|, |b| \leq 1$. For $N \geq 1$, raise the Bezout relation to the $N$-th power:
\begin{align*}
1 = (ap + bq)^N = \sum_{k=0}^{N} \binom{N}{k} a^k p^k b^{N-k} q^{N-k}.
\end{align*}
Taking absolute values and applying the triangle inequality:
\begin{align*}
1 = |1| \leq \sum_{k=0}^{N} \binom{N}{k} |a|^k |p|^k |b|^{N-k} |q|^{N-k} \leq \sum_{k=0}^{N} \binom{N}{k} |p|^k |q|^{N-k}.
\end{align*}
Set $\alpha = \max(|p|, |q|) < 1$. Then each term satisfies $|p|^k |q|^{N-k} \leq \alpha^N$ (since $k + (N-k) = N$ and both $|p|, |q| \leq \alpha$). So
\begin{align*}
1 \leq (N+1) \cdot \alpha^N.
\end{align*}
But $(N+1)\alpha^N \to 0$ as $N \to \infty$ (since $0 < \alpha < 1$), giving a contradiction for $N$ sufficiently large. Therefore $|q| = 1$.
[proof]
The proof is given inline above.
[/proof]
[/claim]
[/step]
[step:Case 1 (continued): show $|\cdot| = |\cdot|_p^s$ for some $s > 0$]
Every positive integer $m$ has a unique prime factorisation $m = p^k \cdot m'$ where $\gcd(m', p) = 1$. Since $m'$ is a product of primes $q \neq p$, the preceding claim gives $|m'| = 1$. Therefore
\begin{align*}
|m| = |p|^k \cdot |m'| = |p|^k.
\end{align*}
The $p$-adic absolute value gives $|m|_p = p^{-k}$, so $|m| = |p|^k = (p^{-k})^{-\log_p|p|} = |m|_p^s$ where $s := -\log_p |p| = \log_p(1/|p|) > 0$ (since $0 < |p| < 1$).
Since the absolute values of all positive integers determine the absolute value on $\mathbb{Q}$ (via $|a/b| = |a|/|b|$), we have $|x| = |x|_p^s$ for all $x \in \mathbb{Q}$. By the theorem [Characterisation of Equivalent Absolute Values](/theorems/???), this means $|\cdot|$ is equivalent to $|\cdot|_p$.
[/step]
[step:Case 2 (unbounded): bound $|m|$ above using base-$n$ expansion]
Assume there exists $n \in \mathbb{Z}_{>1}$ with $|n| > 1$. Let $m \in \mathbb{Z}_{>1}$ be arbitrary. Write $m$ in base $n$:
\begin{align*}
m = a_0 + a_1 n + a_2 n^2 + \cdots + a_r n^r, \quad 0 \leq a_i \leq n - 1, \quad a_r \neq 0,
\end{align*}
where $r = \lfloor \log_n m \rfloor$, so $n^r \leq m < n^{r+1}$ and $r \leq \log_n m$.
For each digit $a_i \in \{0, 1, \ldots, n-1\}$, since $a_i = 1 + 1 + \cdots + 1$ ($a_i$ times), the triangle inequality gives $|a_i| \leq a_i \cdot |1| = a_i \leq n - 1$. Therefore
\begin{align*}
|m| \leq \sum_{i=0}^{r} |a_i| \cdot |n|^i \leq (n-1) \sum_{i=0}^{r} |n|^i \leq (n-1)(r+1)|n|^r,
\end{align*}
where we used $|n| \geq 1$ (in fact $|n| > 1$) to bound the geometric sum by $(r+1)|n|^r$.
Setting $C_n = (n-1)(\log_n m + 2)$ (since $r + 1 \leq \log_n m + 1 + 1$) and using $|n|^r \leq |n|^{\log_n m} = m^{\log_n |n|}$, we obtain
\begin{align*}
|m| \leq C_n \cdot m^{\log_n |n|}.
\end{align*}
[/step]
[step:Eliminate the constant by the $k$-th power trick]
For any $k \geq 1$, apply the bound from the previous step to $m^k$ in place of $m$:
\begin{align*}
|m|^k = |m^k| \leq C_n(k) \cdot (m^k)^{\log_n |n|},
\end{align*}
where $C_n(k) = (n-1)(k \log_n m + 2)$. Taking $k$-th roots:
\begin{align*}
|m| \leq C_n(k)^{1/k} \cdot m^{\log_n |n|}.
\end{align*}
Since $C_n(k)$ is polynomial in $k$, $C_n(k)^{1/k} \to 1$ as $k \to \infty$. Letting $k \to \infty$:
\begin{align*}
|m| \leq m^{\log_n |n|} \quad \text{for all } m \in \mathbb{Z}_{> 1}.
\end{align*}
[guided]
The purpose of the $k$-th power trick is to eliminate the polynomial factor $C_n$ from the upper bound. The idea is that for an absolute value, $|m^k| = |m|^k$ (multiplicativity), so exponentiating $m$ amplifies the left-hand side exponentially while the polynomial correction grows only polynomially. After taking $k$-th roots and sending $k \to \infty$, the polynomial factor disappears.
This is a standard technique in the theory of absolute values. The key inputs are: (1) multiplicativity $|m^k| = |m|^k$, which is exact (no inequality); (2) the triangle inequality, which produces the polynomial overhead; and (3) the fact that polynomial growth is negligible compared to exponential growth.
After applying the bound to $m^k$: $|m|^k \leq (n-1)(k\log_n m + 2) \cdot m^{k \log_n |n|}$. Taking $k$-th roots: $|m| \leq [(n-1)(k\log_n m + 2)]^{1/k} \cdot m^{\log_n |n|}$. As $k \to \infty$, the factor $[(n-1)(k\log_n m + 2)]^{1/k} \to 1$ (since $\lim_{k \to \infty} k^{1/k} = 1$ and constants raised to the $1/k$ power also tend to $1$). We conclude $|m| \leq m^{\log_n |n|}$ for every integer $m > 1$.
[/guided]
[/step]
[step:Interchange $m$ and $n$ to obtain the exact power relation]
The bound $|m| \leq m^{\log_n |n|}$ holds for every pair of integers $m, n > 1$ with $|n| > 1$. We now apply the same argument with the roles of $m$ and $n$ interchanged.
We first show $|m| > 1$ for all $m > 1$. Writing $n$ in base $m$, the same analysis gives $|n| \leq |n|^{\text{something}}$... Instead, we apply the bound directly: for the fixed $n$ with $|n| > 1$, write $n$ in base $m$. The bound gives $|n| \leq n^{\log_m |m|}$. Since $|n| > 1$, we need $\log_m |m| > 0$, which requires $|m| > 1$ (otherwise $|m| \leq 1$ would give $n^{\log_m |m|} \leq 1 < |n|$, but $\log_m |m|$ is not well-defined when $|m| = 1$). So assume for now $|m| > 1$ and we verify this below.
Claim: $|m| > 1$ for all integers $m > 1$. If $|m| \leq 1$ for some $m > 1$, then write $n$ in base $m$: $n = b_0 + b_1 m + \cdots + b_s m^s$ with $0 \leq b_i \leq m-1$ and $s \leq \log_m n$. Then
\begin{align*}
|n| \leq \sum_{i=0}^{s} |b_i| |m|^i \leq (m-1)(s+1) \cdot 1 = (m-1)(\lfloor \log_m n \rfloor + 1).
\end{align*}
Applying this to $n^k$: $|n|^k = |n^k| \leq (m-1)(k \log_m n + 1)$. Taking $k$-th roots and letting $k \to \infty$ gives $|n| \leq 1$, contradicting $|n| > 1$. So $|m| > 1$ for all $m > 1$.
Now with $|m| > 1$, the bound from the previous step applied with $m$ and $n$ interchanged gives
\begin{align*}
|n| \leq n^{\log_m |m|}.
\end{align*}
Taking logarithms of the two bounds:
\begin{align*}
\log |m| &\leq \log_n |n| \cdot \log m, \\
\log |n| &\leq \log_m |m| \cdot \log n.
\end{align*}
The first gives $\frac{\log |m|}{\log m} \leq \frac{\log |n|}{\log n}$, and the second gives $\frac{\log |n|}{\log n} \leq \frac{\log |m|}{\log m}$. Therefore
\begin{align*}
\frac{\log |m|}{\log m} = \frac{\log |n|}{\log n} \quad \text{for all } m, n > 1 \text{ with } |m|, |n| > 1.
\end{align*}
Since this ratio is independent of $m$, call it $s > 0$. Then $\log|m| = s \log m$, giving $|m| = m^s$ for all integers $m > 1$. Since $|{-}m| = |m|$ and $|1/m| = 1/|m|$, this extends to $|x| = |x|_\infty^s$ for all $x \in \mathbb{Q}$.
[guided]
The heart of the argument is the symmetry trick: the bound $|m| \leq m^{\log_n |n|}$ was derived by writing $m$ in base $n$. By writing $n$ in base $m$, we get the reverse bound $|n| \leq n^{\log_m |m|}$. Together, these force $\frac{\log |m|}{\log m} = \frac{\log |n|}{\log n}$, showing the ratio is a universal constant.
But we first need to verify that $|m| > 1$ for all $m > 1$, since the reverse bound requires expanding $n$ in base $m$ and applying the same limiting argument. If $|m| \leq 1$, the bound on $|n|$ becomes $|n| \leq (m-1)(k\log_m n + 1)^{1/k} \to 1$ as $k \to \infty$, contradicting $|n| > 1$. So the assumption $|n| > 1$ for one $n$ forces $|m| > 1$ for all $m > 1$.
Now set $s = \log|m|/\log m$, which is independent of $m$. For any $x = \pm p_1^{e_1} \cdots p_r^{e_r} / (q_1^{f_1} \cdots q_t^{f_t})$ in $\mathbb{Q}$:
\begin{align*}
|x| = \frac{|p_1|^{e_1} \cdots |p_r|^{e_r}}{|q_1|^{f_1} \cdots |q_t|^{f_t}} = \frac{p_1^{se_1} \cdots p_r^{se_r}}{q_1^{sf_1} \cdots q_t^{sf_t}} = |x|_\infty^s.
\end{align*}
So $|\cdot| = |\cdot|_\infty^s$, which is equivalent to $|\cdot|_\infty$ by the [Characterisation of Equivalent Absolute Values](/theorems/???).
[/guided]
[/step]
[step:Conclude by combining both cases]
In Case 1, $|\cdot|$ is equivalent to $|\cdot|_p$ for a unique prime $p$ (the prime is unique because distinct primes give inequivalent absolute values: $|p|_p < 1$ but $|p|_q = 1$ for $q \neq p$). In Case 2, $|\cdot|$ is equivalent to $|\cdot|_\infty$.
The two cases are mutually exclusive: if $|\cdot|$ is equivalent to $|\cdot|_\infty$, then $|2| = 2^s > 1$ for some $s > 0$, so $|\cdot|$ is unbounded on $\mathbb{Z}$, placing it in Case 2.
Therefore every non-trivial absolute value on $\mathbb{Q}$ is equivalent to exactly one of $|\cdot|_\infty$ or $|\cdot|_p$ for a unique prime $p$.
[/step]
