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.
$\displaystyle 0.626262.....=0.62+0.0062+0.000062+......=\frac{62 }{100}+\frac{62}{10000}+\frac{62}{1000000}+.......$
$\displaystyle 0.62626262.....=\frac{62}{10^2}+\frac{62}{10^4}+\f rac{62}{10^6}+....=\frac{62}{10^2}\left(1+\frac{1} {10^2}+\frac{1}{10^4}+.......\right)$
$\displaystyle 0.626262......=\frac{62}{10^2}\left(\displaystyle{ \lim_{n\to\infty}\frac{1-\frac{1}{100^n}}{1-\frac{1}{100}}}\right)=\frac{62}{100}\times{\frac{ 100}{99}}=\frac{62}{99}$
Hopefully using the same method you will be able to do your two problems.
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.
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S = 52.31272727.... eqn (1)
100S = 5231.272727.... eqn (2)
10000S = 523127.2727..... eqn (3)
Eqn (3) - eqn (2): etc.
1. Let n denote the number of the repeated decimals. Then you can use the following 2 equations:
$\displaystyle \begin{array}{rcr}x&=&0.929292... \\ 10^2 x&=&92.929292...\end{array}$
The exponent of the base 10 corresponds to n.
3. Now subtract the first equation from the second cloumnwise and you'll get:
$\displaystyle 99x = 92$
Solve for x and simplify if possible: $\displaystyle x = \dfrac{92}{99}$
4. With your 2nd example you have to multiply first to get an immediately repeating decimal before you can use the described method:
$\displaystyle \begin{array}{rcr}x&=&52.3127272727... \\ 100x&=&5231.27272727... \\ 10000x&=&523127.272727...\end{array}$
Subtract the last two equations:
$\displaystyle 9900x = 517896$
I'll leave the rest for you.
EDIT: ... too late - again.
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.