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Simple Continued Fractions Question (I Think My Book Has a Mistake)

  1. Apr 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that if the simple continued fraction expression of the rational number [itex] \alpha [/itex], [itex] \alpha > 1 [/itex], is [itex] [a_0; a_1, a_2, \dotsc, a_k] [/itex], then the simple continued fraction expression of [itex] \frac{1}{\alpha} [/itex] is [itex] [0; a_1, a_2, \dotsc, a_k] [/itex].

    2. Relevant equations



    3. The attempt at a solution


    Suppose [itex] \alpha [/itex] is a rational number greater than [itex] 1 [/itex] with simple continued fraction expression
    [itex] \alpha = [a_0; a_1, a_2, \dotsc, a_k] = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{\ddots a_{k - 1} + \cfrac{1}{a_k}}}}. [/itex]
    Then we see that [itex] \frac{1}{\alpha} [/itex] is also a rational number with simple continued fraction expression
    [itex] \frac{1}{\alpha} = \cfrac{1}{a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{\ddots a_{k - 1} + \cfrac{1}{a_k}}}}} = [0; a_0, a_1, \dotsc, a_k]. [/itex]

    This is, of course, different than what the book says the solution should be. However, I don't see where I went wrong, if I did go wrong. Any help would be greatly appreciated!
     
  2. jcsd
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