Alphabeta Math
DefinitionDefinition: AI-adaptedProof: Not applicableSession-authored (Fable 5 assisted)judge pass (z-ai/glm-5.2)audited 2026-07-28
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.

Divisibility in Z\mathbb{Z}: dad \mid a when a=dqa = dq for some integer qq

Definition

Let d,aZd, a \in \mathbb{Z} (The integers as equivalence classes of pairs of naturals). We say dd divides aa, and write dad \mid a, when

a  =  dqfor some qZ,a \;=\; d q \qquad \text{for some } q \in \mathbb{Z},

the product being that of Arithmetic on the integers. We write dad \nmid a when this fails. In this situation dd is called a divisor, or a factor, of aa, and aa is called a multiple of dd.

This is the relation the library already has, not a second one. The published Division with remainder in Z\mathbb{Z}: for aZa \in \mathbb{Z} and b>0b > 0 there are unique q,rZq, r \in \mathbb{Z} with a=qb+ra = qb + r and 0r<b0 \le r < b introduces it in its own Statement, in these words: "We say bb divides aa, written bab \mid a, when a=qba = qb for some qZq \in \mathbb{Z}." Since multiplication on Z\mathbb{Z} is commutative (The integers form a commutative ring), a=qda = qd and a=dqa = dq are the same condition, so the definition above is that relation verbatim and the two usages agree everywhere. The theorem defined it for use on its own page and left the systematic theory to a later page; this is that page, and this item records the agreement rather than introducing a rival notion.

The remainder test. For b>0b > 0 the same Statement records that bab \mid a holds exactly when the remainder rr in a=qb+ra = qb + r, 0r<b0 \le r < b, is 00.

Boundary values. Each is one line from the ring axioms, and each is used below, so all three are recorded here rather than assumed:

  • d0d \mid 0 for every integer dd, including d=0d = 0, since 0=d00 = d \cdot 0;
  • 0a0 \mid a only for a=0a = 0, since a=0qa = 0 \cdot q forces a=0a = 0;
  • 1a1 \mid a and aaa \mid a for every aa, since a=1aa = 1 \cdot a and a=a1a = a \cdot 1.

Remarks

Depends on

Used by

…and 21 more results.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 39 results over 15 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