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Motivation

We all know that:

$$ .\bar{9} =.999 \dots= 1$$

I was wondering if the following (obviously not rigorous) statement could be defined on the same footing?

Question

$$ x = \bar{9} $$
$$ \implies x/10 = \bar{9}.9 $$

Subtracting the above equations:

$$ .9 x = -.9 $$ $$ \implies x = -1 $$

Hence, $ x=\bar{9} = -1 $

Does this already exist? Does this representation of negative numbers lead to any apparent contradictions? What are the operations one can preserve with this kind of representation which still make sense?

P.S: I know this bit of a wild idea ...

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  • $\begingroup$ the sequence. .9, .99, .999,.... converges and it is the limit that is equal to 1. the sequence 9,99, 999,.... does not converge and approaches infinite and is undefined. So you are subtracting infinity from infinity and manipulating it to get -1. This is equivalent to $\infty + 1 = \infty \implies \infty - \infty = \infty - (\infty + 1) = -1$ $\endgroup$
    – fleablood
    Commented Dec 9, 2015 at 10:17
  • $\begingroup$ Same would be true 66666 or 77777 etc. $\endgroup$
    – fleablood
    Commented Dec 9, 2015 at 10:18
  • $\begingroup$ so 1111111.... = -1 and 7777777... = -1 so -1 = -7 is a contradiction. $\endgroup$
    – fleablood
    Commented Dec 9, 2015 at 10:21
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    $\begingroup$ Hmm. It's late. You're right. I made an error there. But my point still holds that the term doesn't converge and is unacceptable. $\endgroup$
    – fleablood
    Commented Dec 9, 2015 at 10:54
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    $\begingroup$ Hmm, as often occurs I go to bed and then realize an entirely different way of thinking about things. I still maintain 999999.... diverges so isn't acceptable but that does depend on the metric. If we define another metric where it does then the limit could be defined and ... well, I guess it would converge to -1. However such a metric would redefine the reals (as irrationals are not longer defined). So, yes, it can be thought of as such and I imagine it wouldn't really have contradictions. But it wouldn't be the reals we know. So, sorry, for being pig headed. $\endgroup$
    – fleablood
    Commented Dec 9, 2015 at 17:38

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