# Thread: Application with division theorem problem.

1. ## 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

2. ## 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.