You don't need the real numbers at all to show that 0.999... = 1, because it's true in the rationals as well!
0.999... is defined as the sum to infinity of 9/10 + 9/100 + 9/1000 + ..., and you can show just by summing geometric progressions that that's equal to 9/10 × 1/(1-1/10) = 9/10 × 10/9 = 1.
A proof without so much notation, but which follows the same line of argument and is in fact rigorous in spite of looking like one of those facile non-proofs, is to say: suppose x = 0.999... . Then 10x = 9.999... . Subtract the former from the latter to get 9x = 9, whence x = 1.
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timeasmymeasure
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Date: 2009-02-19 05:09 pm (UTC)0.999... is defined as the sum to infinity of 9/10 + 9/100 + 9/1000 + ..., and you can show just by summing geometric progressions that that's equal to 9/10 × 1/(1-1/10) = 9/10 × 10/9 = 1.
A proof without so much notation, but which follows the same line of argument and is in fact rigorous in spite of looking like one of those facile non-proofs, is to say: suppose x = 0.999... . Then 10x = 9.999... . Subtract the former from the latter to get 9x = 9, whence x = 1.