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Math Help - 0.9999..=1 proof

  1. #1
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    Dobrich
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    0.9999..=1 proof

    Hire some DIFIRENT numbers:
    1
    0.9...9=x
    0.9...8
    0.9...7
    .......
    0.9...1
    0.9...0
    0.9..-1
    .........
    0.0...1=y
    0
    x=0.9.. / lets *10
    0.9..+0.9..+0.9.. ten times =(1-y)+(1-y).. ten times SO
    10x= 10-10y / -0.9...
    10x-x=(10-10y)-(1-y)
    9x=9-9y
    9x=9-9y
    x=(9-9y)/9 and this is NOT 9 !
    x=9(1-y)/9
    x=0.9....

    This last one clearly shows that0.9...9 is not some "magic" number its number like anyother. This 0.00...1 is more "special" but is stillordinary number.
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  2. #2
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    Re: 0.9999..=1 proof

    Quote Originally Posted by rt20 View Post
    Hire some DIFIRENT numbers:
    1
    0.9...9=x
    0.9...8
    0.9...7
    .......
    0.9...1
    0.9...0
    0.9..-1
    .........
    0.0...1=y
    I have no idea what any of these mean. I know that "0.999..." means the decimal number whose only digits are "9". But I don't know what you mean by "0...9". How many digits does that "..." skip over? It can't be an infinite number since there is no "last" digit in that case.

    0
    x=0.9.. / lets *10
    0.9..+0.9..+0.9.. ten times =(1-y)+(1-y).. ten times SO
    "ten times"? Do you mean 0.9+ 0.9+ 0.9+ 0.9+ 0.9+ 0.9+ 0.9+ 0.9+ 0.9= 10(0.9)= 9? I don't know what you mean by 1- y because I don't know what you mean by "0.0...1" which is apparently how you have defined y
    And so nothing beyond here makes sense.

    10x= 10-10y / -0.9...
    10x-x=(10-10y)-(1-y)
    9x=9-9y
    9x=9-9y
    x=(9-9y)/9 and this is NOT 9 !
    x=9(1-y)/9
    x=0.9....

    This last one clearly shows that0.9...9 is not some "magic" number its number like anyother. This 0.00...1 is more "special" but is stillordinary number.
    Should I conclude that you do not understand how "decimal notation" works?

    If you really want a proof that 0.999...= 1, do this: 0.999...= 0.9+ 0.09+ 0.009+ ...= 0.9+ 0.9(.1)+ 0.9(.01)+ 0.9(.001)+ ... In other words, it is a geometric series, of the form \sum ar^n[ with a= 0.9 and r= 0.1. It is easy to show that the sum of such a geometric series is \frac{a}{1- r} which, here, is \frac{0.9}{1- 0.1}= \frac{0.9}{0.9}= 1.
    Last edited by HallsofIvy; April 22nd 2012 at 03:51 PM.
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  3. #3
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    Dobrich
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    Re: 0.9999..=1 proof

    "It can't be an infinite number since there is no "last" digit in that case."
    Well there is. I just write it. In mathematics you can have any number you can think of! Why not ?

    Saying that that there is not "last digit" is like saying 0=1 witch is not true.
    0= 0 + 0 + 0 + ...
    = (1 - 1) + (1 - 1) + (1 - 1) + ...
    = 1 - 1 + 1 - 1 + 1 - 1 + ...
    = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + ...
    = 1 + 0 + 0 + 0 + ...
    = 1


    I am saying that 0.0..1 is the closest to zero number. By definition. I don't have to prove it - i just to define it : )
    0.9.. is the closest to 1, with make 1-0.0...1 =0.9.... Its very simple.
    Last edited by rt20; April 22nd 2012 at 05:04 PM.
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