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

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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.

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$v_{p} (a b) = v_{p} (a) + v_{p} (b)$ 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

Statement

Let $p$ be a prime (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ) and let $a, b \in Z$ be nonzero, with $v_{p}$ as in The $p$ -adic valuation $v_{p} (a)$ of a nonzero integer: the greatest $k \in N$ with $p^{k} ∣ a$ . Then $a b \neq 0$ and

$v_{p} (a b) = v_{p} (a) + v_{p} (b),$

the sum taken in $N$ (Addition of natural numbers). If moreover $a + b \neq 0$ , then

$v_{p} (a + b) \geq \min {v_{p} (a), v_{p} (b)},$

the minimum of two natural numbers, which exists because the order on $N$ is total ( $\leq$ is a linear order on $N$ ).

Facts & Assumptions

Given: A prime $p$ and nonzero integers $a, b$ ; $α : = v_{p} (a)$ and $β : = v_{p} (b)$ .

[L1]

For a nonzero $u$ : $p^{v_{p} (u)} ∣ u$ , and $u = p^{v_{p} (u)} u^{'}$ for some nonzero $u^{'}$ with $p ∤ u^{'}$ ; moreover $p^{k} ∣ u$ holds exactly for $k \leq v_{p} (u)$ (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 $p^{k} ∣ a$ ).

[L2]

Exponent law in a monoid for natural exponents: $g^{m + n} = g^{m} g^{n}$ (Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute, Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ , Semigroup and monoid, $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ ).

[L3]

If $p$ is prime and $p ∣ u v$ then $p ∣ u$ or $p ∣ v$ (Euclid's lemma: if $p$ is prime and $p ∣ a b$ then $p ∣ a$ or $p ∣ b$ ).

[L4]

Divisibility is linear: $d ∣ u$ and $d ∣ v$ give $d ∣ u x + v y$ , in particular $d ∣ u + v$ (Divisibility is reflexive and transitive on $Z$ , and is linear: if $d ∣ a$ and $d ∣ b$ then $d ∣ a x + b y$ for all integers $x, y$ ; also $d ∣ a$ implies $d ∣ a c$ , $- d ∣ a$ and $d ∣ - a$ ); $d ∣ u$ means $u = d c$ for some $c$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ).

[L5]

A product of two nonzero integers is nonzero, and $x z = y z$ with $z \neq 0$ gives $x = y$ (The integers have no zero divisors; multiplicative cancellation).

[L6]

$Z$ is a commutative ring: multiplication is associative and commutative and $x \cdot 1 = x$ (The integers form a commutative ring, Arithmetic on the integers, The integers as equivalence classes of pairs of naturals); its order is total, antisymmetric and transitive (The integers form a totally ordered ring, Order on the integers, The naturals embed in the integers).

[L7]

On $N$ : $\leq$ is a linear order, so any two naturals are comparable and have a minimum ( $\leq$ is a linear order on $N$ ); $m \leq n$ means $m + c = n$ for some $c$ (Order on the natural numbers); $σ (k) = k + 1$ (Addition of natural numbers, The natural numbers $N$ (von Neumann)); $m < n$ exactly when $σ (m) \leq n$ (Discreteness: $σ (n)$ is the immediate successor), and $m < σ (n)$ exactly when $m \leq n$ (On $N$ the order is membership: $m < n ⟺ m \in n$ ).

Proof

technique · direct

1.1L5

$a b \neq 0$ , so $v_{p} (a b)$ is defined.

1.2L1choose

Fix $a^{'}$ and $b^{'}$ with $a = p^{α} a^{'}$ , $b = p^{β} b^{'}$ , both nonzero, and $p ∤ a^{'}$ , $p ∤ b^{'}$ .

1.3L7

Now assume also $a + b \neq 0$ , and put $m : = \min {α, β}$ , which exists because $\leq$ is total on $N$ ; then $m \leq α$ and $m \leq β$ .

2.1step 1.2L2L6

$a b = (p^{α} a^{'}) (p^{β} b^{'}) = (p^{α} p^{β}) (a^{'} b^{'}) = p^{α + β} (a^{'} b^{'})$ , using commutativity, associativity and the exponent law.

2.2step 1.2L3

$p ∤ a^{'} b^{'}$ : otherwise [L3] would give $p ∣ a^{'}$ or $p ∣ b^{'}$ , both excluded by step 1.2.

2.3step 1.3L1L4

By [L1], $p^{m} ∣ a$ and $p^{m} ∣ b$ , so $p^{m} ∣ a + b$ by linearity; since $a + b \neq 0$ , [L1] applied to $a + b$ gives $m \leq v_{p} (a + b)$ , which is the second assertion.

3.1step 1.1step 2.1L6

$p^{α + β} \neq 0$ , since $a b \neq 0$ and $a b = p^{α + β} (a^{'} b^{'})$ would otherwise be $0$ .

3.2step 1.1step 2.1L1L4

$p^{α + β} ∣ a b$ by step 2.1, so $α + β \leq v_{p} (a b)$ .

4.1step 2.1step 2.2step 3.1L1L2L4L5L7

Suppose $α + β < v_{p} (a b)$ . Then $α + β + 1 \leq v_{p} (a b)$ , so $p^{α + β + 1} ∣ a b$ ; fix $c$ with $a b = p^{α + β + 1} c = p^{α + β} (p c)$ , using the exponent law. Cancelling $p^{α + β} \neq 0$ against step 2.1 gives $a^{'} b^{'} = p c$ , that is $p ∣ a^{'} b^{'}$ , contradicting step 2.2.

5.1step 3.2step 4.1L7

Hence $v_{p} (a b) = α + β$ by totality and antisymmetry of the order on $N$ , which is the first assertion.

6.1step 5.1step 2.3∎

Both assertions are established.

Remarks

Primality is used exactly once, at step 2.2, and it is used through Euclid's lemma: if $p$ is prime and $p ∣ a b$ then $p ∣ a$ or $p ∣ b$ . That is the whole reason the additivity of $v_{p}$ is a theorem about primes and not about arbitrary integers greater than $1$ : for a composite base the analogue fails, and the companion page of the divisibility page already carries the failure of the product principle ( $6 ∣ 4 \cdot 9$ while $6 ∤ 4$ and $6 ∤ 9$ : dividing a product does not force dividing a factor, and the coprimality hypothesis is what fails).
The second clause is an inequality and cannot be an equality. Take $p = 2$ , $a = 2$ , $b = 2$ : then $v_{2} (a) = v_{2} (b) = 1$ while $v_{2} (a + b) = v_{2} (4) = 2$ . The companion page's worked example computes such valuations explicitly ( $360 = 2^{3} \cdot 3^{2} \cdot 5$ and $84 = 2^{2} \cdot 3 \cdot 7$ , with $\gcd (360, 84) = 12$ and $lcm (360, 84) = 2520$ read off the exponents ↗).
The minimum is taken in $N$ , through $\leq$ is a linear order on $N$ , not through the library's Maximum and minimum of a set, which is stated for real numbers. The values of $v_{p}$ are natural numbers, so no passage to $R$ is needed or wanted here.

$v_{p} (a b) = v_{p} (a) + v_{p} (b)$ 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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

vp(ab)=vp(a)+vp(b)vp​(ab)=vp​(a)+vp​(b) for nonzero integers a,ba,b, and vp(a+b)≥min⁡{vp(a),vp(b)}vp​(a+b)≥min{vp​(a),vp​(b)} whenever aa, bb and a+ba+b are all nonzero

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

$v_{p} (a b) = v_{p} (a) + v_{p} (b)$ 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