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

    <br />
\frac{{187}}<br />
{{57}} = 3 + \frac{1}<br />
{{3 + \tfrac{1}<br />
{{1 + \tfrac{7}<br />
{9}}}}} = 3 + \frac{1}<br />
{{3 + \tfrac{1}<br />
{{1 + \tfrac{1}<br />
{{1 + \tfrac{2}<br />
{7}}}}}}} = 3 + \frac{1}<br />
{{3 + \tfrac{1}<br />
{{1 + \tfrac{1}<br />
{{1 + \tfrac{1}<br />
{{3 + \tfrac{1}<br />
{2}}}}}}}}}<br /> <br />

    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 187,57:

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

    Thus, 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
    2 = 1\cdot 1+0
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