06-25-2004, 07:43 PM | #1 |
FFR Player
Join Date: Apr 2004
Posts: 28
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.999 = 1
hey saw this on the HTS messageboard.... thought it was really interesting.
Here are the proofs that 0.999~ (representing 0.9 recurring) = 1 Here are a few proofs. The last couple are easiest to understand if you don't understand more advance maths: 999~ = sigma(.9*[.1]^[n-1]). .9999~ = .9 + .09 + .009 + .0009 ... ad infinum sigma[i:0->inf.](.9*.1^i) omega->sigma = .9/(1 - .1) = .9/.9 = 1 .999~ = 1 .999~ = .9 + .09 + .009 .... (ad infinum) Infinite geometric progression: a + ar + ar^2... (ad infinum) Let a, 1st term = .9 Let r, common ratio = 10^-1 Sum to infinity = a / (1 - r) = .9 / (1 - .1) = .9/.9 = 1 .999~ = 1 sigma(n=1 ; n->inf.) 9/(10^n) = (def. of geometric series) 9 * sigma(n=1 ; n->inf.) (1/10)^n = (property of a series) 9 * 1/9 = (r/[1-r], r = .1) 3/9 = 1/3 .999~ = 1 The sequences (.9, .99, .999... ad infinum), and (1, 1, 1... ad infinum) are equivalent, so they have the same limit, .999~. S = .999~ S = 0.9 + 0.09 + 0.009 + 0.0009... ad infinum S = 0.9 + (1/10)(0.9 + 0.09 + 0.009 + .0009... ad infinum) S = 0.9 + (1/10)S (9/10)S = .9 S = 1 .999~ = 1 .333~ = sigma(n=1 ; n->inf.) 3/(10^n) = (def. of geometric series) 3 * sigma(n=1 ; n->inf.) (1/10)^n = (property of a series) 3 * 1/9 = (common ratio, r/[1-r], r = .1) 3/9 = 1/3 1/3 = .333~ 1/3 * 2 = 2/3 .333~ * 2 = .666~ 2/3 = .666~ .333~ + .666~ = .999~ 1/3 + 2/3 = 3/3 3/3 = .999~ 3/3 = 1 .999~ = 1 .000~ = 0/9 .111~ = 1/9 .222~ = 2/9 .333~ = 3/9 .444~ = 4/9 .555~ = 5/9 .666~ = 6/9 .777~ = 7/9 .888~ = 8/9 .999~ = 9/9 9/9 = 1 .999~ = 1 x = 0.999~ 10x = 9.999~ 10x - x = 9.999~ - 0.999~ 9x = 9 x = 1 0.999~ = 1 1/3=.33333~ 3(.33333~)=.99999~ 3(1/3)=1 therefor, 1=.99999~ Makes you question reality dosent it? |
06-25-2004, 07:45 PM | #2 |
FFR Player
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That's way more advanced than Blizzard's april fools joke... wow, I barely get it...
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06-25-2004, 07:46 PM | #3 |
FFR Player
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yea it does, thats the 3rd time for meh.
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06-26-2004, 12:12 AM | #4 |
(The Fat's Sabobah)
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Been posted before.
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