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For any real number $x$ we define its decimal expansion as $N\cdot x_1x_2x_2\cdots$ where $N=\lfloor x\rfloor$ and $$x_i=\left\lfloor 10^i \left(x- \left(N+\sum_{j=1}^{i-1}\frac{x_j}{10^j}\right) \right)\right\rfloor.$$

Now I have two questions regarding this definition:

  1. Why will each $x_k$ be a digit between $0$ and $9$? That is clear in the case of $x_1$ since $x-N$ being the fractional part of $x$ will be in $[0,1)$ and so $10(x-N)\in[0,10)$. In the case of $x_2$ it is not so clear. Intuitively, if from the fractional part we subtract one tenth's of the "first digit decimal point" so we must getting something like $0.0x_2x_3\cdots$ and hence multiplying by $100$ (and taking floor) is the correct thing to do, to recover $x_2$. However I cannot seem to make this idea rigorous.

  2. Why can't the decimal expansion end in a string of $9's$? I think if we presumed that it did then, after some $k$ the difference between $x$ and $N.x_1\cdots x_k$ would be zero. That will be a contradiction because clearly each $x_i$ is unique. But how to justify that such a difference ultimately becomes zero?


Update: The answers posted below both use induction to prove (1). Is it correct to do it without induction as follows: Suppose $i\ge 3$ (the cases $i=1,2$ being similar). Now, $$10^{i-1}\left(x-\left(N+\sum_{j=1}^{i-2}\frac{x_j}{10^j}\right)\right)<1+x_{i-1}$$ by definition of the floor function. Hence $10^{i}(x-(N+\sum_{j=1}^{i-1}\frac{x_j}{10^j}))<10$ and so $x_i\le 9$. Similarly, since $$10^{i-1} \left(x- \left(N+\sum_{j=1}^{i-2}\frac{x_j}{10^j} \right)\right)\ge x_{i-1}$$ so $10^{i}\left(x-\left(N+\sum_{j=1}^{i-1}\frac{x_j}{10^j}\right)\right)\ge 0$ following which $x_i\ge 0$.

Thank you.

(Just to clarify bounty will be given to the best posted answer, even if above is correct)

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  • $\begingroup$ regarding 1, what is not rigorous about the idea ? $\endgroup$
    – jimjim
    Apr 30 '20 at 12:13
  • $\begingroup$ We cannot a priori think of first decimal point before establishing that the definition is well defined. $\endgroup$
    – Shahab
    Apr 30 '20 at 12:36
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    $\begingroup$ Although conceptually you are indeed subtracting the previous decimals, and multiplying by a power of 10 to shift the next decimal before the decimal point, you don't really need the concept of a decimal point to prove it. The only thing you need is that $r-\lfloor r \rfloor$ is a real number in $[0,1)$ for any $r$, that is to say that you can split any real into an integer and a fractional part. If you have that, then you can prove it using induction. Note that the concept of a fractional part of a real is not dependent on the representation of a real in decimal notation. $\endgroup$ Apr 30 '20 at 13:01
  • $\begingroup$ Regarding your argument in the edited question, how did you conclude the first inequality? $\endgroup$ May 12 '20 at 8:37
  • $\begingroup$ @Clement Yung $X-\lfloor X\rfloor<1$ for all $X$. $\endgroup$
    – Shahab
    May 12 '20 at 10:22
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$$ \newcommand{\bb}[1]{\left( #1 \right)} \newcommand{\f}[1]{\left\lfloor #1 \right\rfloor} $$ Write $x_0 := N$. Note that your expression becomes the following: $$ x_i = \f{10^i\bb{x - \sum_{j=0}^{i-1}\frac{x_j}{10^j}}} $$


For (1), we can make use of the following lemma:

Lemma: For any $k \in \mathbb{N}$, we have: $$ \sum_{i=0}^k \frac{x_i}{10^i} = \frac{\f{10^{k}x}}{10^{k}} $$

Proof. We prove by induction. The case is clear for $k = 0$, as by definition $x_0 = \f{x}$. Now suppose $\sum_{i=0}^k \frac{x_i}{10^i} = \frac{\f{10^kx}}{10^k}$. Then: \begin{align*} \sum_{i=0}^{k+1} \frac{x_i}{10^i} &= \frac{\f{10^kx}}{10^k} + \frac{x_{k+1}}{10^{k+1}} \\ &= \frac{\f{10^kx}}{10^k} + \frac{1}{10^{k+1}}\f{10^{k+1}\bb{x - \sum_{j=0}^{k}\frac{x_j}{10^j}}} \\ &= \frac{\f{10^kx}}{10^k} + \frac{1}{10^{k+1}}\f{10^{k+1}x - 10^{k+1}\frac{\f{10^kx}}{10^k}} \\ &= \frac{\f{10^kx}}{10^k} + \frac{1}{10^{k+1}}\f{10^{k+1}x - \underbrace{10\f{10^kx}}_\text{integer}} \\ &= \frac{\f{10^kx}}{10^k} + \frac{1}{10^{k+1}}\bb{\f{10^{k+1}x} - 10\f{10^kx}} \\ &= \frac{\f{10^kx}}{10^k} + \frac{\f{10^{k+1}x}}{10^{k+1}} - \frac{\f{10^kx}}{10^k} \\ &= \frac{\f{10^{k+1}x}}{10^{k+1}} \end{align*} Now, it's simple to prove that $0 \leq x_i \leq 9$. We observe that: \begin{align*} x_i = \f{10^ix - 10^i\frac{\f{10^{i-1}x}}{10^{i-1}}} = \f{10^ix - 10\f{10^{i-1}x}} = \f{10\bb{10^{i-1}x - \f{10^{i-1}x}}} \end{align*} We know that for any integer $n$, $0 \leq n - \f{n} < 1$. Thus: \begin{align*} 0 \leq 10^{i-1}x - \f{10^{i-1}x} < 1 &\implies 0 \leq 10\bb{10^{i-1}x - \f{10^{i-1}x}} < 10 \\ &\implies 0 \leq \f{10\bb{10^{i-1}x - \f{10^{i-1}x}}} \leq 9 \end{align*} So $0 \leq x_i \leq 9$.


For (2), we shall show that there is no $M \in \mathbb{Z}^+$ such that for $i > M$, $x_i = 9$. Suppose such an $M \geq 1$ exists, and suppose $x_{M} = n$. We observe that for $M' > M$: \begin{align*} 10^{M}\bb{x - \sum_{j=0}^{M-1} \frac{x_j}{10^j}} - (n + 1) &= 10^{M}\bb{x - \sum_{j=0}^{M-1} \frac{x_j}{10^j}} - 1 - n \\ &\geq^* 10^{M}\bb{\sum_{j=0}^{M'}\frac{x_j}{10^j} - \sum_{j=0}^{M-1} \frac{x_j}{10^j}} - 1 - n \\ &= 10^M\sum_{j=M}^{M'} \frac{x_j}{10^j} - 1 - n\\ &= 10^M\sum_{j=M+1}^{M'} \frac{x_j}{10^j} - 1 \\ &= 10^M\sum_{j=M+1}^{M'} \frac{9}{10^j} - 1 \\ &= 10^M\frac{\frac{9}{10^{M+1}}\bb{1 - \frac{1}{10^{M' - M}}}}{1 - \frac{1}{10}} - 1\\ &= - \frac{1}{10^{M' - M}} \end{align*} We can let $M' \to +\infty$, and we have that $10^{M}\bb{x - \sum_{j=1}^{M-1} \frac{x_j}{10^j}} - (n + 1) \geq 0$. Thus: $$ \f{10^{M}\bb{x - \sum_{j=0}^{M-1} \frac{x_j}{10^j}} - (n + 1)} \geq 0 \implies x_M \geq n + 1 $$ which contradicts that $x_M = n$. Note that the starred inequality can be easily proven as follows: $$ x - \sum_{i=0}^{M'} \frac{x_i}{10^i} = x - \frac{\f{10^{M'}x}}{10^{M'}} \geq x - \frac{10^{M'}x}{10^{M'}} = 0 $$

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  • $\begingroup$ Regarding (2), the definition of $x_i$ in this question is such that the decimal expansion is unique, (as each $x_i$ is obviously uniquely defined). Fixing this definition, it is not obvious to me why this particular definition does not permit a string of terminating $9$'s. $\endgroup$
    – Shahab
    May 12 '20 at 7:56
  • $\begingroup$ @Shahab I've amended my answer. $\endgroup$ May 12 '20 at 8:27
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  1. WLOG, $N=0$ (you can rescale $x$), and $$0\le(x-0.)<1$$ starts the induction. Then $$0\le10^n(x-0.x_1x_2\cdots x_n)<1\implies0\le10^{n+1}(x-0.x_1x_2\cdots x_n)<10$$ so that taking the floor, the next digit is one of $0,1,\cdots 9$. And in turn $$0\le10^{n+1}(x-0.x_1x_2\cdots x_nx_{n+1})<1$$ because this is the fractional part of $10^{n+1}(x-0.x_1x_2\cdots x_nx_{n+1})$, i.e. what remains of a number after you removed the integer part.

  2. applying this definition, you will never get an infinite repetition of $9$, because such repetitions tend to a number with a finite expansion ($0.234999\cdots=0.234\bar9=0.235$), and by the definition, the computed digits will be zeroes, not nines.

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  • $\begingroup$ Thank you. Can you also please check the updated question? $\endgroup$
    – Shahab
    May 12 '20 at 10:53
  • $\begingroup$ @Shahab: the same, in a slightly different format. $\endgroup$
    – user65203
    May 12 '20 at 10:54
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  1. Your attempt it correct. Alternatively, you can proove it by induction. Let's proove that $x_i$ is a digit between 0 and 9 and if you remove first $i$ digits (i.e. $x-\bigl(N+\sum_{j=1}^{i}\frac{x_j}{10^j}\bigr)$), you get something in the interval $[0,10^{-i})$. The basic step for $i=1$ is clear, you already said it. The induction step looks like this: Let's suppose that it holds for $n\in\Bbb{N}$. Then $x-\bigl(N+\sum_{j=1}^{i}\frac{x_j}{10^j}\bigr)$ is from the inductional propostition in the interval $[0,10^{-i})$ Therefore $10^{i+1}\bigl(x-\bigl(N+\sum_{j=1}^{i}\frac{x_j}{10^j}\bigr)\bigr)\in[0,10)$ and $\left\lfloor10^{i+1}\bigl(x-\bigl(N+\sum_{j=1}^{i}\frac{x_j}{10^j}\bigr)\bigr)\right\rfloor = x_{i+1}$ is a digit between 0 and 9. Also holds: $$ \left\lfloor10^{i+1}\bigl(x-\bigl(N+\sum_{j=1}^{i}\frac{x_j}{10^j}\bigr)\bigr)\right\rfloor \leq 10^{i+1}\bigl(x-\bigl(N+\sum_{j=1}^{i}\frac{x_j}{10^j}\bigr)\bigr) $$ so $$ 10^{-i-1}\left\lfloor10^{i+1}\bigl(x-\bigl(N+\sum_{j=1}^{i}\frac{x_j}{10^j}\bigr)\bigr)\right\rfloor \leq \bigl(x-\bigl(N+\sum_{j=1}^{i}\frac{x_j}{10^j}\bigr)\bigr) \\ 10^{-i-1}x_{i+1} \leq \bigl(x-\bigl(N+\sum_{j=1}^{i}\frac{x_j}{10^j}\bigr)\bigr) \\ \bigl(x-\bigl(N+\sum_{j=1}^{i}\frac{x_j}{10^j}\bigr)\bigr) - 10^{-i-1}x_{i+1} \geq 0 \\ \bigl(x-\bigl(N+\sum_{j=1}^{i+1}\frac{x_j}{10^j}\bigr)\bigr) \geq 0 $$ And also: $$ 10^{i+1}\bigl(x-\bigl(N+\sum_{j=1}^{i}\frac{x_j}{10^j}\bigr)\bigr) - \left\lfloor10^{i+1}\bigl(x-\bigl(N+\sum_{j=1}^{i}\frac{x_j}{10^j}\bigr)\bigr)\right\rfloor < 1 \\ 10^{i+1}\bigl(x-\bigl(N+\sum_{j=1}^{i}\frac{x_j}{10^j}\bigr)\bigr) - x_{i+1} < 1 \\ \bigl(x-\bigl(N+\sum_{j=1}^{i}\frac{x_j}{10^j}\bigr)\bigr) - 10^{-i-1}x_{i+1} < 10^{-i-1} \\ \bigl(x-\bigl(N+\sum_{j=1}^{i+1}\frac{x_j}{10^j}\bigr)\bigr) < 10^{-i-1} $$ This means that $x_{i+1}$ is a digit between 0 and 9 and $\bigl(x-\bigl(N+\sum_{j=1}^{i+1}\frac{x_j}{10^j}\bigr)\bigr) \in [0,10^{-n-1})$. Therefore the statement holds for any $i\in\Bbb{N}$. QED

  2. Proof by contradiction. Let's suppose that there is an $i\in\Bbb{N}$ such that all digits starting by $x_i$ are nines. Then you have: $$ x_i=9\\ x-\bigl(N+\sum_{j=1}^{i}\frac{x_j}{10^j}\bigr)=0.\bar{9}\cdot10^{-i}=10^{-i} $$ However, $10^{-1} \notin [0,10^{-i})$ And this is a contradiction with the statement in the answer to question 1.

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  • $\begingroup$ Can you modify your argument for (2) to a general real number? $\endgroup$
    – Shahab
    May 12 '20 at 8:06
  • $\begingroup$ @Shahab I made a proper proof for the second answer. $\endgroup$
    – Hume2
    May 12 '20 at 8:19
  • $\begingroup$ Thank you. Can you also please check the updated question? $\endgroup$
    – Shahab
    May 12 '20 at 8:20
  • $\begingroup$ @Shahab Your attempt is also an induction in a way. You start at $i=1,2$ and then you apply the same rules to prove it for higher $i$. $\endgroup$
    – Hume2
    May 12 '20 at 8:31
  • $\begingroup$ Technically I don't use induction. The cases $i=1,2$ have a similar proof but they stand alone as does the case $i>2$. My question is: is this correct? $\endgroup$
    – Shahab
    May 12 '20 at 10:31
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Proof by example:

Let the number be $\pi$.

At a certain stage of the discovery of the decimals, assume that we have established

$$0\le10^5(\pi-3.14159)<1.$$

This implies

$$0\le10^6(\pi-3.14159)<10$$ and it turns out that

$$\lfloor10^6(\pi-3.14159)\rfloor=2.$$

Due to the bracketing in $[0,10)$, the new digit is perforce one of $0,1,\cdots9$.

Then by subtracting $2$ we get the fractional part of the above number, such that

$$0\le10^6(\pi-3.14159)-2=10^6(\pi-3.141592)<1$$ and we can iterate.

The process is started with $0\le\pi-3<1.$

This is indeed an inductive proof, as you can replace the concrete decimals by variables throughout.


For the second question, notice that you will never get an infinite representation like $3.1415\bar9$ because the series is numerically equal to the number $3.1416$, with the representation $3.1416\bar0$. Hence uniqueness is guaranteed.

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