The $p$-adic valuation extends to the nonzero rationals by $v_p(a/b) := v_p(a) - v_p(b) \in \mathbb{Z}$, independently of the representation; it satisfies $v_p(xy) = v_p(x) + v_p(y)$, and $v_p(x+y) \ge \min\{v_p(x), v_p(y)\}$ whenever $x$, $y$ and $x+y$ are nonzero

Let $p$ be a prime (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$). A rational is a class $[(a,b)]$ of pairs of integers with $b \ne 0$, written $a/b$ (The rationals as equivalence classes of pairs of integers), and $[(a,b)] \ne 0$ holds exactly when $a \ne 0$ (Arithmetic on the rationals). Write $\iota : \mathbb{N} \to \mathbb{Z}$ for the embedding of The naturals embed in the integers and $j : \mathbb{Z} \to \mathbb{Q}$, $j(k) = [(k,1)]$, for that of The integers embed in the rationals.

For a nonzero rational $x = [(a,b)]$ set

$v_p(x) \;:=\; \iota\bigl(v_p(a)\bigr) - \iota\bigl(v_p(b)\bigr) \;\in\; \mathbb{Z},$

with $v_p$ on nonzero integers as in The $p$-adic valuation $v_p(a)$ of a nonzero integer: the greatest $k \in \mathbb{N}$ with $p^{k} \mid a$. Then:

1. The value does not depend on the representative, so $v_p$ is a well defined function from the nonzero rationals to $\mathbb{Z}$.
2. It extends the integer valuation: $v_p(j(a)) = \iota(v_p(a))$ for every nonzero integer $a$.
3. $v_p(xy) = v_p(x) + v_p(y)$ for all nonzero rationals $x, y$.
4. $v_p(x+y) \ge \min\{\, v_p(x),\, v_p(y) \,\}$ whenever $x$, $y$ and $x+y$ are all nonzero, the minimum being taken in the totally ordered $\mathbb{Z}$.

Unlike its restriction to $\mathbb{Z}$, this valuation takes integer values, which is why the difference is formed after transporting the two natural numbers into $\mathbb{Z}$ along $\iota$.

• Relation to the published $2$-adic example. The published The $2$-adic absolute value gives an ultrametric on $\mathbb{Q}$, in which every triangle is isosceles and every point of a ball is a centre ↗ records that the general $p$-adic machinery is available, but nevertheless develops $p=2$ from parity alone. The present lemma supplies the general algebraic extension: representation-independence is exactly the assertion that $ad=cb$ forces the two candidate values to agree.

• Nothing metric is stated here, deliberately. The $p$-adic absolute value $|x|_p = p^{-v_p(x)}$ and the ultrametric it induces need real powers with integer exponents and the definition of a metric space, all of which live far above this page in the library's order; they are not defined here and nothing on this page depends on them. What is proved is the algebra: a homomorphism from the nonzero rationals under multiplication to $\mathbb{Z}$ under addition, satisfying the ultrametric inequality on valuations.

• The values are integers, not naturals. $v_p(1/p) = -1$, so the extension genuinely leaves $\mathbb{N}$; that is why the two integer valuations are transported along $\iota$ before being subtracted. As on $\mathbb{Z}$, the value at $0$ is left undefined (The $p$-adic valuation $v_p(a)$ of a nonzero integer: the greatest $k \in \mathbb{N}$ with $p^{k} \mid a$).