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byrnema comments on Artificial Addition - Less Wrong
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Ok. Interesting.
I can see and agree that 0.999... can in the limit equal two, whereas in any finite representation would still be less than 2.
I don't consider them to be "the same number" in that sense... even though they algebraically equate (once the limit is reached) in a theoretical framework that can encompass infinities.
ie, in maths, I'd equate them but in the "real world" - I'd treat them separately.
Edit: and reading further... it seems I'm wrong again. Of course, the whole point of putting "..." is to represent the fact that this is the limit of the decimal expansion of 0.999... to infinity.
therefore yep, 1.999... = 2
Where my understanding failed me is that 1.999... does not in fact represent the summation of the infinite set of 1 + 0.9 + 0.09 + ... which summation could, in fact, simply not be taken to its full limit. The representation "1.999..." can only represent either the set or the limit of the set, and mathematical convention has it as the latter, not the former.
Another argument that may be more convincing on a gut level:
9x(1/9) is exactly equal to 1, correct?
Find the decimal representation of 1/9 using long division: 1/9=0.11111111... (note there is no different or superior way to represent this number as a decimal)
9x(1/9) = 9x(0.11111111...)=0.9999999... which we already agreed was exactly equal to 1.
Yes :)
See my previous (edited) comment above.
Oh sorry, my bad. I should have read the thread. Or the link.
No problem. It is a great proof (there aren't many so simple and succinct). Just bad luck on timing ;)