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Suppose I construct a number starting 0. (followed by an infinite number of digits that are 3) (followed by an infinite number of digits that are 1), like $0.\bar{3}\bar{1}$.

Is this a real number?

Can I do calculations with it?

Is it real if I simply follow it by a single digit as in $0.\bar{3}1$

What are the limitations here in combining sets of overbar statements to create numbers?

Math Man
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2 Answers2

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There is no object matching your description. In fact, the notation "$0.\bar{3}\bar{1}$" is not defined. More specifically, the phrasing "(followed by an infinite number of digits that are 3)(followed by an infinite number of digits that are 1)" does not make sense in any formalism I've heard of. Similarly, $0.\bar{3}1$ is undefined (and again, there is not really any sensible definition for it).

For reference, we define: $$0.a_{1}a_{2}\ldots a_{n}\bar{b} = 0.a_{1}a_{2}\ldots a_{n}+\sum_{k=n}^{\infty}\frac{b}{10^{k+1}},$$ and we may give a similar definition for when $b$ is replaced by a string of digits $b_{1},b_{2},\ldots, b_{m}$: $$0.a_{1}a_{2}\ldots a_{n}\overline{b_{1}b_{2}\ldots b_{m}} = 0.a_{1}a_{2}\ldots a_{n}+\sum_{k=n}^{\infty}\frac{\sum_{i=1}^{m}10^{m-i}b_{i}}{10^{k+m}}.$$ We simply neglect to define $0.\bar{3}1$ because the concept itself makes no sense.

Will R
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  • Can't we make sense of this concept with an equation like this: $1.\bar{2}-.\bar{9}=.\bar{2}3$? When you do the subtraction each digit pulls a carry from the left except the very last one infinitely far away. – Math Man Feb 24 '16 at 17:40
  • @MathMan: The idea of "infinitely far away" is imprecise, and there is no "very last digit". Ultimately these words make no sense together. When you have infinitely many digits, you cannot have both a "first" digit and a "last" digit. As for that "equation", probably the easiest rigorous way to perform a calculation like the one indicated, besides expressing everything in terms of rational numbers and doing the calculation that way (which is what I would advocate in any other situation), is to notice that $0.\bar{9}=1,$ and so $1.\bar{2}-0.\bar{9} = 1.\bar{2}-1 = 0.\bar{2}$. – Will R Feb 24 '16 at 19:36
  • Interesting thoughts. "When you have infinitely many digits, you cannot have both a 'first' digit and a 'last' digit." Why not? The reals from 0 - 1 are an infinite set with a first and last element? Second, How do we know $0.\bar{9}=1$? Consider the difference $0.\bar{0}1$? – Math Man Feb 24 '16 at 21:26
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If your number existed, it could be defined as

$\begin{array}\\ \lim_{n \to \infty} \left( \sum_{k=0}^n 3\cdot 10^{-k} + \sum_{k=0}^n 10^{-k-n-1} \right) &=\lim_{n \to \infty} \left( 3\sum_{k=0}^n \cdot 10^{-k} + 10^{-n-1}\sum_{k=0}^n 10^{-k} \right)\\ &=\lim_{n \to \infty} (3+10^{-n-1})\sum_{k=0}^n \cdot 10^{-k}\\ &=\lim_{n \to \infty} (3+10^{-n-1})\frac{1-10^{-n-1}}{9}\\ &=\lim_{n \to \infty} (\frac13 -\frac29 10^{-n-1}+\frac19 10^{-2n-2})\\ &= \frac13\ \end{array} $

which is the same as $\lim_{n \to \infty} \sum_{k=0}^n 3\cdot 10^{-k} = \frac13 $.

marty cohen
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    Which eliminates all meaning to the infinite ones (which never get a chance to startl. Hmmm.... I like this answer. Like it a lot, but it could be misleading to the OP as implying the concept is legitimate. – fleablood Feb 24 '16 at 03:29
  • From the answers an comments I'm understanding that mathematicians consider a number of this form to be "not the description of any real number" "not defined", and "does not make sense". You also offer an interesting proof that if it were defined it would be equal to $1/3$. All that makes sense. Isn't it possible to get numbers like this as the solution to some equations though? Consider $1.\bar{2}-.\bar{9}=.\bar{2}3$ or $1-.\bar{9}=.\bar{0}1$. – Math Man Feb 24 '16 at 17:58
  • Nope. Once you have an infinite number of digits, there can be nothing following. If anything followed, the sequence of digits would have had to come to an end and therefore could not be infinite. – marty cohen Feb 24 '16 at 23:06