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Math Help - Repeating Decimals As Fraction

  1. #1
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    Repeating Decimals As Fraction

    Express the given repeating decimals as a fraction.(1) 0.929292...(2) 52.31272727...
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    Quote Originally Posted by fdrhs1984 View Post
    Express the given repeating decimals as a fraction.(1) 0.929292...(2) 52.31272727...
    Dear fdrhs1984,

    I shall show you an example similar to the ones that you have given. Suppose you have to express 0.62626262....as a fraction.

    0.626262.....=0.62+0.0062+0.000062+......=\frac{62  }{100}+\frac{62}{10000}+\frac{62}{1000000}+.......

    0.62626262.....=\frac{62}{10^2}+\frac{62}{10^4}+\f  rac{62}{10^6}+....=\frac{62}{10^2}\left(1+\frac{1}  {10^2}+\frac{1}{10^4}+.......\right)

    0.626262......=\frac{62}{10^2}\left(\displaystyle{  \lim_{n\to\infty}\frac{1-\frac{1}{100^n}}{1-\frac{1}{100}}}\right)=\frac{62}{100}\times{\frac{  100}{99}}=\frac{62}{99}

    Hopefully using the same method you will be able to do your two problems.
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    Quote Originally Posted by fdrhs1984 View Post
    Express the given repeating decimals as a fraction.(1) 0.929292...(2) 52.31272727...
    Here is the slacker's way:

    S = 0.929292..... eqn (1)

    100S = 92.9292.... eqn (2)

    Eqn (2) - eqn (1): 99S = 92 => S = 92/99.

    --------------------------------------------------------------------------------------

    S = 52.31272727.... eqn (1)

    100S = 5231.272727.... eqn (2)

    10000S = 523127.2727..... eqn (3)

    Eqn (3) - eqn (2): etc.
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    Quote Originally Posted by fdrhs1984 View Post
    Express the given repeating decimals as a fraction.(1) 0.929292...(2) 52.31272727...
    1. Let n denote the number of the repeated decimals. Then you can use the following 2 equations:

    \begin{array}{rcr}x&=&0.929292... \\ 10^2 x&=&92.929292...\end{array}

    The exponent of the base 10 corresponds to n.

    3. Now subtract the first equation from the second cloumnwise and you'll get:

    99x = 92

    Solve for x and simplify if possible: x = \dfrac{92}{99}

    4. With your 2nd example you have to multiply first to get an immediately repeating decimal before you can use the described method:

    \begin{array}{rcr}x&=&52.3127272727... \\ 100x&=&5231.27272727... \\ 10000x&=&523127.272727...\end{array}

    Subtract the last two equations:

    9900x = 517896

    I'll leave the rest for you.

    EDIT: ... too late - again.
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    Quote Originally Posted by Sudharaka View Post
    Dear fdrhs1984,

    I shall show you an example similar to the ones that you have given. Suppose you have to express 0.62626262....as a fraction.

    0.626262.....=0.62+0.0062+0.000062+......=\frac{62  }{100}+\frac{62}{10000}+\frac{62}{1000000}+.......

    0.62626262.....=\frac{62}{10^2}+\frac{62}{10^4}+\f  rac{62}{10^6}+....=\frac{62}{10^2}\left(1+\frac{1}  {10^2}+\frac{1}{10^4}+.......\right)

    0.626262......=\frac{62}{10^2}\left(\displaystyle{  \lim_{n\to\infty}\frac{1-\frac{1}{100^n}}{1-\frac{1}{100}}}\right)=\frac{62}{100}\times{\frac{  100}{99}}=\frac{62}{99}

    Hopefully using the same method you will be able to do your two problems.
    This method is a bit too advanced since we have not covered limits in class.

    Quote Originally Posted by mr fantastic View Post
    Here is the slacker's way:

    S = 0.929292..... eqn (1)

    100S = 92.9292.... eqn (2)

    Eqn (2) - eqn (1): 99S = 92 => S = 92/99.

    --------------------------------------------------------------------------------------

    S = 52.31272727.... eqn (1)

    100S = 5231.272727.... eqn (2)

    10000S = 523127.2727..... eqn (3)

    Eqn (3) - eqn (2): etc.
    This method is slightly better than the first reply involving limits.
    Last edited by mr fantastic; August 13th 2010 at 03:35 PM. Reason: Merged posts
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  6. #6
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    Quote Originally Posted by fdrhs1984 View Post
    This method is slightly better than the first reply involving limits.
    Simpler maybe, but I don't think it's fair to call it better. I think they're equally good.

    Quote Originally Posted by fdrhs1984 View Post
    This method is a bit too advanced since we have not covered limits in class.
    More info on the Wikipedia page on geometric series.
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    Quote Originally Posted by earboth View Post
    1. Let n denote the number of the repeated decimals. Then you can use the following 2 equations:

    \begin{array}{rcr}x&=&0.929292... \\ 10^2 x&=&92.929292...\end{array}

    The exponent of the base 10 corresponds to n.

    3. Now subtract the first equation from the second cloumnwise and you'll get:

    99x = 92

    Solve for x and simplify if possible: x = \dfrac{92}{99}

    4. With your 2nd example you have to multiply first to get an immediately repeating decimal before you can use the described method:

    \begin{array}{rcr}x&=&52.3127272727... \\ 100x&=&5231.27272727... \\ 10000x&=&523127.272727...\end{array}

    Subtract the last two equations:

    9900x = 517896

    I'll leave the rest for you.

    EDIT: ... too late - again.
    Nice break down of the questions. I'll do the rest.

    Quote Originally Posted by undefined View Post
    Simpler maybe, but I don't think it's fair to call it better. I think they're equally good.



    More info on the Wikipedia page on geometric series.
    Thanks for the link.
    Last edited by mr fantastic; August 13th 2010 at 03:34 PM. Reason: Merged posts
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