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  "body": "\n## Proof\n\n1. Let $x \\ge 0$ be real. For each natural number $k$, let $f_k = \\floor{10^k x}$, the [[floor|0YVRNEDSVG]] of $10^k x$.\n2. Let $a_0 = f_0 = \\floor{x}$. Since $0$ is an [[integer|0C4JS9RF5M]] with $0 \\le x$, we get $a_0 \\ge 0$. So $a_0$ is a natural number.\n3. For $k \\ge 1$, let $d_k = f_k - 10 f_{k-1}$. This is an integer.\n4. We show $0 \\le d_k \\le 9$. Let $u = 10^{k-1} x$ and $m = f_{k-1}$. By the [[characterization of the floor|0R9B0B5458]], $m \\le u \u003c m + 1$.\n5. Multiply by $10$: $10 m \\le 10^k x \u003c 10 m + 10$.\n6. The integer $10 m$ is at most $10^k x$, so $10 m \\le f_k$, because $f_k$ is the greatest such integer. Hence $d_k \\ge 0$.\n7. Also $f_k \\le 10^k x \u003c 10 m + 10$. The integer $f_k$ is less than the integer $10m + 10$, so $f_k \\le 10 m + 9$. Hence $d_k \\le 9$.\n8. For $K \\ge 0$, let $s_K = a_0 + \\sum_{k=1}^{K} d_k 10^{-k}$. We show $s_K = f_K / 10^K$ by induction on $K$.\n9. For $K = 0$, $s_0 = a_0 = f_0$. If $s_{K-1} = f_{K-1}/10^{K-1}$, then $s_K = f_{K-1}/10^{K-1} + (f_K - 10 f_{K-1})/10^K = f_K / 10^K$.\n10. By the characterization of the floor, $f_K \\le 10^K x \u003c f_K + 1$. Divide by $10^K$: $s_K \\le x \u003c s_K + 10^{-K}$. So $0 \\le x - s_K \u003c 10^{-K}$.\n11. By [[Bernoulli's inequality|0M1MX28VXK]] with $9$ in place of $x$, $10^K = (1 + 9)^K \\ge 1 + 9K \u003e K$ for each $K$.\n12. Let $\\varepsilon \u003e 0$. By the [[Archimedean property|0WH916TJNY]], there is $n \\ge 1$ with $1/n \u003c \\varepsilon$.\n13. For $K \\ge n$, step 11 gives $10^K \u003e K \\ge n$, so $10^{-K} \u003c 1/n \u003c \\varepsilon$. With step 10, $\\abs{x - s_K} \u003c \\varepsilon$.\n14. By the [[epsilon-N form of the limit|09G6KHS6GY]], $s_K \\to x$ as $K \\to \\infty$.\n15. The partial sums of the series $\\sum_{k \\ge 1} d_k 10^{-k}$ are $s_K - a_0$. By the [[limit of a sum|0RAM4E07NR]], with the constant sequence $-a_0$, they tend to $x - a_0$.\n16. So the series [[converges|0CVVKY7GY2]], and its [[sum|0K0H5E3Z5N]] is $x - a_0$. That is, $x = a_0 + \\sum_{k=1}^{\\infty} d_k 10^{-k}$, with digits $d_k \\in \\set{0, \\dots, 9}$ by steps 6 and 7.\n\n## Idea\n\nThe truncation of $x$ to $K$ decimal places is $\\floor{10^K x}/10^K$. Consecutive truncations differ by one digit, which lies between $0$ and $9$. The truncations are within $10^{-K}$ of $x$, so the series converges to $x$.\n",
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