ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31

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The Chinese remainder bijection $Z / 12 \to Z / 3 \times Z / 4$ computed on all standard representatives

Example

For standard representatives, the Chinese remainder map is

$a$	$([a]_{3}, [a]_{4})$	$a$	$([a]_{3}, [a]_{4})$
$0$	$([0]_{3}, [0]_{4})$	$6$	$([0]_{3}, [2]_{4})$
$1$	$([1]_{3}, [1]_{4})$	$7$	$([1]_{3}, [3]_{4})$
$2$	$([2]_{3}, [2]_{4})$	$8$	$([2]_{3}, [0]_{4})$
$3$	$([0]_{3}, [3]_{4})$	$9$	$([0]_{3}, [1]_{4})$
$4$	$([1]_{3}, [0]_{4})$	$10$	$([1]_{3}, [2]_{4})$
$5$	$([2]_{3}, [1]_{4})$	$11$	$([2]_{3}, [3]_{4})$

Facts & Assumptions

Given: The map $Φ ([a]_{12}) : = ([a]_{3}, [a]_{4})$ .

[L1]

Every class modulo a positive integer has one standard representative (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 $Z$ ).

[L2]

Since $3, 4$ are coprime, $Φ$ is a bijection and preserves addition and multiplication componentwise (Chinese remainder theorem for a finite pairwise-coprime list: simultaneous residues determine one class modulo the product, and the resulting bijection preserves addition and multiplication).

[L3]

Modular addition and multiplication are computed by adding and multiplying representatives (Addition and multiplication on $Z / n$ by $[a]_{n} + [b]_{n} = [a + b]_{n}$ and $[a]_{n} [b]_{n} = [a b]_{n}$ ).

Verification

technique · direct

1.1

Reducing each integer $0, \dots, 11$ modulo $3$ and modulo $4$ gives the displayed table. Its entries are all distinct, and [L1] shows that they exhaust the domain and codomain standard pairs, in agreement with [L2].

L1L2

1.2

For a nontrivial addition check, $[7]_{12} + [8]_{12} = [3]_{12}$ , and the table gives $([1]_{3}, [3]_{4}) + ([2]_{3}, [0]_{4}) = ([0]_{3}, [3]_{4})$ .

1.3

For a nontrivial multiplication check, $[5]_{12} [7]_{12} = [11]_{12}$ , and the table gives $([2]_{3}, [1]_{4}) ([1]_{3}, [3]_{4}) = ([2]_{3}, [3]_{4})$ .

2.1

The table computes the whole bijection, while steps 1.2 and 1.3 instantiate its componentwise arithmetic preservation.

step 1.1step 1.2step 1.3L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

17 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, The Chinese Remainder Theorem (standard reference, not scraped)

The Chinese remainder bijection Z/12→Z/3×Z/4Z/12→Z/3×Z/4 computed on all standard representatives