False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31

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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.
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False: $[a]_n[b]_n=[0]_n$ always implies $[a]_n=[0]_n$ or $[b]_n=[0]_n$

Statement

False claim: for every natural modulus $n$ ,

$[a]_n[b]_n=[0]_n\quad\Longrightarrow\quad[a]_n=[0]_n\ \text{or}\ [b]_n=[0]_n.$

Facts & Assumptions

Given: The quotient $\mathbb Z/6$ and the classes $[2]_6,[3]_6$ .

[L1]

Products in $\mathbb Z/n$ are computed by multiplying representatives (Addition and multiplication on $\mathbb{Z}/n$ by $[a]_n+[b]_n=[a+b]_n$ and $[a]_n[b]_n=[ab]_n$ ).

[L2]

Composite moduli admit nonzero classes with zero product; moreover $[a]_n=[b]_n$ exactly when $a\equiv b\pmod n$ , which means $n\mid(a-b)$ (A modulus $n>1$ is composite exactly when two nonzero classes in $\mathbb{Z}/n$ have product $[0]_n$ , The congruence class $[a]_n$ and the quotient set $\mathbb{Z}/n$ , Congruence modulo an integer: $a\equiv b\pmod n$ when $n\mid(a-b)$ , including the moduli $0$ and $1$ ).

Refutation

technique · direct

1.1

In $\mathbb Z/6$ , $[2]_6[3]_6=[6]_6=[0]_6$ by [L1].

1.2

Neither $6\mid2$ nor $6\mid3$ , so $[2]_6\ne[0]_6$ and $[3]_6\ne[0]_6$ .

2.1

Thus the antecedent of the claimed implication holds while both alternatives in its conclusion fail, so the claim is false.

step 1.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 47 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

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

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