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Essentially, I am wondering if you can represent a repeating decimal and have a number. For example, perhaps 0.9 repeating with an 8 at the end, the largest decimal less than one (because 0.9 repeating is equal to one.) Logic would dictate that you would have 0.98 with a bar only above the 9. However, I do not know if a number like this exists and if so, how I can represent it. Sorry for my lack of formatting skills.

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  • $\begingroup$ Do you mean a limited number of repeated $9$s followed by an $8$, or an infinite number of $9$ digits followed by an $8$? In the latter case, such a number is not particularly useful to talk about, since it is not possible to use the $8$ digit... $\endgroup$ – abiessu Jul 15 '14 at 17:33
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    $\begingroup$ Separately, there is no "largest real number less than one"... $\endgroup$ – abiessu Jul 15 '14 at 17:34
  • $\begingroup$ A similar question $\endgroup$ – abiessu Jul 15 '14 at 17:35
  • $\begingroup$ I think this particular question has come up more than once before. The idea of an infinite sequence of 9s followed by an 8 is coherent by itself, but then you can't turn it into a usable number system. $\endgroup$ – MJD Jul 15 '14 at 17:41
  • $\begingroup$ My question was similar, but worded differently. $\endgroup$ – Erisin Jul 15 '14 at 17:43
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$$.99999.....8=\lim_{n \to \infty}\left( \sum_{k=1}^n {9\over 10^k} \right)+{8\over 10^{n+1}}$$ $$=\lim_{n \to \infty} 1-10^{-n}+.8*10^{-n}$$ $$=\lim_{n \to \infty} 1-.2*10^{-n}$$ So $.99999...8$ is actually five times closer to one than $.9999...$ (yes, I'm aware that I'm being inconsistent with my infinities, but the result is the same) and therefore if the second is equal to one, so is the first. Of course, you can argue about whether $.9999...$ really is $1$, but I don't want to get into that.

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  • $\begingroup$ This just went over my head. $\endgroup$ – Erisin Jul 15 '14 at 17:45
  • $\begingroup$ the limit as n approaches infinity of $10^{-n}$ is 0. So your method tells you .99999.... 8 is exactly the same as .999999...9 $\endgroup$ – Jorge Fernández Hidalgo Jul 15 '14 at 17:49
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    $\begingroup$ @Bananarama: Yes, exactly, but we can see that without taking the limit, which is more clarifying. I was trying to evaluate the whole expression without taking the limit because if you actually take the limit of the whole expression, it evaluates to $1$, which is an uninteresting and unhelpful result. However, if we find the value of this expression in terms of $n$ rather than evaluating as it goes to infinity, we can see that it is, in fact, just as useless an expression as we initially thought it was. $\endgroup$ – Laertes Jul 15 '14 at 17:51
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There is not such thing as the real number closest to another real number, this is because suppose $a'$ is the number closest to $a$. Then define $a''=\frac{a'+a}{2}$ then $a''$ is "between" $a$ and $a''$ so $a'$ was not the closest number to $a$.

Since there is no number "between" $.\overline{9}$ and $1$ we have $1=.\overline{9}$

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  • $\begingroup$ I thought it was because we could multiply the whole thing (.9 repeating is equal to x) by 10, get 9.9 repeating = 10x, subtract x and .9 repeating from both sides and get 9 = 9x, and divide both sides by 9 and get x=1, thus 0.9 repeating = 1 $\endgroup$ – Erisin Jul 15 '14 at 18:01
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$0.\bar 98$ is nonsense, since $0.\bar 9$ informs us that $9$, and $9$ only repeats indefinitely (infinitely), and hence does not change to another digit, ever. In order to have an $8$ tagged on to a series of $9$, the series of nines would have to be finite (terminate).

With respect referencing "the largest decimal less than $1$," there is no such number.

If we write $$0.99999999999998$$ and conclude that is the largest decimal $\lt 1$, I can "come back at you" to offer the counterexample $$0.999999999999998$$ Then $$0.9999999999999998$$ And so on, each new number greater than the previous.

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When you write $.\bar 9$ you are indicating that there is nothing more to the decimal expansion than $.99999999...$. Placing anything after the bar doesn't make sense with how the repeating part is defined. Particularly, this is because it does not indicate where the "8" will go.

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We can use the exact same proof as the one for $0.\bar{9}=1$ to prove that $0.\bar{9}8=1$. (Except that $0.\bar{9}8$ doesn't exist.

The simple reason that $0.\bar{9}8$ doesn't exist: The infinite sequence of decimals is too long to be a number.

The more advanced reason: Real numbers can be defined as a sequence of integers optionally including one decimal point. More specifically these sequences are of ordinal length $\omega$. The sequence of integers given by $0.\bar{9}8$ is of length $\omega +1$ and is thus not a real number. For more info see http://en.wikipedia.org/wiki/Ordinal_number

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  • $\begingroup$ You could simply say that in this number, all digits after the decimal point up to digit $\omega-1$ are equal to 9, and digit $\omega$ is equal to 8. $\endgroup$ – Laertes Jul 15 '14 at 18:07
  • $\begingroup$ Read through the link I provided and I think you will become confident that $\omega -1$ does not exist. $\endgroup$ – Dylan Stephano-Shachter Jul 15 '14 at 18:12
  • $\begingroup$ Point taken; after reading through I modify my objection to say that although your answer is correct, demonstrating that $0.\bar 9 8 \not \in \mathbb R$ is not equivalent to saying that it does not exist. $\endgroup$ – Laertes Jul 15 '14 at 18:19
  • $\begingroup$ Well it does exist, it is just not really a number. It is just the sequence 0,'.',9,9,9,...,9,8. $\endgroup$ – Dylan Stephano-Shachter Jul 15 '14 at 18:35
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Contrary to some of the answers in this thread, your idea of an infinite sequence of 9s followed by an 8 is perfectly coherent. Normally when we consider infinite sequences, we imagine the positions in the sequence indexed by positive integers, and we that each infinite decimal is associated with a function that takes one of these positions and tells us the element of the decimal at that position. For a number like $0.14159\ldots$ we have $f(1) = 1, f(2) = 4, f(3) = 1, f(4) = 5, $ and so on. The function $f$ is defined on the domain $\omega=\{1, 2, 3, 4, \ldots\}$.

There is indeed a coherent notion of $\omega + 1 = \{1, 2, 3, \ldots , \omega\}$, which is just like $\omega$ itself, but with one extra element on the end. We could consider an “infinity-plus-one” decimal which is associated with a function $f$ that takes a position, which is now an element of $\omega+1$, and tells you the decimal digit at that position. Such sequence have an infinite number of elements, and then one extra element at the end.

The problem is that these sequences do not behave very much like numbers. With regular decimals, we can associate the sequence defined by $f$ with the real number $\frac1{10}f(1) + \frac1{100}f(2) + \frac1{1000}f(3) + \ldots$ and then perform arithmetic operations on the sequences in the usual way. The important operations are $+, \times, $ and $<$.

But there doesn't seem to be any reasonable way to extend the meanings of $+, \times, $ and $<$ to apply to the $\omega+1$ sequences. There is one problem after another. Here is one especially simple example: Consider the number $\epsilon = 0.\bar01$. What is $\epsilon \div 2$?

Here is another example: It might be possible to define $\times$ so that $\epsilon \times 9 = 0.\bar09$. Then $\epsilon + \epsilon\times 9$ ought to be equal to $\epsilon \times 10$, but what do you get when you do that? There should be a carry into the position just left of the last position, but there is no position just left of the last position!

You can find all sorts of similar perplexities. So we don't do this, because the objects we get can't be made to behave like numbers.

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  • $\begingroup$ How does it follow that, given an infinite sequence with two endpoints, the places second from the end and so forth don't exist? The decimal places second from the beginning and so forth certainly do. $\endgroup$ – Laertes Jul 15 '14 at 18:09
  • $\begingroup$ If the sequence is indexed by $\{1, 2, 3, \ldots, \omega\}$, then every position is either an ordinary number, one of $1, 2, 3, \ldots$, or else it's $\omega$. The position $\omega$ is not the next-to-last because it is the very last. None of the positions $1, 2, 3, \ldots $ is next-to-last either. (You tell me why not.) So none of them are next-to-last. $\endgroup$ – MJD Jul 15 '14 at 18:13
  • $\begingroup$ That proves only that the next-to-last digit does not have a finite or ordinal index. Granted I can't think of what form that index would be, but this doesn't prove that it doesn't exist. For that matter, it could be inexpressable and still exist. $\endgroup$ – Laertes Jul 15 '14 at 18:23
  • $\begingroup$ Every element of the sequence has an index in $\{1, 2, 3, \ldots, \omega \}$ because that's the sequence I said I was discussing. $\endgroup$ – MJD Jul 15 '14 at 18:31
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I've seen explanations of number systems in which the decimal point is followed by infinitely many digits, which in turn are followed by more digits, but (aside from not understanding these systems well enough to say anything useful about them) I don't think I've ever seen a number system in which there is both an infinite number of digits after the decimal point and a final digit.

If you can write $0.\bar{9}8$ and say it is somehow meaningful, why can't I write $0.\bar{9}81$ and claim that $0.\bar{9}81 > 0.\bar{9}8$?

I don't think there's any notation that will do quite what you want. If $x < 1$ then $x < x + \frac12(1 - x) < 1$, so the only way to have a "greatest decimal less than one" is to posit that there are decimals whose sum or difference isn't a decimal or where half the decimal isn't another decimal. In other words some operation that is closed over all the decimals we normally talk about (including infinite decimals) would not be closed over your decimals, which seems to defeat the purpose of trying to define such a numbering system.

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