The question defines $x \in \mathbb{R}$ where x>0 and a sequence of integers with $a_0: [x], a_1=[10^1(x-a_0)]$ until $a_n=[10^n(x-(a_0+10^{-1}a_1+ ... + 10^{-n}a_{n-1}))]$. I want to prove that $0 \leq a_n \leq 9$ for each $n \in \mathbb{N}$.

I am completely stumped at what to do. I feel like the Archimdean property would be useful. I also feel that a proof by induction would be useful, but I can't even think of how to solve the base case. I would really, really appreciate it if someone could help me out.

Here's my attempt at proving the base case $(0 \geq d_1 \geq 9)$ for an induction-style proof:

Since $a_0=[x]$, we can assume WLOG that $x-a_0\geq0$. Hence, this implies that $a_1\geq 0$. So that's one side of the proof done. But I'm not sure how to prove $a_1 \leq 9$.

  • $\begingroup$ prove that 0 ≤ d_n ≤ 9 That will be difficult since you don't define $d_n$ anywhere in the question. $\endgroup$ – dxiv Oct 5 '16 at 3:56
  • $\begingroup$ Unfortunately $d_n$ isn't defined anywhere in the question I have, but I am assuming $d_n$ must be an integer since we're looking for a decimal expansion and also since we're taking the floor of numbers, if that makes sense. $\endgroup$ – Nikitau Oct 5 '16 at 4:00
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    $\begingroup$ dxiv is hinting that you want to use $a_n$ in place of $d_n$ (or vice versa). $\endgroup$ – Mario Carneiro Oct 5 '16 at 4:04
  • $\begingroup$ @MarioCarneiro Oh, how silly of me. Thank you! $\endgroup$ – Nikitau Oct 5 '16 at 4:06
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    $\begingroup$ Hint: if $\;0 \le x \lt 1\;$ and $\;d = \lfloor 10 \cdot x \rfloor\;$ then $\; 0 \le d \lt 10\;$ and $\; 0 \le 10 x - d \lt 1\;$. $\endgroup$ – dxiv Oct 5 '16 at 4:20

Your intuition is perfectly correct.

Let's start with a stating and review of the archimedean property, which I will prove at the end of this post..

In $\mathbb R$

a) for $x > 0$ and $y$ there exists an $n; n \in \mathbb Z$ so that $nx > y$.

a') for $x > 0$ and $y$ there exist a unique $n \in \mathbb Z; z \in \mathbb R; 0\le z < x$ so that $y = nx + z$. Or in other words there is a unique $n$ so that $nx \le y < (n+1)x$.

b) for $x < y$ there is a $q \in \mathbb Q$ so that $x < q < y$.

and I'm going to add a third equivalent condition.

c) for $b > 1$ and $x > 0$ there exist a unique $n \in \mathbb Z$ where $b^n \le b^n < b^{n+1}$.


Okay. So $x > 0$.

So there exists a unique $n$ so that $10^n \le x < 10^{n+1}$.

There exist a unique $b_n \in \mathbb Z$ and $x_n < 10^n$ so that $x = b_n*10^n + x_n$. So $x \ge 10^n$ and $x_n < 10^n$ we know $b_n \ge 0$. As $x < 10^{n+1}$ we know $b_n < 10$.

Let $x_n = b_{n-1}*10^{n-1} + x_{n-1}$. $b_{n-1}$ might be $0$ this time around but $b_{n-1}$ by the same argument is $0\le b_{n-1} < 10$.

Apply induction indefinitely to derive $\sum_{k=n; -1}^{-\infty}b_i*10^{k}$ where $0\le b_k < 10; b_k \in \mathbb Z$.

That's it.


So proof of archemedian property:

In all of these we know $\mathbb R$ has the least upper bound/greatest lower bound property.

a) Let $S=\{xn|n \in Z\}$. If there is no $nx > y$ then $y$ is an upper bound of $S$ and $S$ is bounded above, so $m = \sup S$ exists. So $m-x$ which is less than $m$ is not an upper bound. So there exists an $n$ so that $m-x < nx < m$. So adding $x$ to that inequality yields $m < (n+1)x$. But $(n+1)x \in S$ which means $m$ is not an upper bound. This is a contradiction. So our assumption that there was no $nx > y$ was false.

Note: this also means that the $S$ is not bounded above at all as $y$ was arbitrary.

Also note: We can also prove there exists an $n$ so that $nx < y$ the same way.

a') Let $S = \{kx|k\in \mathbb Z; kx \le y\}$. By definition $S$ is bounded above by $y$. And we know by proving a) that $S$ is not empty.

Let $m = \sup S$. Then $m - x$ is not an upper bound, there is an $n$ so that $m-x < nx < m$ so $(n+1)x > m$ so $(n+1)x \not \in S$ so $(n+1)x > y$ sot $nx < y < (n+1)x$. Obviously $n$ is unique ($k < n \implies k \le n-1 \implies kx < (k+1)x < y$ and similarly $k > n \implies y < kx < (k+1)x$). Let $z = y - nx$ and ... that's it.

b) Let $v = (y-x) > 0$. So by a) there is an $n$ so that $n(y-x) > 1$. So $ny > 1+ nx$. Note: $n > 0$ as $n \le 0 \implies n(y-x) \le 0$.

By a') let $m \in \mathbb Z$ so that $(m-1)*1 \le nx < m*1$. And therefore $m \le nx + 1$.

Sooooo $nx < m \le nx + 1 < ny$. Dividing by $n$ we get $x < m/n < y$.

c) Okay, we do this much as we did a).

$b > 1$ so $n > m \iff b^n > b^m$. So let $S = \{b^n|n\in \mathbb Z\}$.

Assume $S$ is bounded above. Then let $m = \sup S$. For $.9m < m$ there exists a $10^n$ so that $.9m < 10^n \le m$. So $m < 9m < 10^{n+1}$ which contradicts $m$ being an upper bound. So $S$ is not bounded above and $x $ is not an upper bound so there is an $n+1$ so that $x < 10^{n+1}$.

Let $T = \{10^n| 10^n > x\}$. By definition $T$ is bounded below by $x$ and by the argument above $T$ is not empty. Let $m = \inf T$.

Then there is and $10^{n+1}$ so that $m < 10^{n+1} < 10m$ So $10^n < m$. So $10^n \not \in T$ so $10^n \le x \le m \le 10^{n+1}$.


Actually that was all rather long and messy.

In short: For non-empty $S = \{a_n|n \in \mathbb Z\}$ with the condition that $a_i < a_{i+1}$ then for any x; $\inf S < x < \sup S$ there is an $n$ so that $a_n \le x < a_{n+1}$. (If $S$ is not bounded above/below, let $\sup/\inf S = \pm \infty.)

Proof: Let $T = \{a_i| a_i \le x\}$ and $U = \{a_i|a_i < x\}$.

As $\inf S < x < \sup S$, x is neither an upper or lower bound of $S$ neither $T$ nor $U$ are empty. As $T$ and $U$ are countable $\sup T = \max T \in T$ and $\max T \le x$ and $ \inf U = \min U \in U$ and $min U > x$. (Simply by induction go through $n = 0,1,-1, 2,-2...$ until you find $n$ so that $a_n \in T$ and $a_{n+1} \in U$. This is doable only for countable monotonic sets.) $\max T = a_n$ for some $n$ and $\min U = a_{n+1}$.

Furthermore if $a_{i+1} - a_i \ge N > 0$ for some $N$ then $N$ is not bounded. Because: if it were then would be an $\sup S - N < a_n < \sup S < a_{n+1}$. Similar for bounded above.

Then by letting $S = \{nx\}$ and note $(i+1)x - ix = x$, the archemedian property as you know it.

By letting $S = \{b^n\}$ and noting for $n > 0$ $b^n \ge b$ then c) is shown for positive $x$.

  • $\begingroup$ Quick question, but how did you get $10^{n} \leq x < 10^{n+1}$ from the archimedian prop? I understand that by the prop that for any $x<y$ $\exists n \in \mathbb{N}$ s.t. $nx > y$. $\endgroup$ – Nikitau Oct 6 '16 at 17:53
  • $\begingroup$ Yeah, I typed and editted as I went along. It's equivalent to the archimedian prop and proven the same way but is a different statement. I'll edit my answer to reflect that. $\endgroup$ – fleablood Oct 6 '16 at 18:47
  • $\begingroup$ Wow I was not expecting this. Thank you so much! I think the proof will definitely be helpful for me for other questions. Thank you! $\endgroup$ – Nikitau Oct 7 '16 at 0:22

I can only show you the logic.

Note that all square bracket [.] means the least integer function.

Let 75.2609 be an instance of x. Then, $a_0 = [x] = 75$. Its action is equivalent to stripping off the integral part.

$x – a_0$ is to get the decimal part (i.e. .2609).

$10^1(x – a_0)$ is 10 times the above giving 2.609.

$[10^1(x – a_0)]$ is to get the corresponding integral part (2). Therefore, $a_1 = 2$.

The formula is just doing the stripping of all the decimal digits in the non-integral portion repeatedly until it ends.

Every decimal digit stripped off ($a_1, a_2, ... a_n$) can only be chosen from 0 ~ 9 (inclusive) and therefore lies within the said range.

  • $\begingroup$ Thank you. I do understand the logic, I'm just having a tough time with the formal proof of it. This was a very clear explanation though! $\endgroup$ – Nikitau Oct 5 '16 at 4:37

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