## 0.999... not equal 1. Its Math BS | |

bbbUser ID: 130546 United States 10/25/2006 05:29 PM Report Abusive Post Report Copyright Violation | |

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Anonymous Coward User ID: 150446 United States 10/25/2006 07:13 PM Report Abusive Post Report Copyright Violation | 1/9 = .111... Quoting: 2/9 = .222... 3/9 = .333... 4/9 = .444... 5/9 = .555... 6/9 = .666... 7/9 = .777... 8/9 = .888... 9/9 = .999... I see a pattern here. And if 9/9 = .999..., then .999... = 1, because 9/9 = 1. I don't believe the people that posted here are beyond a third grade level. 1/9th < .111... 9/9th > .999... bbb.1111... is an INADEQUATE representation of 1/9, so in effect, 1/9 > .1111..., because we KNOW that 9/9 ( or 1 ) is > .9999... |

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Anonymous Coward User ID: 158633 Russian Federation 10/25/2006 07:35 PM Report Abusive Post Report Copyright Violation | Have you ever heard about Selector Calculus ? [link to en.wikipedia.org] or Finite Difference [link to en.wikipedia.org] Kind regards. |

John Gabriel User ID: 33838289 United States 02/06/2013 08:19 PM Report Abusive Post Report Copyright Violation | 0.999... is an ill-defined concept and one runs into problems with arithmetic when treating 0.999... as a number. Arithmetic is guaranteed only to function correctly when the objects are rational numbers. Real numbers do not exist. thenewcalculus.weebly.com/uploads/5/6/7/4/5674177/proof_that_0.999_not_equal_1.pdf |

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Anonymous Coward User ID: 12680882 United States 02/08/2013 04:09 PM Report Abusive Post Report Copyright Violation | Consider \sum_{n=0}^{\infty}\frac{9}{10}\left(\frac{1}{10}\right)^{n} It is known, and rather obvious, that the series converges to one, as
with r = \frac{1}{10} and a = \frac{9}{10} instead of 1. |

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