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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 ...

drewdles
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  • 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 $\infty + 1 = \infty \implies \infty - \infty = \infty - (\infty + 1) = -1$ – fleablood Dec 09 '15 at 10:17
  • Same would be true 66666 or 77777 etc. – fleablood Dec 09 '15 at 10:18
  • so 1111111.... = -1 and 7777777... = -1 so -1 = -7 is a contradiction. – fleablood Dec 09 '15 at 10:21
  • It's kind of a "loose" representation in the same way we can (sorta ... ) define: $ -1 = 1 +2 + 4+ 8 + ... $ (Its a geometric series ... extended by analytic continuation) – drewdles Dec 09 '15 at 10:25
  • Further I do not think u can map $-1$ to another infinite number by the means of using the distributive property alone. Note: I am not saying $1111... = \infty $ – drewdles Dec 09 '15 at 10:27
  • Except we can't define -1 = 1 + 2 + 4 + 8 + .... for the exact same reasons. It's outside the radius of converges. It's more than loose. It is wrong. – fleablood Dec 09 '15 at 10:27
  • You can't break basic laws of convergence and claim $\sum_{i=1}^{\infty}910^i = -1$ in one sentence and then refuse to break the law of distribution in the next. Your* broke the laws of distribution when you dividide by 10, after all. – fleablood Dec 09 '15 at 10:31
  • So when I say "loose" representation I mean we can in a sense meaningfully map the geometric series outside the radius of convergence by the means of analytic continuation. – drewdles Dec 09 '15 at 10:31
  • If I can't claim 9*1111111.... = 9999999.... then you can not claim 9999999..../10 = 99999......9.9. (which neither of us can do.) – fleablood Dec 09 '15 at 10:32
  • Let us continue this discussion in chat. – drewdles Dec 09 '15 at 10:32
  • U can claim $9 \times 1111.... = 9999.... $ – drewdles Dec 09 '15 at 10:34
  • Then if 9999...... = -1 then so does 111111.... and thus -9 = -1. – fleablood Dec 09 '15 at 10:36
  • Let: $ x = 111... $ Then $ .1 x= \bar{1}.1 $ subtracting the both of them: $ .9 x = -.1 \implies x = -1/9 $ ... As we would expect by the distributive property. I do not see how u equate 9999.... and 1111.... – drewdles Dec 09 '15 at 10:39
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    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. – fleablood Dec 09 '15 at 10:54
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    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. – fleablood Dec 09 '15 at 17:38
  • Related: http://math.stackexchange.com/questions/1623917/why-does-an-argument-similiar-to-0-999-1-show-999-1/ – Simply Beautiful Art Jan 25 '16 at 23:35

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