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Math Help - Converting Decimals (Non-ending + Repeating) to Fractions

  1. #1
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    Converting Decimals (Non-ending + Repeating) to Fractions

    Just wondering what is the best way of converting a non ending repeating decimal number into a fraction.

    Ex:

    Convert 0.0315 into a fraction (315 is repeating. Sorry don't know how to dot numbers in latex!)

    Cheers

    BIOS
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  2. #2
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    Quote Originally Posted by BIOS View Post
    Just wondering what is the best way of converting a non ending repeating decimal number into a fraction.

    Ex:

    Convert 0.0315 into a fraction (315 is repeating. Sorry don't know how to dot numbers in latex!)

    Cheers

    BIOS
    let x = 0.0315315315...

    1000x = 31.5315315...

    1000x - x = 999x = 31.5

    x = 31.5/999 = 7/222
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  3. #3
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    Can you explain how this works exactly? Would be greatly appreciated. E.g i can perform this but not sure exactly how it works.

    Thanks.
    Last edited by BIOS; December 21st 2010 at 05:04 PM. Reason: elaboration
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  4. #4
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    0.0\overline {315}  = \dfrac{1}{{10}} + \sum\limits_{k = 0}^\infty  {\dfrac{{315}}{{10^{4 + 3k} }}}
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  5. #5
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    Err thanks Plato. Care to break that down a little further?
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  6. #6
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    Hello, BIOS!

    \text{Convert }0.0\overline{315}\text{ into a fraction.}

    I'll show you what skeeter did . . .


    Let x \,=\,0.0315315\hots

    \begin{array}{cccccc}\text{Multiply by 10,000:} & 10,\!000x &=& 315.315315\hdots \\<br />
\text{Multiply by 10:} & \quad\;\; 10x &=& \;\;\;0.315315\hdots \\<br />
\text{Subtract:} & \;\; 9990x &=& 315\qquad\qquad\;\;  \end{array}


    Hence: . x \;=\;\dfrac{315}{9990} \;=\;\dfrac{7}{222}

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  7. #7
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    Hi Soroban. Thanks for the reply

    I understand the method in terms of applying it. I'm just looking for an explanation as to why it works.

    BIOS
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  8. #8
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    Quote Originally Posted by BIOS View Post
    Hi Soroban. Thanks for the reply

    I understand the method in terms of applying it. I'm just looking for an explanation as to why it works.

    BIOS
    It works because it works - there's not really any other way to explain it. You need to find which multiple of your original number has the same repeating digits, then subtract to eliminate the repeating digits. Division gives you the fraction...


    Anyway, Plato's method is using infinite geometric series...

    \displaystyle 0.0\overline{315} = 0.0315315315315\dots

    \displaystyle = \frac{315}{10\,000} + \frac{315}{10\,000\,000} + \frac{315}{10\,000\,000} + \dots

    \displaystyle = \frac{315}{10}\left(\frac{1}{1000} + \frac{1}{1\,000\,000} + \frac{1}{1\,000\,000\,000} + \dots\right)

    \displaystyle = \frac{315}{10}\left[\left(\frac{1}{1000}\right)^1 + \left(\frac{1}{1000}\right)^2 + \left(\frac{1}{1000}\right)^3 + \dots\right].


    The stuff in brackets is a convergent infinite geometric series with \displaystyle a = \frac{1}{1000} and \displaystyle r = \frac{1}{1000}. So its sum is \displaystyle \frac{a}{1 - r} = \frac{\frac{1}{1000}}{1 - \frac{1}{1000}} = \frac{\frac{1}{1000}}{\frac{999}{1000}} = \frac{1}{999}.

    So \displaystyle \frac{315}{10}\left[\left(\frac{1}{1000}\right)^1 + \left(\frac{1}{1000}\right)^2 + \left(\frac{1}{1000}\right)^3 + \dots\right]=\frac{315}{10}\cdot \frac{1}{999}

    \displaystyle = \frac{7}{222}.
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  9. #9
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    Thanks Prove it! Makes it alot clearer
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