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Thread: Application with division theorem problem.

  1. #1
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    Application with division theorem problem.

    by repeated use of the division theorem, find the infinite decimal representing the rational number 5/7.


    so we know r= 5/7. and in this form: a=bq+r.

    that b must be greater than r. So we can assume that b equals say 8/10.

    then 7/10 < 5/7 < 8/10.

    I do not know where to go from here.

    From this, do we use that since, r=5/7, b=8/10.

    a=(8/10)(q) + 5/7
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  2. #2
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    Re: Application with division theorem problem.

    $\dfrac 5 7 = q d + r$

    $d_1 = 0.1$

    $\dfrac 5 7 = q(0.1) + r$

    ignore $r$ a second

    $q(0.1) \leq \dfrac 5 7 < (q+1)(0.1)$

    $q \leq \dfrac{50}{7} < q+1$

    $q = 7$

    $r = \dfrac 5 7 - 7(0.1) = \dfrac{1}{70}$

    Now we repeat starting with $r, ~d=0.01$

    $q(0.01) \leq \dfrac{1}{70} < (q+1)(0.01)$

    $q \leq \dfrac{100}{70} < q+1$

    $q = 1$

    $r = \dfrac{1}{70} - \dfrac{1}{100} = \dfrac{3}{700}$

    rinse and repeat with $d$ reduced by a factor of 10 each iteration.

    $4(0.001) < \dfrac{3}{700} < 5(0.0001)$

    $r = \dfrac{3}{700} - \dfrac{4}{1000} = \dfrac{1}{3500}$

    $2(0.0001) < \dfrac{1}{3500} < 3(0.0001)$

    $r = \dfrac{1}{3500} - \dfrac{2}{10000} = \dfrac{3}{35000}$

    etc. until you see repetition.

    you can simplify this whole procedure by just using long division.
    Thanks from topsquark
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