{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe rearrange finite double sums and use $\\sum_{d \\mid k} \\mu(d) = [k = 1]$. Here $[k = 1]$ is $1$ if $k = 1$ and $0$ otherwise.\n\n1. By the [[sum of the Möbius function over divisors|0ASC73QEQT]], for each positive integer $k$ we have $\\sum_{d \\mid k} \\mu(d) = [k = 1]$. All divisors here are [[positive divisors|0FCWA16FZE]].\n2. Fix a positive integer $n$. Let $T$ be the set of pairs $(d, e)$ of positive integers with $de \\mid n$.\n3. For positive integers $d$ and $e$, the conditions \"$d \\mid n$ and $e \\mid n/d$\", \"$de \\mid n$\" and \"$e \\mid n$ and $d \\mid n/e$\" are equivalent. Each one says that $n = dek$ for a positive integer $k$.\n4. Assume first that $g(m) = \\sum_{e \\mid m} f(e)$ for all positive integers $m$. Then\n$$\\sum_{d \\mid n} \\mu(d) g\\!\\left(\\frac{n}{d}\\right) = \\sum_{d \\mid n} \\sum_{e \\mid n/d} \\mu(d) f(e) = \\sum_{(d, e) \\in T} \\mu(d) f(e).$$\n5. By step 3, we can sum over $e$ first:\n$$\\sum_{(d, e) \\in T} \\mu(d) f(e) = \\sum_{e \\mid n} f(e) \\sum_{d \\mid n/e} \\mu(d) = \\sum_{e \\mid n} f(e) \\, [n/e = 1].$$\nThe last equality is step 1 with $k = n/e$.\n6. Only $e = n$ has $n/e = 1$. So the sum equals $f(n)$. This proves the formula for $f$.\n7. Conversely, assume that $f(m) = \\sum_{e \\mid m} \\mu(e) g(m/e)$ for all positive integers $m$. Then\n$$\\sum_{d \\mid n} f(d) = \\sum_{d \\mid n} \\sum_{e \\mid d} \\mu(e) \\, g\\!\\left(\\frac{d}{e}\\right).$$\n8. The pairs $(d, e)$ with $d \\mid n$ and $e \\mid d$ correspond one-to-one to the pairs $(e, k)$ with $ek \\mid n$, by $k = d/e$ and $d = ek$.\n9. So, using step 3 with $(e, k)$ in place of $(d, e)$,\n$$\\sum_{d \\mid n} f(d) = \\sum_{(e, k) :\\, ek \\mid n} \\mu(e) g(k) = \\sum_{k \\mid n} g(k) \\sum_{e \\mid n/k} \\mu(e) = \\sum_{k \\mid n} g(k) \\, [n/k = 1].$$\n10. Only $k = n$ contributes, so $\\sum_{d \\mid n} f(d) = g(n)$. Since $n$ was arbitrary, both directions hold.\n\n## Idea\n\nSubstitute one relation into the other and swap the order of the two divisor sums. The inner sum then becomes a sum of $\\mu$ over the divisors of a number. That sum is $1$ for the number $1$ and $0$ otherwise, so only one term survives.\n",
  "commit": "68a8c3dc73fd55004b2796431c607c30960d1a85",
  "front": {
    "kind": "proof",
    "msc": [
      "11A25"
    ],
    "name": "Proof of the Möbius inversion formula",
    "proves": "07Q4TBTFVB",
    "sources": [
      {
        "locator": "Proofs of generalizations",
        "url": "https://en.wikipedia.org/wiki/M%C3%B6bius_inversion_formula"
      }
    ],
    "tag": "S9N8VV57EJ",
    "uses": [
      "0C8QENHYZW",
      "0ASC73QEQT",
      "0FCWA16FZE"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/S9N8VV57EJ/",
  "links": [
    {
      "field": "proves",
      "target": "07Q4TBTFVB"
    },
    {
      "field": "uses",
      "target": "0C8QENHYZW"
    },
    {
      "field": "uses",
      "target": "0ASC73QEQT"
    },
    {
      "field": "uses",
      "target": "0FCWA16FZE"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/S9N8VV57EJ.md",
  "msc": [
    "11A25"
  ],
  "name": "Proof of the Möbius inversion formula",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/S9N/S9N8VV57EJ.md",
  "status": null,
  "tag": "S9N8VV57EJ",
  "uri": "m2://S9N8VV57EJ",
  "url": "https://math2.tamnd.com/wiki/proof-of-the-mobius-inversion-formula/",
  "writing": {
    "tag": "S9N8VV57EJ",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}