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It's easy to represent (via fractions) numbers like $2,34$ or $2,\overline{34}$ and even $2,3\overline{4}$. But what about $2,\overline{3}4$?

$2,3333333333333333333333333333333333....$ and I'll never be able to write the number $4$!!

Does it exist with this notation?

Is a rational number?

Notation: The top bar represents periodic numbers.

user26857
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    I think I know what you're asking and the answer is "no". No notation exists for that. It's an interesting question to ask because you would think that fractions with decimal places that never end are irrational. But the definition of a rational number is $p/q$ for $q\neq 0$ so something like $1/3$ is indeed rational. –  Aug 10 '16 at 14:47
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    The position of the $4$ is not defined, therefore a number cannot be defined this way. – Peter Aug 10 '16 at 14:49
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    @KingDuken: "fractions with decimal places that never end are irrational"??? – barak manos Aug 10 '16 at 14:49
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    @barakmanos I was explaining that his assumption for never ending decimals are irrational is interesting, even though it's not true. –  Aug 10 '16 at 14:51
  • @KingDuken: Oh, sorry, must have missed that part. – barak manos Aug 10 '16 at 14:52
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    As an aside, there are uses for generalizations of the notion of 'sequence' where you can have an infinite list of places, and then more places after all of those. However, none of those are decimals, since, by definition, all of the places in a decimal numeral are indexed by integers. –  Aug 10 '16 at 15:19
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    To expand on Hurkyl's comment, you might be interested in a brief example of that sort of number system as in http://math.stackexchange.com/a/483582/26369 – Mark S. Aug 11 '16 at 11:13

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No such number exists. There is no infinitieth digit in the decimal representation of real numbers, because the notation $$0.A_1A_2A_3\cdots$$ is a shorthand for $$\sum_{k=1}^\infty A_k10^{-k}=\sup_{n\in\Bbb N}\left(\frac{A_1}{10}+\frac{A_2}{10^2}+\cdots+\frac{A_n}{10^n}\right)$$

  • But the OP said nothing about real numbers in his question. Try to be open-minded. – Mikhail Katz Aug 10 '16 at 15:25
  • @mikhailkatz I deemed helpful stating clearly the meaning of the notation he was working with and why it does not allow what the post suggests. –  Aug 10 '16 at 19:34
  • G., what you stated clearly is your interpretation of what the OP asked. I provided a different interpretation. – Mikhail Katz Aug 11 '16 at 07:53
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If you interpret (or define) $2.\overline{3}4$ as the limit of the sequence $2.34,2.334,2.3334,2.33334,\ldots$, then

$$2.\overline{3}4=2.\overline{3}={7\over3}$$

Barry Cipra
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I'd say: yes the number does exist. It is however equal to $2.\overline{3}$. The final "4" is a limit of $\frac{4}{10^n}$ for $n \rightarrow \infty$, which is zero. The number exists the same way $0.\overline{9}$ exists.

don-joe
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Nice question! Such a number exists in the hyperreals. Simply place the digit 4 at a rank represented by an infinite hyperinteger $H$. See Elementary Calculus. The number in question is a hyperrational number.

Mikhail Katz
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  • But that would interpret $2.\bar{3}$ as notating a terminating decimal, not a periodic one! –  Aug 10 '16 at 15:21
  • @Hurkyl, the notation is ambiguous. Without a digit after it, there is no reason to assume that the 3s should stop, so you get the rational 7/3. Once a 4 appears afterwards, it is clear that the 3s stop at an infinite rank. – Mikhail Katz Aug 10 '16 at 15:23
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    Your suggestion is not well-defined. There are infinitely many possible numbers, all of which distinct, which can be interpreted as your suggestion. Not to mention that sending someone asking such a question to the hyperreals is like giving a thirsty man in the desert a Klein bottle of water. – Asaf Karagila Aug 10 '16 at 15:30
  • @Asaf I am aware of your opinions and many of them with regard to the hyperreals and needless to say I disagree with them. – Mikhail Katz Aug 11 '16 at 07:52
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    (1) You are imagining you know my opinions, and you are wrong. I do, however, think that everything has time and place, and shoving infinitesimals and the hyperreals everywhere is plain wrong. (2) Am I also wrong about your interpretation not being well defined? – Asaf Karagila Aug 11 '16 at 08:25
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    @AsafKaragila, I think the language of "shoving" should be avoided in this type of venue if minimal norms of conduct are, for a change, to be observed, though perhaps my hope is vain. – Mikhail Katz Aug 11 '16 at 08:28
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    Irrespectively of the venial/grave/mortal sins @Asaf's point (1) may or may not be a proof of, what about his point (2)? You know, sticking to the maths and all... – Did Aug 12 '16 at 11:56