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  "body": "\n## Statement\n\nLet $\\mathbf{1}$ be the arithmetic function with $\\mathbf{1}(n) = 1$ for all $n$, and let $\\varepsilon$ be the function with $\\varepsilon(1) = 1$ and $\\varepsilon(n) = 0$ for $n \u003e 1$.\nThen $\\mu * \\mathbf{1} = \\varepsilon$, where $\\mu$ is the [[Möbius function|0C8QENHYZW]] and $*$ is the [[Dirichlet convolution|0KH1N66QWW]].\nThat is, for each positive integer $n$,\n$$\\sum_{d \\mid n} \\mu(d) = \\begin{cases} 1 \u0026 \\text{if } n = 1, \\\\ 0 \u0026 \\text{if } n \u003e 1. \\end{cases}$$\n\n## Remarks\n\nThe proof for $n \u003e 1$: let $k = \\omega(n) \\ge 1$. Only squarefree divisors count, so the sum is $\\sum_{j=0}^{k} \\binom{k}{j} (-1)^j = (1 - 1)^k = 0$.\nFor example, the divisors of $12$ give $\\mu(1) + \\mu(2) + \\mu(3) + \\mu(6) = 1 - 1 - 1 + 1 = 0$.\nThis identity is the key step in the [[Möbius inversion formula|07Q4TBTFVB]].\n\n## Explanations\n\n### L1 human\n\nThe signs $\\mu(d)$ over the divisors of $n$ cancel out, except for $n = 1$.\nThis is the same cancellation as in the principle of inclusion and exclusion.\n",
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