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This is 'more intuitive' if you think about it this way: If any two real numbers are not equal, then you can take the average and get a third number that is ha

This is 'more intuitive' if you think about it this way: If any two real numbers are not equal, then you can take the average and get a third number that is half way between them. Conversely, if the average of two numbers is equal to either of the numbers, then the two numbers are equal. (this isn't a proof, just a way to convince yourself of this) What's the average of .9999... and 1?

saagarjha 7y ago
0.999…5 obviously.
- fsflover 7y ago
  But at which position that "5" is? Tell the number please.
  - saagarjha 7y ago
    Position 999…, duh.
- kaetemi 7y ago
  Okay, now multiply that by 2.
  - saagarjha 7y ago
    0.999…
edflsafoiewq 7y ago
Or: if they're different, what is their difference?
- peanutz454 7y ago
  0.000...1 obviously
  - edflsafoiewq 7y ago
    Why not 0.000...2?
  - TallGuyShort 7y ago
    Thinking of it this way actually shows you that x.999... (ad infinitum) is the same as y.000 (ad infinitum) if x + 1 = y.
deleted 7y ago
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