LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-02 (claude-opus-5)

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.

For integers $a$ and $b$ the following are equivalent: $a \mid b$ and $b \mid a$ ; $b = ua$ for a unit $u$ ; $|a| = |b|$ . Being associates is an equivalence relation whose class of $a$ is $\{a, -a\}$

Statement

Let $a, b \in \mathbb{Z}$ . The following are equivalent:

$a \mid b$ and $b \mid a$ , that is, $a \sim b$ (Associates in $\mathbb{Z}$ : integers each of which divides the other);
$b = ua$ for some unit $u \in \mathbb{Z}^{\times} = \{1,-1\}$ ( $(\mathbb{Z}, \cdot, 1)$ is a commutative monoid whose group of units is $\{1, -1\}$ ; equivalently $u \mid 1$ holds exactly for $u = 1$ and $u = -1$ );
$|a| = |b|$ (The absolute value $|a|$ of an integer).

Moreover $\sim$ is an equivalence relation on $\mathbb{Z}$ (Equivalence relation, equivalence class, and the quotient set $A/{\sim}$ ), and the class of $a$ is

$[a] \;=\; \{\, a, \, -a \,\},$

which is $\{0\}$ when $a = 0$ and has exactly two elements otherwise.

Facts & Assumptions

Given: Integers $a$ and $b$ , and the relation $\sim$ of Associates in $\mathbb{Z}$ : integers each of which divides the other.

[L1]

$\mathbb{Z}$ is a commutative ring, with $-(-x) = x$ , $(-x)y = -(xy)$ , $x \cdot 1 = x$ and $x \cdot 0 = 0$ (The integers form a commutative ring, Arithmetic on the integers).

[L2]

The order on $\mathbb{Z}$ is total, antisymmetric and transitive, is compatible with addition ( $x \le y$ implies $x + z \le y + z$ ), and positives are closed under multiplication ( $0 < x$ and $0 < y$ imply $0 < xy$ ) (The integers form a totally ordered ring, Order on the integers).

[L3]

$d \mid a$ means $a = dq$ for some $q$ ; $0 \mid a$ only for $a = 0$ (Divisibility in $\mathbb{Z}$ : $d \mid a$ when $a = dq$ for some integer $q$ ).

[L4]

Divisibility is reflexive and transitive, and $d \mid a$ implies $d \mid -a$ and $-d \mid a$ (Divisibility is reflexive and transitive on $\mathbb{Z}$ , and is linear: if $d \mid a$ and $d \mid b$ then $d \mid ax + by$ for all integers $x, y$ ; also $d \mid a$ implies $d \mid ac$ , $-d \mid a$ and $d \mid -a$ ).

[L5]

If $d \mid a$ and $a \ne 0$ then $d \ne 0$ and $|d| \le |a|$ (If $d \mid a$ and $a \ne 0$ then $d \ne 0$ and $|d| \le |a|$ ; hence the set of divisors of a nonzero integer is bounded above by $|a|$ ).

[L6]

$\mathbb{Z}^{\times} = \{1,-1\}$ : an integer $u$ is a unit exactly when $u = 1$ or $u = -1$ , and $1 \ne -1$ ( $(\mathbb{Z}, \cdot, 1)$ is a commutative monoid whose group of units is $\{1, -1\}$ ; equivalently $u \mid 1$ holds exactly for $u = 1$ and $u = -1$ ).

[L7]

$|x| = x$ when $x \ge 0$ and $|x| = -x$ when $x \le 0$ (The absolute value $|a|$ of an integer); $|x| \ge 0$ , $|x| = 0$ exactly when $x = 0$ , and $|xy| = |x|\,|y|$ (Absolute value in $\mathbb{Z}$ : $|a| \ge 0$ ; $|a| = 0$ exactly when $a = 0$ ; $|-a| = |a|$ ; $|ab| = |a|\,|b|$ ; $-|a| \le a \le |a|$ ; and $|a| \le c$ exactly when $-c \le a \le c$ ).

[L8]

A relation is an equivalence relation when it is reflexive, symmetric and transitive, and then the class of $a$ is $[a] = \{\, b : a \sim b \,\}$ (Equivalence relation, equivalence class, and the quotient set $A/{\sim}$ ).

[L9]

The classes of an equivalence relation are nonempty, cover the set, and are pairwise equal or disjoint (The equivalence classes of an equivalence relation are nonempty, cover $A$ , and are pairwise equal or disjoint; conversely every such cover arises from exactly one equivalence relation).

Proof

technique · direct

1.1

Every integer $x$ satisfies $x = |x|$ or $x = -|x|$ : by totality either $x \ge 0$ , when $|x| = x$ , or $x \le 0$ , when $|x| = -x$ and so $x = -(-x) = -|x|$ .

L1L2L7

1.2

$1 \ne 0$ , $0 < 1$ , $|1| = 1$ and $|-1| = 1$ . Indeed $1 = 0$ would give $-1 = -0 = 0 = 1$ , contradicting $1 \ne -1$ in [L6]. By totality either $0 \le 1$ or $1 \le 0$ ; in the second case adding $-1$ gives $0 \le -1$ , and $-1 \ne 0$ (else $1 = 0$ ), so $0 < -1$ , hence $0 < (-1)(-1) = 1$ , which with $1 \le 0$ contradicts antisymmetry. So $0 \le 1$ , and $0 < 1$ since $1 \ne 0$ ; then $|1| = 1$ , while $0 \le 1$ gives $-1 \le 0$ by compatibility with addition, so $|-1| = -(-1) = 1$ .

L1L2L6L7

1.3

Claim 1 implies claim 3. Suppose $a \mid b$ and $b \mid a$ . If $a = 0$ then $0 \mid b$ forces $b = 0$ , so $|a| = 0 = |b|$ . If $a \ne 0$ , then $b \mid a$ with $a \ne 0$ gives $b \ne 0$ and $|b| \le |a|$ , and $a \mid b$ with $b \ne 0$ gives $|a| \le |b|$ ; antisymmetry gives $|a| = |b|$ .

L2L3L5L7

1.4

$\sim$ is reflexive, since $a \mid a$ ; symmetric, since its defining condition is unchanged when $a$ and $b$ are interchanged; and transitive, since $a \mid b$ , $b \mid c$ give $a \mid c$ and $c \mid b$ , $b \mid a$ give $c \mid a$ . So it is an equivalence relation.

L4L8

2.1

Claim 3 implies claim 1. Suppose $|a| = |b|$ . By step 1.1, $b$ is $|b|$ or $-|b|$ , that is $|a|$ or $-|a|$ ; and, again by step 1.1, $|a|$ is $a$ or $-a$ . Hence $b = a$ or $b = -a$ . If $b = a$ then $a \mid b$ and $b \mid a$ by reflexivity. If $b = -a$ then $a \mid -a = b$ by [L4]; and $b = -a$ gives $a = -(-a) = -b$ , so $b \mid -b = a$ , again by [L4].

step 1.1L1L4

2.2

Claim 2 implies claim 3. If $b = ua$ with $u$ a unit then $u = 1$ or $u = -1$ , so $|u| = 1$ by step 1.2, and $|b| = |u|\,|a| = |a|$ .

step 1.2L6L7

3.1

Claim 3 implies claim 2. By step 2.1, $|a| = |b|$ gives $b = a = 1 \cdot a$ or $b = -a = (-1)a$ , and $1$ and $-1$ are units.

step 2.1L1L6

4.1

The three claims are equivalent: claim 3 implies claim 1 by step 2.1 and claim 1 implies claim 3 by step 1.3, while claim 3 implies claim 2 by step 3.1 and claim 2 implies claim 3 by step 2.2.

step 2.1step 1.3step 2.2step 3.1

5.1

The class of $a$ is $[a] = \{\, b : a \sim b \,\} = \{\, b : |b| = |a| \,\} = \{a, -a\}$ : the middle equality is the equivalence of claims 1 and 3, and the last holds because $|{-a}| = |a|$ by [L7], while conversely $|b| = |a|$ gives $b = a$ or $b = -a$ by step 2.1. At $a = 0$ this set is $\{0\}$ , since $-0 = 0$ ; and for $a \ne 0$ it has exactly two elements, since by totality either $0 < a$ , when adding $-a$ gives $-a < 0 < a$ , or $a < 0$ , when adding $-a$ gives $a < 0 < -a$ ; in both cases $a \ne -a$ .

step 2.1step 4.1L1L2L7L8

6.1

By [L9] the classes $\{a,-a\}$ are nonempty, cover $\mathbb{Z}$ , and any two are equal or disjoint; together with steps 4.1, 1.4 and 5.1 this is the full statement.

step 4.1step 1.4step 5.1L9∎

Remarks

This is the "up to sign" of elementary number theory made precise. Every statement below that fixes a sign — $\gcd(a,b) \ge 0$ , $\operatorname{lcm}(a,b) \ge 0$ , the nonnegative generator of a subgroup of $(\mathbb{Z},+)$ — is choosing one representative from a class $\{a,-a\}$ , and claim 3 is what says the choice is between exactly two candidates.
The equivalence of claims 1 and 2 is the general ring-theoretic statement, and it is the reason associates are defined by mutual divisibility rather than by "differ by a sign": mutual divisibility is the formulation that survives when the unit group is larger than $\{1,-1\}$ .

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 79 results over 26 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Divisibility (ring theory) (Wikipedia) (standard reference, not scraped)
Divisor (Wikipedia) (standard reference, not scraped)
Unit (ring theory) (Wikipedia) (standard reference, not scraped)

For integers aaa and bbb the following are equivalent: a∣ba \mid ba∣b and b∣ab \mid ab∣a; b=uab = uab=ua for a unit uuu; ∣a∣=∣b∣|a| = |b|∣a∣=∣b∣. Being associates is an equivalence relation whose class of aaa is {a,−a}\{a, -a\}{a,−a}