[proofplan] We construct the decimal digits by a greedy recursion. At stage $n$, the digit $d_n$ is chosen so that the $n$-th partial sum stays below $x$ while making the remaining error smaller than $10^{-n}$. The partial sums then form a convergent increasing [sequence](/page/Sequence), and the error estimate forces their [limit](/page/Limit) to be exactly $x$. [/proofplan] [step:Choose each digit by a greedy remainder estimate] Let $D := \{0,1,\ldots,9\}$ denote the set of decimal digits. Define $S_0 := 0$ and $R_0 := x$. Since $0 \le x < 1$, we have $0 \le R_0 < 1$. Assume for some $n \in \mathbb{N}$ that digits $d_1,\ldots,d_{n-1} \in D$ have been chosen and that \begin{align*} S_{n-1} := \sum_{j=1}^{n-1} \frac{d_j}{10^j}, \qquad R_{n-1} := x - S_{n-1}, \qquad 0 \le R_{n-1} < \frac{1}{10^{n-1}}, \end{align*} where for $n=1$ the empty sum is $S_0=0$. Choose $d_n \in D$ maximal subject to \begin{align*} \frac{d_n}{10^n} \le R_{n-1}. \end{align*} Such a digit exists because $0 \in D$ satisfies the inequality and $D$ is finite. Define \begin{align*} S_n := S_{n-1} + \frac{d_n}{10^n} = \sum_{j=1}^{n} \frac{d_j}{10^j}, \qquad R_n := x - S_n. \end{align*} The choice of $d_n$ gives $R_n \ge 0$. To prove the upper bound, consider two cases. If $d_n \le 8$, maximality gives \begin{align*} R_{n-1} < \frac{d_n+1}{10^n}, \end{align*} and hence \begin{align*} R_n = R_{n-1} - \frac{d_n}{10^n} < \frac{1}{10^n}. \end{align*} If $d_n = 9$, the induction hypothesis gives \begin{align*} R_n = R_{n-1} - \frac{9}{10^n} < \frac{1}{10^{n-1}} - \frac{9}{10^n} = \frac{1}{10^n}. \end{align*} Thus, for every $n \in \mathbb{N}$, \begin{align*} 0 \le x - S_n < \frac{1}{10^n}. \end{align*} [guided] The goal of the construction is to keep track not only of the digits but also of the error after each chosen digit. Let $D := \{0,1,\ldots,9\}$ be the set of possible decimal digits. Define the initial partial sum and remainder by \begin{align*} S_0 := 0, \qquad R_0 := x. \end{align*} The hypothesis $0 \le x < 1$ gives \begin{align*} 0 \le R_0 < 1. \end{align*} Now suppose that $d_1,\ldots,d_{n-1} \in D$ have already been chosen. Define the current partial sum and current remainder by \begin{align*} S_{n-1} := \sum_{j=1}^{n-1} \frac{d_j}{10^j}, \qquad R_{n-1} := x - S_{n-1}. \end{align*} For $n=1$, this means $S_0=0$. The induction invariant is \begin{align*} 0 \le R_{n-1} < \frac{1}{10^{n-1}}. \end{align*} We choose $d_n$ greedily: among all digits $k \in D$ satisfying \begin{align*} \frac{k}{10^n} \le R_{n-1}, \end{align*} take the largest one. This maximum exists because the admissible set is a nonempty finite subset of $D$: it is nonempty since $0 \in D$ and $0 \le R_{n-1}$. After choosing $d_n$, define \begin{align*} S_n := S_{n-1} + \frac{d_n}{10^n}, \qquad R_n := x - S_n. \end{align*} Since $d_n$ was chosen to satisfy $d_n/10^n \le R_{n-1}$, we get \begin{align*} R_n = R_{n-1} - \frac{d_n}{10^n} \ge 0. \end{align*} It remains to prove that the new remainder is smaller than $10^{-n}$. If $d_n \le 8$, then $d_n+1$ is also a digit. Because $d_n$ was maximal, $d_n+1$ cannot satisfy the admissibility inequality, so \begin{align*} R_{n-1} < \frac{d_n+1}{10^n}. \end{align*} Subtracting $d_n/10^n$ gives \begin{align*} R_n = R_{n-1} - \frac{d_n}{10^n} < \frac{d_n+1}{10^n} - \frac{d_n}{10^n} = \frac{1}{10^n}. \end{align*} If $d_n=9$, there is no next digit in $D$, so maximality alone does not give the same comparison. Instead we use the induction hypothesis: \begin{align*} R_{n-1} < \frac{1}{10^{n-1}} = \frac{10}{10^n}. \end{align*} Therefore \begin{align*} R_n = R_{n-1} - \frac{9}{10^n} < \frac{10}{10^n} - \frac{9}{10^n} = \frac{1}{10^n}. \end{align*} In both cases, \begin{align*} 0 \le R_n < \frac{1}{10^n}. \end{align*} Since $R_n=x-S_n$, this proves that for every $n \in \mathbb{N}$, \begin{align*} 0 \le x - S_n < \frac{1}{10^n}. \end{align*} [/guided] [/step] [step:Show the greedy partial sums converge] The sequence $S:\mathbb{N}\to\mathbb{R}$ is defined by \begin{align*} S(n) := S_n = \sum_{j=1}^{n} \frac{d_j}{10^j}. \end{align*} For every $n \in \mathbb{N}$, \begin{align*} S_{n+1}-S_n = \frac{d_{n+1}}{10^{n+1}} \ge 0, \end{align*} so $(S_n)_{n=1}^{\infty}$ is increasing. The estimate from the previous step gives $S_n \le x$ for every $n$, so $(S_n)_{n=1}^{\infty}$ is bounded above. By the [Monotone Convergence Theorem for Sequences](/theorems/743), there exists $L \in \mathbb{R}$ such that \begin{align*} \lim_{n\to\infty} S_n = L. \end{align*} [/step] [step:Identify the limit with the original number] From the greedy remainder estimate, for every $n \in \mathbb{N}$, \begin{align*} 0 \le x - S_n < \frac{1}{10^n}. \end{align*} The sequence $a:\mathbb{N}\to\mathbb{R}$ defined by $a(n):=10^{-n}$ satisfies \begin{align*} \lim_{n\to\infty} \frac{1}{10^n}=0. \end{align*} Applying the [Squeeze Theorem](/theorems/627) for Sequences to the sequence $(x-S_n)_{n=1}^{\infty}$ gives \begin{align*} \lim_{n\to\infty} (x-S_n)=0. \end{align*} Since $S_n \to L$, the limit laws for sequences give \begin{align*} x-L = 0. \end{align*} Hence $L=x$. [guided] The construction gives a quantitative error bound: \begin{align*} 0 \le x - S_n < \frac{1}{10^n} \end{align*} for every $n \in \mathbb{N}$. The right-hand side tends to zero: \begin{align*} \lim_{n\to\infty} \frac{1}{10^n}=0. \end{align*} We now apply the Squeeze Theorem for Sequences to the sequence $(x-S_n)_{n=1}^{\infty}$. Its lower comparison sequence is constantly $0$, its upper comparison sequence is $10^{-n}$, and both comparison sequences converge to $0$. Therefore \begin{align*} \lim_{n\to\infty} (x-S_n)=0. \end{align*} From the previous step, there is a real number $L$ such that \begin{align*} \lim_{n\to\infty} S_n=L. \end{align*} Using the limit law for the difference of convergent sequences, we obtain \begin{align*} 0 = \lim_{n\to\infty} (x-S_n) = x - \lim_{n\to\infty} S_n = x-L. \end{align*} Thus $L=x$. The reason this completes the identification is that the partial sums were already known to converge; the error estimate forces the only possible limit to be the original number $x$. [/guided] [/step] [step:Convert convergence of partial sums into a decimal expansion] By definition of an infinite [series](/page/Series), the convergence of the partial sums to $x$ means \begin{align*} \sum_{n=1}^{\infty} \frac{d_n}{10^n} = \lim_{n\to\infty} \sum_{j=1}^{n} \frac{d_j}{10^j} = x. \end{align*} Each $d_n$ belongs to $D=\{0,1,\ldots,9\}$ by construction. Therefore the digit sequence $(d_n)_{n=1}^{\infty}$ is a decimal expansion of $x$, written \begin{align*} x = 0.d_1d_2d_3\ldots. \end{align*} This proves the theorem. [/step]