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  "body": "\n## Proof\n\n1. Let $f$ and $g$ be [[arithmetic functions|0H2EPRZC0D]] and $n$ a positive integer. Let $D$ be the set of [[positive divisors|0FCWA16FZE]] of $n$, and $P$ the set of ordered pairs $(a, b)$ of positive integers with $ab = n$.\n2. By the definition of the [[Dirichlet convolution|0KH1N66QWW]], $(f * g)(n) = \\sum_{d \\in D} f(d) \\, g(n/d)$.\n3. For $d \\in D$, the number $n/d$ is a positive integer and $d \\cdot (n/d) = n$. So $\\beta(d) = (d, n/d)$ defines a map $\\beta : D \\to P$.\n4. For $(a, b) \\in P$, the equation $n = ab$ shows $a \\in D$, and $n/a = b$. So $\\beta(a) = (a, b)$, and $\\beta$ is surjective.\n5. If $\\beta(d) = \\beta(d')$, then the first coordinates give $d = d'$. So $\\beta$ is injective, hence a bijection.\n6. Reindexing the finite sum of step 2 along $\\beta$ gives $(f * g)(n) = \\sum_{(a, b) \\in P} f(a) \\, g(b)$.\n\n## Idea\n\nA divisor $d$ of $n$ and the pair $(d, n/d)$ carry the same information. So summing over divisors is the same as summing over the ordered factorizations $n = ab$.\n",
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