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Recently I stumbled across a, to me, rather strange idea. I was messing around with the proof of $0.999... = 1$, when I figured that what $0.999...$ means is that those are all nines. That way I came upon a weird idea. Say $a = 0.999...$, then $a = 1 - x$. Yet, how do we define what $x$ is? It should say $x = 0$, but my theory did not. If $0.999...$ is just an endless sequence of nines, then why can't we say $x$ just is an endless sequence of zeroes, ending with a 1, like $\frac{1}{10^\infty}$? If we take the equation $$n = 0.9$$ for example, then, what would $1 - n$ be? Yes indeed, $0.1$. Following that theory, can't we say that $$0.999... = 1 - \frac{1}{10^\infty}$$Now, if $0.999... = 1$ this would be impossible. Go figure. $$1 = 1 - \frac{1}{10^\infty}$$ then $$\frac{1}{10^\infty} = 0$$ but that is impossible, because we cannot say $$10^\infty * 0 = 1$$When I came to this point, I really got stuck, because, in my head everything I did was right, however, it is impossible. Can somebody please explain to me what mistakes I may have made, and enlighten me about what else I did wrong?

Thanks in advance. Sjoerd Dorrestijn.

EDT: I prefer $\frac{1}{10^\infty}$ to use as an indication of $0.000...01$, even though $10^\infty = \infty$ in some way, I just seem to find this more clear.

EDT2: Just to be clear, I read a proof that $0.999...9 = 1$ because it'd be $1 - 0.000...0 = 1 - 0 = 1$. What I tried to prove here is that it is not equal to $1$, because otherwise maths would collapse. My question was whether I am right or wrong. Since the original statement uses infinity (an infinite amount of nines) I think it is a must to use infinity as well. So, the question is if either the original statement is false, or if I made a mistake somewhere.

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marked as duplicate by Namaste, Brian Borchers, Misha Lavrov, erfink, Demophilus Nov 16 '17 at 3:11

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    $\begingroup$ Infinity is not a number. What you have written is mathematical nonsense. $\endgroup$ – Dan Rust Nov 15 '17 at 22:16
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    $\begingroup$ ... but $\lim_{n\to \infty}\frac1{10^n}=0$ $\endgroup$ – Hagen von Eitzen Nov 15 '17 at 22:19
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    $\begingroup$ If you think this is cool, wait til you learn about -1/12. $\endgroup$ – Randall Nov 15 '17 at 22:19
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    $\begingroup$ It would be worth, I think, to read this question, my answer to it, and the comments on my answer. $\endgroup$ – Asaf Karagila Nov 15 '17 at 22:21
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    $\begingroup$ @Randall One of the best smiles I've had on a maths site ever, but please don't. The misunderstandings there are out of this league. $\endgroup$ – Mark Bennet Nov 15 '17 at 22:22
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The (proven true) statement that $0.999\ldots=1$ means that $\lim_{n\to\infty}0.\underbrace{999\ldots9}_n=1$. That in turn is equivalent to $\lim_{n\to\infty}\left(1-0.\underbrace{999\ldots9}_n\right)=0$. Now note that $1-0.\underbrace{999\ldots9}_n=\frac{1}{10^n}$, so we have $\lim_{n\to\infty}\frac{1}{10^n}=0$.

However, although the above reasoning works, it is easier and more straightforward to conclude that $\lim_{n\to\infty}\frac{1}{10^n}=0$ directly from the definition of the limit, without ever considering $0.999\ldots$.

Finally, it is advisable to avoid the notation $\frac{1}{10^\infty}$ altogether, as it is not defined given the usual definition of its constituent symbols.

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  • $\begingroup$ Looking at the first line of your comment, can I conclude that $0.999...$ itself isn't even equal to $1$ at all (without the usage of limits)? That'd explain a lot. Thanks! $\endgroup$ – SuperSjoerdie Nov 15 '17 at 22:52
  • $\begingroup$ I'm not sure if I understand the question. $0.999\ldots=1$. Even if you develop a theory of infinite decimals that represent real numbers without the usage of limits -- and I've seen textbooks that do that -- you still end up with $0.999\ldots=1$. (Not surprising, since those theories are equivalent to the limit theory, by design.) Now, if you want to introduce some alternative notation for an alternative expression that evaluates to a number different from $1$ in some alternative number system... well, you can do that, but to avoid confusion, it is important not to call it $0.999\ldots$. $\endgroup$ – Chris Culter Nov 15 '17 at 23:01
  • $\begingroup$ It came from a discussion I had with my brother earlier. I did not believe the statement, but he saw a video about it or smth (numberphile posted it, I guess), and my question was if either that is false or my thoughts were false. Thanks for the answer! $\endgroup$ – SuperSjoerdie Nov 16 '17 at 0:16
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can't we say that $0.9999...={1\over 10^\infty}$

Not if we don't define our terms. Obviously $10^\infty$ is a term that needs defining, but let me go one step back - how do we define $0.999...$? Or rather, how do we define decimal notation in general?

The answer is via limits. Think of an expression like "$0.999...$" as a name for a number - it's not the number itself, it's how we write it down, and there may be other names for the same number. Decimal notation describes a number as a limit of simpler numbers: e.g. when we say "$\pi=3.14159...$" what we mean is that $\pi$ is the limit of the sequence $$3, 3.1, 3.14, 3.141, ...$$

Now, precisely defining what a "limit" is takes serious work, but let me sidestep that for the moment. What you've observed is

For each $n$, $1-0.99...99$ ($n$ many $9$s) equals $1\over 10^n$.

From this, we can argue$^*$ that the limits are the same: $$\lim_{n\rightarrow\infty} (1-0.99...99)=\lim_{n\rightarrow\infty}{1\over 10^n}.$$ Let's look at the left hand side for a moment; we can argue$^*$ that $\lim (A-B)=\lim A-\lim B$, and $\lim_{n\rightarrow\infty}1=1$ clearly$^*$ and $\lim_{n\rightarrow\infty}=0.9999...$ since that's what "$0.9999...$" means. Finally, it's not hard to show$^*$ that $\lim_{n\rightarrow\infty}{1\over 10^n}=0$. So this all translates to:

$$1-0.99999...=0.$$ This handles most of your question; the rest boils down to essentially: why can't we say $10^\infty\cdot {1\over 10^\infty}=1$? Well, you're basically asking why we can't divide by zero. Even though zero is a limit of things we can divide by (we can divide by $1$, we can divide by $1\over 10$, we can divide by $1\over 100$, ...), that doesn't mean that it itself is something we can divide by. That is, sometimes we can't swap arithmetic operations and limits: it can happen that $\lim {A\over B}$ is well-defined but $\lim A\over \lim B$ is not, and this is just something we have to be careful of - whenever you do something with limits, you have to prove that you actually can.

Finally, let me point out that notation like "${1\over 10^\infty}$" is highly discouraged, precisely because it suggests that we can manipulate $\infty$ just like a real number. And this is wildly false, and the source of many confusions.


$^*$We can argue these things, or prove these things, or these things become clear, once we've worked with limits a bit. But I want to give the big picture rather than obscure the details.

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  • $\begingroup$ I just posted an edit about why I use $\frac{1}{10^\infty}$, but thanks for the answer! $\endgroup$ – SuperSjoerdie Nov 15 '17 at 23:04
  • $\begingroup$ @SuperSjoerdie The notation "$0.000...0001$" is also misleading, since it suggests that there is a "last digit" in any sense (there isn't, and that notation is in fact meaningless). $\endgroup$ – Noah Schweber Nov 15 '17 at 23:36
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The problem you have with $10^{\infty}$ arises if we treat somewhat imprecise notation as precise mathematical statements. What people (mathematicians) usually mean when they say that $0.99\dots=1$ is that the sequence $0.9 , 0.99 , 0.999 , \dots$ approaches $1$, i.e. the limit $\lim_{n\to\infty}{\sum_{i=0}^{n}0.9\cdot 10^{-i}}$ is equal to $1$. Turning this equation around we get $\lim_{n\to\infty}(1-\frac{1}{10^n})=1$, or in other words that $\lim_{n\to\infty}{\frac{1}{10^n}}=0$. The notation $10^\infty$ supresses this limiting process and suggests that one can use this in the same way as usual real numbers. However limits only behave nicely together with multiplication and addition if they exist. The sequence $10^2, 10^3, 10^4,\dots$ does not have a limit (in the real numbers), so computing with $10^\infty$ leads to mathematical nonsense. A more imprecise way of seeing this would be to say that $10^\infty=\infty$ and $\infty\cdot 0 $ is undefined. This notation usually leads to wrong calculations and/or confusion as you have noticed yourself and should generally be avoided.

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  • $\begingroup$ What you said in the middle is true but saying that "$10^{\infty}$ is imprecise" is not true, $10^{\infty}$ is meaningless, it is important to understand that!! You can not use $\infty$ as a number, the reason we look at it as a sequence is to avoid this. This is why even if the sequence converge we will use limit and not $\infty$ $\endgroup$ – ℋolo Nov 15 '17 at 23:02
  • $\begingroup$ @Holo I just posted an edit, I used $\frac{1}{10^\infty}$ to make clear the number is 0.000....01, because it is divided by a power of 10, to avoid confusion. $\endgroup$ – SuperSjoerdie Nov 15 '17 at 23:07
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    $\begingroup$ I disagree with Holo in the sense that it is only meaningless until we give it a meaning. however if we chose to give it a meaning we have to be very very very careful to be precise what we mean when we say "$\frac{1}{{10}^\infty}$" otherwise it just leads to wrong statements. The expression $0.0000\dots 01$ does indeed make no sense at all and is further proof that one should avoid notation such as "$10^{\infty}$" like the plague. $\endgroup$ – Aufenthaltsraum Nov 15 '17 at 23:14
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What is $\infty$ really? It is not a number, you can't put it in equation and expect to get something, but what you can do is to take a number, let's say $n$ and make it grow larger and larger till it approach $\infty$(we say $\lim\limits_{n\to\infty}$ in this case). Now what happened when you take $h=\frac1n$? $h$ get smaller and smaller till it approuch $0$(we say $\lim\limits_{h\to0}$ in this case), so:

When you say $\lim\limits_{n\to\infty}\frac1n=0$ what you really have $\lim\limits_{n\to\infty}\frac1n=\lim\limits_{h\to 0}h$ and multiplying by $n$ will give you an indeterminate form.

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