ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-07-31

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.

A modulus $n > 1$ is composite exactly when two nonzero classes in $Z / n$ have product $[0]_{n}$

Example

For every integer $n > 1$ , the following are equivalent:

$n$ is composite.
There are nonzero classes $u, v \in Z / n$ with $u v = [0]_{n}$ .

Facts & Assumptions

Given: An integer $n > 1$ .

[L1]

A positive integer greater than $1$ is composite exactly when it has a divisor $d$ with $1 < d < n$ ; otherwise it is prime (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[L2]

Products in $Z / n$ are computed on representatives, and the quotient has standard representatives $0, \dots, n - 1$ (Addition and multiplication on $Z / n$ by $[a]_{n} + [b]_{n} = [a + b]_{n}$ and $[a]_{n} [b]_{n} = [a b]_{n}$ , For $n \geq 1$ , every class in $Z / n$ has one representative $r$ with $0 \leq r < n$ , so $∣ Z / n ∣ = n$ ; while $Z / 0$ is in bijection with $Z$ ).

[L3]

A class $[a]_{n}$ is a unit exactly when $\gcd (a, n) = 1$ , and the gcd is a nonnegative common divisor that is positive when its arguments are not both zero (For $n \geq 1$ , $[a]_{n}$ is a unit if and only if $\gcd (a, n) = 1$ , Common divisor, and the greatest common divisor $\gcd (a, b)$ , with the convention $\gcd (0, 0) : = 0$ ).

Verification

technique · direct

1.1

If $n$ is composite, write $n = a b$ with $1 < a < n$ and $1 < b < n$ . Then $[a]_{n}$ and $[b]_{n}$ are nonzero standard classes, but $[a]_{n} [b]_{n} = [a b]_{n} = [n]_{n} = [0]_{n}$ .

L1L2

1.2

Conversely, suppose nonzero classes have product zero and choose their standard representatives $a, b$ , so $0 < a, b < n$ and $n ∣ a b$ . The class $[a]_{n}$ cannot be a unit, because multiplying an equation $[a]_{n} [b]_{n} = [0]_{n}$ by its inverse would give $[b]_{n} = [0]_{n}$ . Hence $d : = \gcd (a, n) > 1$ by [L3]. Since $d ∣ a$ and $0 < a < n$ , one also has $d < n$ , so $d$ is a nontrivial divisor of $n$ and $n$ is composite by [L1].

L1L2L3

2.1

Steps 1.1 and 1.2 prove the equivalence.

step 1.1step 1.2∎

Depends on

Used by

False: [a]ₙ[b]ₙ=[0]ₙ always implies [a]ₙ=[0]ₙ or [b]ₙ=[0]ₙ False statement

Dependency tree · two levels

30 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

K. Conrad, Modular Arithmetic (standard reference, not scraped)
University of Colorado, proof practice on zero divisors (standard reference, not scraped)

A modulus n>1n>1 is composite exactly when two nonzero classes in Z/nZ/n have product [0]n[0]n​