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Thread: finite continued fractions ?

  1. #1
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    finite continued fractions ?

    How to express the rational numbers as finite continued fractions ?
    with last term >1

    1. 118/303

    2. 187/57

    3. 10001/10101

    4. 12/240005
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  2. #2
    Super Member PaulRS's Avatar
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    $\displaystyle

    \frac{{187}}
    {{57}} = 3 + \frac{{16}}
    {{57}} = 3 + \frac{1}
    {{{\raise0.5ex\hbox{$\scriptstyle {57}$}
    \kern-0.1em/\kern-0.15em
    \lower0.25ex\hbox{$\scriptstyle {16}$}}}} = 3 + \frac{1}
    {{3 + {\raise0.5ex\hbox{$\scriptstyle 9$}
    \kern-0.1em/\kern-0.15em
    \lower0.25ex\hbox{$\scriptstyle {16}$}}}} = 3 + \frac{1}
    {{3 + \tfrac{1}
    {{1 + \tfrac{7}
    {9}}}}}

    $

    $\displaystyle
    \frac{{187}}
    {{57}} = 3 + \frac{1}
    {{3 + \tfrac{1}
    {{1 + \tfrac{7}
    {9}}}}} = 3 + \frac{1}
    {{3 + \tfrac{1}
    {{1 + \tfrac{1}
    {{1 + \tfrac{2}
    {7}}}}}}} = 3 + \frac{1}
    {{3 + \tfrac{1}
    {{1 + \tfrac{1}
    {{1 + \tfrac{1}
    {{3 + \tfrac{1}
    {2}}}}}}}}}

    $

    Do you see how I did it?
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  3. #3
    Global Moderator

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    Here is another approach you can take.

    Use the Euclidean algorithm on $\displaystyle 187,57$:

    $\displaystyle 187 = 3\cdot 57 + 16$
    $\displaystyle 57 = 3\cdot 16 + 9$
    $\displaystyle 16 = 1\cdot 9 + 7$
    $\displaystyle 9 = 1\cdot 7+2$
    $\displaystyle 7 = 3\cdot 2 + 1$
    $\displaystyle 2 = 1\cdot 1+0$

    Thus, $\displaystyle 187/57 = [3;3,1,1,3,1]$.

    Now why does this not match what PaulRS did?
    Probably a computational mistake somewhere, I do not have time to check it now.
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  4. #4
    Lord of certain Rings
    Isomorphism's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    $\displaystyle 2 = 1\cdot 1+0$
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