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CorollaryStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31
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The number-theoretic Möbius function is multiplicative on coprime positive integers

Statement

If m,nm,n are coprime positive integers, then

μ(mn)=μ(m)μ(n).\mu(mn)=\mu(m)\mu(n).

Facts & Assumptions

Given: Coprime positive integers m,nm,n (Coprime integers: gcd(a,b)=1\gcd(a,b) = 1).

[L3]
[L4]

Möbius values transport through poset isomorphisms because the diagonal and recurrence conditions uniquely determine them (The Möbius recurrence: μP(x,x)=1\mu_P(x,x)=1 and both interval sums of μP\mu_P vanish when x<yx<y).

Proof

technique · direct
1.1

By [L1], (a,b)ab(a,b)\mapsto ab is a bijection from [1,m]×[1,n][1,m]_{\mid}\times[1,n]_{\mid} to [1,mn][1,mn]_{\mid}. It preserves and reflects divisibility coordinatewise, again by the disjoint prime supports, so it is a poset isomorphism.

L1construct
2.1

Apply the product theorem at the endpoints and transfer along the isomorphism by [L4]: μ(1,mn)=μ(1,m)μ(1,n)\mu_{\mid}(1,mn)=\mu_{\mid}(1,m)\mu_{\mid}(1,n).

step 1.1L2L4
3.1

Replacing each poset value in step 2.1 by its number-theoretic value using [L3] yields μ(mn)=μ(m)μ(n)\mu(mn)=\mu(m)\mu(n).

step 2.1L3

Depends on

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