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It's not a reduction. If you try to find where to put 0.999... on the number line, it has to go exactly where 1 is. For one thing, 1 - 0.999... = 0.000... beca

by iopq 10y ago

It's not a reduction. If you try to find where to put 0.999... on the number line, it has to go exactly where 1 is. For one thing, 1 - 0.999... = 0.000... because you never get to have any remainder since 0.999... is infinite. Or here's another proof: x = 0.999... 10x = 9.999... 10x - x = 9.999... - 0.999... 9x = 9.000... = 9 9x = 9 x = 1

kazinator 10y ago
Your "proofs" simply assumes that 0.999... is a notation denoting 1, without examining the underpinnings which might legitimize that. 9.999.. - 0.999 is 9 no matter how we define .999... just as long as two or more occurrences of the 0.999... notation all denote the same entity, and we understand that the syntax 9.999... is 9 + 0.999... For example, if we define 0.999... as "rubber duck" then 9.999... stands for 9 + "rubber duck", and 9.999... - 0.999... stands for 9 + "rubber duck" - "rubber duck" = 9.
- deleted 10y ago
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- iopq 10y ago
  While that's true, the whole key to the proof is that 0.999... * 10 = 9.999... Because "one less 9 than infinity is still infinity" it's what really closes the loop on the proof.