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This question already has an answer here:

Is $0.\overline{0}1$ possible in mathematical terms? I know that if you have a repeating decimal, the number is infinite and doesn't end. Thus, the 1 at the end here would stop the repeating decimal, am I right? Is this possible?

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marked as duplicate by Asaf Karagila, Najib Idrissi, Jack D'Aurizio, Willie Wong, John Gowers Jan 8 '15 at 16:36

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    $\begingroup$ If such a number existed, you would have to say it equals zero, since such a number $x$ would be equal to $10x$ and for $x=10x$, $x=0$ $\endgroup$ – turkeyhundt Jan 8 '15 at 3:53
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    $\begingroup$ No. It's a bit like the line in The Phantom Tollbooth, where the old man is describing how to find the largest number. "Follow this line forever, then turn left." $\endgroup$ – Thomas Andrews Jan 8 '15 at 3:54
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    $\begingroup$ In a word, no. Surreal numbers or hyperreals are usually what everyone points to when this question gets asked. They formalize a notion of being "infinitely" close to zero without being zero, but they don't do it like that. $\endgroup$ – Zach Effman Jan 8 '15 at 3:54
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    $\begingroup$ Here's an extremely related question. $\endgroup$ – Milo Brandt Jan 8 '15 at 3:56
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    $\begingroup$ Yup. I had accidentally put the bar over the 1 before editing it $\endgroup$ – Scribblenautical Jan 8 '15 at 4:02
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The number $.\overline{0}1$ isn't possible, as multiple users have pointed out. Here is some intuition for why:

If I told you to list all of the counting numbers out from the beginning, you would start to write:

$0, 1, 2, 3, 4, 5, 6 ...$

and you would never stop, because there are infinitely many of them. To get the next one, you just take the previous one and add $1$.

What if I asked you to write all of the counting numbers out from the beginning, and then finish the list off by writing the letter $T$ at the very end? Well, you would never write the $T$ at the end of the list that I'm asking for, because you would need to finish writing all of the counting numbers out first. And you can't finish writing all of the counting numbers out because at each step, you just add one to get to the next step. There is no largest number to stop at. This is because at each number that you might think would be the largest, you just add one and that gives you the next number!

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  • $\begingroup$ You're right that we'll never rich that. But you didn't say it doesn't exist ! Suppose $\omega+1$ as an ordinal number. But Totally you're right as we can approximate real numbers in any desired accuracy $\endgroup$ – Fardad Pouran Jan 8 '15 at 6:50
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    $\begingroup$ @FardadPouran Please remember that the OP is a freshman in high school currently learning Algebra. I wanted to keep my answer at his/her level. $\endgroup$ – layman Jan 8 '15 at 15:19
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No.

However... You can define a new non-zero number – let's call it $\alpha$ – which satisfies $\alpha^2=0$, and then add it to the real numbers in a similar way that you would add $i$ to the reals to obtain the complex numbers. $\alpha$ acts like a "smallest possible quantity", and lets you be very formal and rigorous. Look up dual numbers if you're curious.

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In general, if you say a real number has, say, a $3$ at some position of the base $10$ representation, there has the be an integer that represents the position of that digit. There must be an integer $n$ so that the $3$ contributes $3\cdot 10^{-n}$ to the real number.

Now, for the $1$ in $0.\overline{0}1$. What is the position of the $1$? There is no actual integer that you can give me where this notation would allow us to say "there is a $1$ in the $n$th position." You can say "the $1$ is in the infinity position," but infinity is not an integer.

When we write $0.12\overline{34}$ we actually are talking about an infinite series:

$$\frac{1}{10^1} + \frac{2}{10^2}+ \frac{3}{10^3} + \frac{4}{10^4} + \cdots$$ where the overline now represents what we are repeating. So this can be written as:

$$\frac{1}{10^1} + \frac{2}{10^2} + \sum_{n=2}^\infty\left( \frac{3}{10^{2n-1}}+\frac{4}{10^{2n}}\right)$$

or, even more precisely:

$$\lim_{N\to\infty}\left(\frac{1}{10^1} + \frac{2}{10^2} + \sum_{n=2}^N\left( \frac{3}{10^{2n-1}}+\frac{4}{10^{2n}}\right)\right)$$

That is the rigorous interpretation of $0.12\overline{34}$. There is simply no such meaning to $0.\overline{0}1$ that is remotely useful. We don't even define what this notation means, because notation is always something we define, and we define it for usefulness.

I certainly understand the naive idea that is trying to be expressed here, but it is naive - it is not based in a rigorous understanding of the real numbers nor of notation.

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Let's say $x$ is a number between $0$ and $1$. We know that we can write $x$ out as a decimal, say $x = 0.d_1d_2d_3 \cdots$. Then, in order to describe the number $x$, all we have to do is say which number is in the tenths place, which one is in the hundredths place, which one is in the thousandth place, etc. A description of these digits determines for you exactly ONE number*.

For example, let's say you're describing a number, and I ask you about its decimal places. You say there is a $3$ in the tenths place, another $3$ in the hundredths place, again a $3$ in the thousandths place, and actually in every place there is a $3$. Since I know what each digit is, I know exactly the number you are describing: the number $0.333\cdots = \frac{1}{3}$.

For a given number $0 < x < 1$, it doesn't actually matter if I can give you a nice formula for the $k$th decimal place. For "most" numbers we can't do that. The point is that the values of the decimals $d_1, d_2, ...$ determine exactly ONE number $x$ with the property that $x = 0.d_1d_2d_3 \cdots$.

Back to the number you were describing. If I asked you "What's in the tenths place?" you'd have to say "zero." "What's in the hundreths place?" "Zero." "For any $k$, what's in the $k$th decimal place?" "Zero." Then you are describing to me the number $0.000 \cdots$, which cannot be anything except the number $0$.

The idea of "an infinite number of zeros followed by a one at the end" doesn't make sense because the way a decimal expansion works: if there IS a $1$ in that expansion somewhere, then there should exist a number $k$ such that $1$ is in the $k$th decimal place.

*We said that a given decimal expansion $0.d_1d_2 \cdots$ represents exactly one number. On the other hand, there is no reason why a number cannot be represented by more than one decimal. Indeed, $0.1000 \cdots$ and $0.0999\cdots$ are two decimal expansions which represent the same number: $\frac{1}{10}$. This is a point of confusion for a lot of people, but there's nothing wrong with it logically. Think of how we use nouns in our language to represent things in our world. Each word stands for exactly one thing (let's pretend things like homophones don't exist), but that doesn't mean that given some thing, there is only one word to describe it.

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For "decimal" expansions of real numbers, the places right of the decimal have order type $\omega$. Order type $\omega$ has no last element.

If you like, you can try defining some new numbers, allowing decimal expansions of order-type $\omega+1$. See what strange properties those numbers will have! It could be fun. See that addition is not always possible! Multiplication is weird!

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