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Math Help - periodic decimal expansion

  1. #1
    Member Jason Bourne's Avatar
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    periodic decimal expansion

    How would I express the number with periodic decimal expansion 0.321212121... in the form \frac{m}{n}. m,n \in \mathbb{Z} ?

    I don't know how this is done.
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  2. #2
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    Quote Originally Posted by Jason Bourne View Post
    How would I express the number with periodic decimal expansion 0.321212121... in the form \frac{m}{n}. m,n \in \mathbb{Z} ?

    I don't know how this is done.
    This is fast and loose and not too rigorous (but it gives the general idea and can be made more rigorous):

    S = 0.321212121

    10S = 3.212121... (1)

    1000S = 321.212121 .... (2)

    (2) - (1) and solve for S.
    Last edited by mr fantastic; January 11th 2009 at 05:39 PM. Reason: Fixed a typo and a spelling mistake
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  3. #3
    Member Jason Bourne's Avatar
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    Quote Originally Posted by mr fantastic View Post
    This is fast and loose and not too rigorous (but it gives the geneal idea and can be made more rigorous):

    S = 0.321212121

    10S = 3.212121... (1)

    100S = 321.212121 .... (2)

    (2) - (1) and solve for S.
    thanks. Would this be how to do this generally?
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  4. #4
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    Quote Originally Posted by Jason Bourne View Post
    How would I express the number with periodic decimal expansion 0.321212121... in the form \frac{m}{n}. m,n \in \mathbb{Z} ?
    I like to do this with a geometric series.
    0.321212121... = \frac{3}<br />
{{10}} + \frac{{21}}<br />
{{10^3 }} + \frac{{21}}<br />
{{10^5 }} +  \cdots  = \frac{3}<br />
{{10}} + \sum\limits_{k = 1}^\infty  {\frac{{21}}<br />
{{10^{2k + 1} }}}
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    This is fast and loose and not too rigorous (but it gives the geneal idea and can be made more rigorous):

    S = 0.321212121

    10S = 3.212121... (1)

    100S = 321.212121 .... (2) <-- I think he meant to write 1000S = ...

    (2) - (1) and solve for S.
    Quote Originally Posted by Jason Bourne View Post
    thanks. Would this be how to do this generally?
    Pretty much. The idea is to get rid of the "tail" depend on how many digits are repeating. e.g. In this case there are 2 digit repeating, so the factor will be 1000 and 10.

    So say if the question is asking for a expression for 0.3214214214214...
    Then, what do you think you should do?
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