LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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

Statement

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 $b \neq 0$ , written $a / b$ (The rationals as equivalence classes of pairs of integers), and $[(a, b)] \neq 0$ holds exactly when $a \neq 0$ (Arithmetic on the rationals). Write $ι : N \to Z$ for the embedding of The naturals embed in the integers and $j : Z \to Q$ , $j (k) = [(k, 1)]$ , for that of The integers embed in the rationals.

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

$v_{p} (x) : = ι (v_{p} (a)) - ι (v_{p} (b)) \in Z,$

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

The value does not depend on the representative, so $v_{p}$ is a well defined function from the nonzero rationals to $Z$ .
It extends the integer valuation: $v_{p} (j (a)) = ι (v_{p} (a))$ for every nonzero integer $a$ .
$v_{p} (x y) = v_{p} (x) + v_{p} (y)$ for all nonzero rationals $x, y$ .
$v_{p} (x + y) \geq \min {v_{p} (x), v_{p} (y)}$ whenever $x$ , $y$ and $x + y$ are all nonzero, the minimum being taken in the totally ordered $Z$ .

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

Facts & Assumptions

Given: A prime $p$ ; nonzero rationals with representatives $x = [(a, b)]$ , $y = [(c, d)]$ , where $a, b, c, d \in Z$ are all nonzero.

[L1]

$[(a, b)] = [(c, d)]$ exactly when $a d = c b$ ; $Q$ consists of such classes with $b \neq 0$ (The rationals as equivalence classes of pairs of integers).

[L2]

$[(a, b)] + [(c, d)] = [(a d + c b, b d)]$ , $[(a, b)] \cdot [(c, d)] = [(a c, b d)]$ , $0 = [(0, 1)]$ and $1 = [(1, 1)]$ (Arithmetic on the rationals); $Q$ is a field (The rationals form a field, Field).

[L3]

$j (k) = [(k, 1)]$ is injective and preserves addition, multiplication and order (The integers embed in the rationals).

[L4]

For a prime $p$ and nonzero integers $u, w$ : $u w \neq 0$ and $v_{p} (u w) = v_{p} (u) + v_{p} (w)$ ; and $v_{p} (u + w) \geq \min {v_{p} (u), v_{p} (w)}$ when $u$ , $w$ and $u + w$ are nonzero ( $v_{p} (a b) = v_{p} (a) + v_{p} (b)$ for nonzero integers $a, b$ , and $v_{p} (a + b) \geq \min {v_{p} (a), v_{p} (b)}$ whenever $a$ , $b$ and $a + b$ are all nonzero).

[L5]

$v_{p} (1) = 0$ (For a prime $p$ and a nonzero integer $a$ : $p^{v_{p} (a)} ∣ a$ and $p^{v_{p} (a) + 1} ∤ a$ ; $p^{k} ∣ a$ holds exactly for $k \leq v_{p} (a)$ ; $v_{p} (a) \geq 1$ exactly when $p ∣ a$ ; $v_{p} (1) = v_{p} (- 1) = 0$ ; and $v_{p} (p) = 1$ , The $p$ -adic valuation $v_{p} (a)$ of a nonzero integer: the greatest $k \in N$ with $p^{k} ∣ a$ ).

[L6]

$ι$ is injective and preserves addition, multiplication and order, with image the nonnegative integers and $ι (0) = 0$ , $ι (1) = 1$ (The naturals embed in the integers).

[L7]

A product of two nonzero integers is nonzero (The integers have no zero divisors; multiplicative cancellation).

[L8]

$Z$ is a commutative ring: addition and multiplication are associative and commutative, $x + 0 = x$ , and every $x$ has an additive inverse $- x$ , with $- (- x) = x$ and $- (u + w) = (- u) + (- w)$ ; we write $u - w$ for $u + (- w)$ (The integers form a commutative ring, Arithmetic on the integers, The integers as equivalence classes of pairs of naturals).

[L9]

The order on $Z$ is total, antisymmetric and transitive and is compatible with addition, so $u \leq w$ implies $u + z \leq w + z$ (The integers form a totally ordered ring, Order on the integers).

[L10]

The order on $N$ is total, so any two naturals have a minimum; addition on $N$ is commutative ( $\leq$ is a linear order on $N$ , Addition is commutative, Order on the natural numbers, Addition of natural numbers, The natural numbers $N$ (von Neumann)).

Proof

technique · direct

1.1L1L2

If $x = [(a, b)]$ is a nonzero rational then $a \neq 0$ and $b \neq 0$ , so $v_{p} (a)$ and $v_{p} (b)$ are both defined.

2.1step 1.1L1L4L6L7L8

Clause 1. Suppose $[(a, b)] = [(c, d)]$ with all four entries nonzero. Then $a d = c b$ , and both sides are nonzero, so [L4] gives $v_{p} (a) + v_{p} (d) = v_{p} (c) + v_{p} (b)$ in $N$ . Applying the addition-preserving $ι$ and rearranging in $Z$ gives $ι (v_{p} (a)) - ι (v_{p} (b)) = ι (v_{p} (c)) - ι (v_{p} (d))$ .

2.2step 1.1L1L2L7

Clause 4. Assume $x$ , $y$ and $x + y$ are nonzero. Then $x + y = [(a d + c b, b d)]$ with $b d \neq 0$ , and $a d + c b \neq 0$ because $x + y \neq 0$ ; also $a d \neq 0$ and $c b \neq 0$ .

3.1step 2.1L3L5L6L8

Clause 2. For a nonzero integer $a$ , $j (a) = [(a, 1)]$ , so $v_{p} (j (a)) = ι (v_{p} (a)) - ι (v_{p} (1)) = ι (v_{p} (a)) - 0 = ι (v_{p} (a))$ .

3.2step 1.1step 2.1L2L4L7

Clause 3. $x y = [(a c, b d)]$ , with $a c \neq 0$ and $b d \neq 0$ , so $v_{p} (x y) = ι (v_{p} (a c)) - ι (v_{p} (b d)) = ι (v_{p} (a) + v_{p} (c)) - ι (v_{p} (b) + v_{p} (d))$ .

3.3step 2.2L4L6L10

By [L4], $v_{p} (a d + c b) \geq \min {v_{p} (a d), v_{p} (c b)} = \min {v_{p} (a) + v_{p} (d), v_{p} (c) + v_{p} (b)}$ ; applying the order-preserving injection $ι$ turns this into the same inequality between the corresponding integers.

4.1step 3.2L6L8

Since $ι$ preserves addition, that value is $(ι (v_{p} (a)) + ι (v_{p} (c))) - (ι (v_{p} (b)) + ι (v_{p} (d)))$ , which rearranges in the commutative ring $Z$ to $(ι (v_{p} (a)) - ι (v_{p} (b))) + (ι (v_{p} (c)) - ι (v_{p} (d))) = v_{p} (x) + v_{p} (y)$ .

4.2step 2.2step 3.3L4L6L8L9

Subtracting the integer $ι (v_{p} (b)) + ι (v_{p} (d)) = ι (v_{p} (b d))$ from both sides, which preserves the order, and using that subtraction of a fixed element commutes with taking the smaller of two integers, gives $v_{p} (x + y) \geq \min {ι (v_{p} (a)) - ι (v_{p} (b)), ι (v_{p} (c)) - ι (v_{p} (d))} = \min {v_{p} (x), v_{p} (y)}$ .

5.1step 2.1step 3.1step 4.1step 4.2∎

Clauses 1 to 4 are established.

Remarks

Relation to the published $2$ -adic example. The published The $2$ -adic absolute value gives an ultrametric on $Q$ , 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 $a d = c b$ 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 $Z$ under addition, satisfying the ultrametric inequality on valuations.
The values are integers, not naturals. $v_{p} (1 / p) = - 1$ , so the extension genuinely leaves $N$ ; that is why the two integer valuations are transported along $ι$ before being subtracted. As on $Z$ , the value at $0$ is left undefined (The $p$ -adic valuation $v_{p} (a)$ of a nonzero integer: the greatest $k \in N$ with $p^{k} ∣ a$ ).

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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources