TheoremStatement: AI-adaptedProof: AI-generatedprecheck 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.

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 $Z$

Statement

Let $n$ be a positive integer. Every class in $Z / n$ (The congruence class $[a]_{n}$ and the quotient set $Z / n$ ) contains exactly one integer $r$ with $0 \leq r < n$ . Consequently the map

$r ⟼ [r]_{n} (0 \leq r < n)$

is a bijection from the von Neumann natural $n$ to $Z / n$ , and $∣ Z / n ∣ = n$ . This includes $n = 1$ , where the only representative is $0$ . For $n = 0$ , the map $a \mapsto [a]_{0}$ is a bijection $Z \to Z / 0$ .

Facts & Assumptions

Given: A positive integer $n$ and integers $a, b$ ; separately, the modulus $0$ .

[L1]

For $n > 0$ there are unique integers $q, r$ with $a = q n + r$ and $0 \leq r < n$ (Division with remainder in $Z$ : for $a \in Z$ and $b > 0$ there are unique $q, r \in Z$ with $a = q b + r$ and $0 \leq r < b$ ).

[F1]

$[u]_{n} = [v]_{n}$ exactly when $n ∣ (u - v)$ (The congruence class $[a]_{n}$ and the quotient set $Z / n$ ).

[L2]

The natural-number embedding into $Z$ is injective and has image the nonnegative integers; the von Neumann natural $n$ is the set of naturals $r < n$ (The naturals embed in the integers, The natural numbers $N$ (von Neumann)).

[F2]

A bijection transports finite cardinality, and $∣ n ∣ = n$ (Injection, surjection, bijection, The cardinality $∣ A ∣$ of a finite set).

Proof

technique · direct

1.1L1F1

By [L1], write $a = q n + r$ with $0 \leq r < n$ . Then $n ∣ (a - r)$ , so $[a]_{n} = [r]_{n}$ .

1.2F1L1

If $0 \leq r, s < n$ and $[r]_{n} = [s]_{n}$ , then $r - s = k n$ for some integer $k$ , so $r = k n + s$ . Both $r = 0 n + r$ and $r = k n + s$ express $r$ with a remainder in the range from $0$ to $n - 1$ , so uniqueness in [L1] gives $r = s$ .

1.3F1algebra

At modulus $0$ , [F1] says $[a]_{0} = [b]_{0}$ exactly when $0 ∣ (a - b)$ , exactly when $a = b$ . Hence $a \mapsto [a]_{0}$ is injective, and it is surjective by the definition of the quotient set.

2.1step 1.1step 1.2L2F2

Thus $r \mapsto [r]_{n}$ from the natural $n$ to $Z / n$ is surjective by step 1.1 and injective by step 1.2, hence bijective.

3.1step 2.1F2L2

Since $n$ is finite with cardinality $n$ , the bijection in step 2.1 gives $∣ Z / n ∣ = n$ . At $n = 1$ its domain is $1 = {0}$ , so there is one class.

4.1step 1.1step 1.2step 2.1step 3.1step 1.3∎

Steps 1.1 through 3.1 prove the positive-modulus statement, and step 1.3 proves the bijection at modulus $0$ .

Depends on

Dependency tree · 2 levels

Sources

K. Conrad, Modular Arithmetic

For n≥1n≥1, every class in Z/nZ/n has one representative rr with 0≤r<n0≤r<n, so ∣Z/n∣=n∣Z/n∣=n; while Z/0Z/0 is in bijection with ZZ