Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31
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For n≥1, every class in Z/n has one representative r with 0≤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≤r<n. Consequently the map

r⟼[r]n(0≤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↦[a]0 is a bijection Z→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=qn+r and 0≤r<n (Division with remainder in Z: for a∈Z and b>0 there are unique q,r∈Z with a=qb+r and 0≤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
1.1

By [L1], write a=qn+r with 0≤r<n. Then n∣(a−r), so [a]n=[r]n.

L1F1
1.2

If 0≤r,s<n and [r]n=[s]n, then r−s=kn for some integer k, so r=kn+s. Both r=0n+r and r=kn+s express r with a remainder in the range from 0 to n−1, so uniqueness in [L1] gives r=s.

F1L1
1.3

At modulus 0, [F1] says [a]0=[b]0 exactly when 0∣(a−b), exactly when a=b. Hence a↦[a]0 is injective, and it is surjective by the definition of the quotient set.

F1algebra
2.1

Thus r↦[r]n from the natural n to Z/n is surjective by step 1.1 and injective by step 1.2, hence bijective.

step 1.1step 1.2L2F2
3.1

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.

step 2.1F2L2
4.1

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

step 1.1step 1.2step 2.1step 3.1step 1.3∎

Depends on

Used by

…and 3 more results.

Dependency tree · two levels

33 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