{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\n1. Let $x \\ge 0$ have the [[decimal expansion|06XY4X9430]] $x = a_0 + \\sum_{k=1}^{\\infty} d_k 10^{-k}$. Let $N \\ge 0$ and $p \\ge 1$ with $d_{k+p} = d_k$ for each $k \u003e N$.\n2. For $K \\ge 0$, let $S_K = a_0 + \\sum_{k=1}^{K} d_k 10^{-k}$. By the definition of the [[sum of a series|0K0H5E3Z5N]], $S_K - a_0 \\to x - a_0$. By the [[limit of a sum|0RAM4E07NR]], $S_K \\to x$ as $K \\to \\infty$.\n3. Also $S_{K+p} \\to x$. Indeed, by the [[epsilon-N form of the limit|09G6KHS6GY]], if $\\abs{S_K - x} \u003c \\varepsilon$ for $K \\ge M$, then also $\\abs{S_{K+p} - x} \u003c \\varepsilon$ for $K \\ge M$.\n4. Fix $K \u003e N$. Multiply $S_{K+p}$ by $10^{N+p}$ and split the sum at $k = N + p$:\n$$10^{N+p} S_{K+p} = 10^{N+p} a_0 + \\sum_{k=1}^{N+p} d_k 10^{N+p-k} + \\sum_{k=N+p+1}^{K+p} d_k 10^{N+p-k}.$$\n5. In the last sum, write $k = j + p$ with $N + 1 \\le j \\le K$. Since $j \u003e N$, $d_{j+p} = d_j$. So the last sum is $\\sum_{j=N+1}^{K} d_j 10^{N-j}$.\n6. In the same way, $10^N S_K = 10^N a_0 + \\sum_{k=1}^{N} d_k 10^{N-k} + \\sum_{j=N+1}^{K} d_j 10^{N-j}$.\n7. Subtract step 6 from steps 4 and 5. The last sums cancel, and $10^{N+p} S_{K+p} - 10^N S_K = M$, where\n$$M = 10^{N+p} a_0 + \\sum_{k=1}^{N+p} d_k 10^{N+p-k} - 10^N a_0 - \\sum_{k=1}^{N} d_k 10^{N-k}.$$\n8. The number $M$ does not depend on $K$. It is an [[integer|0C4JS9RF5M]], since $a_0$ and the $d_k$ are integers and all the exponents of $10$ are at least $0$.\n9. Use the [[limit of a product|0734RSVF2V]] with constant sequences and the [[limit of a sum|0RAM4E07NR]]. The left side of step 7 tends to $10^{N+p} x - 10^N x$ as $K \\to \\infty$.\n10. The right side is the constant sequence $M$, which tends to $M$. By the [[uniqueness of limits|0ZXJ2455NJ]], $10^N (10^p - 1) x = M$.\n11. Since $p \\ge 1$, the integer $10^N (10^p - 1)$ is at least $9$, so it is nonzero. Thus $x = M / (10^N (10^p - 1))$ is a [[rational number|0NCJ5WXN6Z]].\n\n## Idea\n\nShifting the decimal point by one period moves the repeating tail onto itself. Subtracting the two shifted copies cancels the tail and leaves an integer. So a nonzero integer multiple of $x$ is an integer.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "11A63"
    ],
    "name": "Proof that repeating decimals are rational",
    "proves": "065S5B9HF1",
    "sources": [
      {
        "locator": "Every repeating or terminating decimal is a rational number",
        "url": "https://en.wikipedia.org/wiki/Repeating_decimal"
      }
    ],
    "tag": "S59QEA5Q6C",
    "uses": [
      "06XY4X9430",
      "0K0H5E3Z5N",
      "09G6KHS6GY",
      "0RAM4E07NR",
      "0734RSVF2V",
      "0ZXJ2455NJ",
      "0C4JS9RF5M",
      "0NCJ5WXN6Z"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/S59QEA5Q6C/",
  "links": [
    {
      "field": "proves",
      "target": "065S5B9HF1"
    },
    {
      "field": "uses",
      "target": "06XY4X9430"
    },
    {
      "field": "uses",
      "target": "0K0H5E3Z5N"
    },
    {
      "field": "uses",
      "target": "09G6KHS6GY"
    },
    {
      "field": "uses",
      "target": "0RAM4E07NR"
    },
    {
      "field": "uses",
      "target": "0734RSVF2V"
    },
    {
      "field": "uses",
      "target": "0ZXJ2455NJ"
    },
    {
      "field": "uses",
      "target": "0C4JS9RF5M"
    },
    {
      "field": "uses",
      "target": "0NCJ5WXN6Z"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/S59QEA5Q6C.md",
  "msc": [
    "11A63"
  ],
  "name": "Proof that repeating decimals are rational",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/S59/S59QEA5Q6C.md",
  "status": null,
  "tag": "S59QEA5Q6C",
  "uri": "m2://S59QEA5Q6C",
  "url": "https://math2.tamnd.com/wiki/proof-that-repeating-decimals-are-rational/",
  "writing": {
    "tag": "S59QEA5Q6C",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}