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Is there any way to make sense of expressions with infinitely many digits to the left of the decimal point?

There's a famous proof that $0.999\dots=1$ which starts with $x=0.999\dots$ and derives $10x=x+9$, therefore $x=1$. We can do the same thing with an infinite string of $9$'s to the right of the decimal point: Assume $x=9999\dots99$

Then $10x=9999\dots990$

Then $10x+9=9999\dots999=x$

Finally we get $10x+9=x$

So $x=-1$
$\qquad$ My question is: Is it possible to make sense of the expressions ''$99999\dots9$'' and ''$99999\dots90$'' and the algebraic manipulations above? If not, why does this work for $x=0.999\dots$ and not for $x=999\dots$?

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    $\begingroup$ Pay attention. You can't work with $9999...$ the same way you work with $0.9999...$ $\endgroup$
    – Or Shahar
    Jul 15, 2022 at 14:03
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    $\begingroup$ $99999$$\cdots$ looks as if it is bigger than any positive integer $\endgroup$
    – Henry
    Jul 15, 2022 at 14:03
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    $\begingroup$ No, we can't... $\endgroup$ Jul 15, 2022 at 14:06
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    $\begingroup$ The sequence of infinitely many 9's does not mean anything at all. Whenever you deal with infinitely long expressions you are really dealing with limits. That's how you define the meaning of $0.999\ldots$. $\endgroup$ Jul 15, 2022 at 14:09
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    $\begingroup$ Note that $0.999...$ is a sum of the form $\Sigma_{k=1}^n 9 \cdot 10^{-k}$, which converges while $999....$ would be the sum $\Sigma_{k=1}^n 9 \cdot 10^{k}$ which diverges to positive infinity. $\endgroup$ Jul 15, 2022 at 14:09

2 Answers 2

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The problem with these equations is the "$\dots$". In infinite expressions such as the infamous $0.9999\dots$, the ellipsis represents a limit. In the decimal case, what it means is the limit of the sequence $$ 0.9, 0.99, 0.999, \text{ and so on} $$ which you can prove converges in various ways, for example it is trivially increasing, and is bounded above by $1$. The manipulations that follow are valid assuming the limit exists. Let $x=0.9999\dots$ (i.e., the limit of the above sequence). Then:\begin{eqnarray} x &=& 0.9999\dots\\ 10x&=&9.9999\dots\\ 10x&=&9+x\\ x &=& 1 \end{eqnarray} Now, we turn to your expression $999\dots$. The ellipsis here would indicate that this is meant to be the limit of the sequence $$ 9, 99, 999, \text{ and so on} $$ Of course, the limit doesn't exist under the standard metric on $\mathbb{R}$, but if you pretend you don't know that, you can do the kind of manipulations you wrote out and say that if the limit $999\dots$ exists, it is $-1$.

It looks very similar to an expression for a $p$-adic number, which are typically represented by an infinite string of base-$p$ digits for some prime $p$ (see this question and its answers for why it should be a prime). In $p$-adic spaces, you can do those sorts of manipulations you wrote down and make meaningful sense of them. Wikipedia presents the nice example of the $5$-adic $$ \dots1313132 = \frac13 $$ Indeed, you can represent any rational number as an infinite string of base $p$ digits. So, for example, in $5$-adics again, you can write $$ \dots 44444 = -1 $$ which follows from the same kind of manipulations you did: Let $x = \dots 4444$. Then (keep in mind I'm in base $5$ here):\begin{eqnarray} x &=& \dots 4444\\ 10x &=& \dots 4440\\ 10x+4&=&\dots 4444 = x\\ 4x +4&=&0\\ x &=&-1 \end{eqnarray}

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  • $\begingroup$ Really appreciate for your edit which makes it more clear... $\endgroup$
    – Andrew Li
    Jul 18, 2022 at 16:01
  • $\begingroup$ It's a big shame your question got such a negative response from some users. It is really a very natural and reasonable question. $\endgroup$ Jul 18, 2022 at 16:48
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No, you cannot express numbers that way. Technically, you could let $x=99999\dots$, but it makes absolutely no sense to write $10x = 99999\dots0$ since there are an infinite number of $9$'s before the $0$. Even if you could, solving $10x+9=x$ involves subtracting $x$ from both sides. Since $x$ is infinite, $10x+9$ is also infinite and you cannot subtract one infinity from another like that.

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    $\begingroup$ "but it makes absolutely no sense to write $10x = 99999...0$ since there are an infinite number of 9's before the 0" -- a priori there is no reason why you couldn't have numbers (or some other mathematical object) given by a sequence of digits indexed by $\omega + 1$. $\endgroup$ Jul 15, 2022 at 14:12
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    $\begingroup$ I guess what I'm trying to say is that in writing $9999\dots0$ we are assuming $99999\dots$ has an ending point which it does not since there an infinite number of $9$'s $\endgroup$
    – bobeyt6
    Jul 15, 2022 at 14:16
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    $\begingroup$ @bobeyt6, it does not imply that "999..." has an ending point, any more than writing ordinals as $0, 1, 2, 3, \ldots, \omega, \omega+1, \omega+2, \ldots, \omega \cdot 2, \ldots$ implies that there are finitely many natural numbers. $\endgroup$ Jul 15, 2022 at 14:17
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    $\begingroup$ @MeesdeVries writing $99999\dots0$ means that there is a $0$ at the end of the infinite number of $9$'s $\endgroup$
    – bobeyt6
    Jul 15, 2022 at 14:18
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    $\begingroup$ This is different since $0.999\dots$ is not infinite even if it has an infinite number of $9$'s $\endgroup$
    – bobeyt6
    Jul 15, 2022 at 14:39

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