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I have seen this problem come up many times and I was wondering if my proof is valid for $0.999\ldots=1$ where $0.999\ldots$ is continuous: $$x=0.999\ldots$$ $$10x=9.999\ldots$$ $$10x-x=9.999\ldots-0.999\ldots$$ $$9x=9$$ $$x=1$$ The reason for people struggling with this idea is because of their incapability to grasp $\infty$. If it is possible, can you agree or disagree with this proof. And in addition, create another proof regarding this same concept of $0.999\ldots=1$.

However, does this work for any other continuous equality? This question really got to me after I expored $\infty$ in a whole different way.

Thanks.

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    Many, many proofs here: http://en.wikipedia.org/wiki/0.999... – foobar1209 Mar 19 '14 at 22:19
  • See also my question here http://math.stackexchange.com/questions/98288/the-difference-between-10-and-9-99999-recurring which I thought was interesting, because there is an intuition involved here which is not quite trivial. But that got closed. – Mark Bennet Mar 19 '14 at 22:25
  • oh ok. i will look it up. thanks! –  Mar 19 '14 at 22:30
  • $0.999\dots = \frac{9}{10} + \frac{9}{100} + \dots = \frac{9/10}{1 - 1/10} = 1$! – MT_ Mar 21 '14 at 00:21

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Your argument is fine. The justifications of individual steps in it bear closer examination, but the bottom line is that they can be justified.

Essentially this same argument is often used to derive the formula for the sum of a geometric series: \begin{align} x & = 1 + r + r^2 + r^3 + \cdots \\[6pt] rx & = r + r^2 + r^3 + r^4 + \cdots \\[6pt] x - rx & = (1 + r + r^2 + r^3 + \cdots) - (r + r^2 + r^3 + \cdots) = 1. \\[6pt] x(1-r) & = 1 \\[6pt] x & = \frac{1}{1-r} \end{align}

This all works only if the series converges; that's one of those things about justification of individual steps.

Michael Hardy
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