# real number and decimal expansions

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)

• regarding 1, what is not rigorous about the idea ? Apr 30 '20 at 12:13
• We cannot a priori think of first decimal point before establishing that the definition is well defined. Apr 30 '20 at 12:36
• 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. Apr 30 '20 at 13:01
• Regarding your argument in the edited question, how did you conclude the first inequality? May 12 '20 at 8:37
• @Clement Yung $X-\lfloor X\rfloor<1$ for all $X$. May 12 '20 at 10:22

$$\newcommand{\bb}{\left( #1 \right)} \newcommand{\f}{\left\lfloor #1 \right\rfloor}$$ 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: $$x - \sum_{i=0}^{M'} \frac{x_i}{10^i} = x - \frac{\f{10^{M'}x}}{10^{M'}} \geq x - \frac{10^{M'}x}{10^{M'}} = 0$$

• 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. May 12 '20 at 7:56
• @Shahab I've amended my answer. May 12 '20 at 8:27
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.

• Thank you. Can you also please check the updated question? May 12 '20 at 10:53
• @Shahab: the same, in a slightly different format.
– user65203
May 12 '20 at 10:54
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.

• Can you modify your argument for (2) to a general real number? May 12 '20 at 8:06
• @Shahab I made a proper proof for the second answer. May 12 '20 at 8:19
• Thank you. Can you also please check the updated question? May 12 '20 at 8:20
• @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 $i$. May 12 '20 at 8:31
• 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? May 12 '20 at 10:31

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.