Any way to represent number that comes after repeating decimal?

Question

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.

Answer 1

$$.99999.....8=\lim_{n \to \infty}\left( \sum_{k=1}^n {9\over 10^k} \right)+{8\over 10^{n+1}}$$ $$=\lim_{n \to \infty} 1-10^{-n}+.8*10^{-n}$$ $$=\lim_{n \to \infty} 1-.2*10^{-n}$$ 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.

Answer 2

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 $1=.\overline{9}$

Answer 3

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.

Answer 4

$0.\bar 98$ 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.

Answer 5

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

Answer 6

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.

Answer 7

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.