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

• Dec 21st 2010, 04:29 PM
BIOS
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
• Dec 21st 2010, 04:49 PM
skeeter
Quote:

Originally Posted by BIOS
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
• Dec 21st 2010, 04:56 PM
BIOS
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.
• Dec 21st 2010, 05:25 PM
Plato
$\displaystyle 0.0\overline {315} = \dfrac{1}{{10}} + \sum\limits_{k = 0}^\infty {\dfrac{{315}}{{10^{4 + 3k} }}}$
• Dec 21st 2010, 05:30 PM
BIOS
Err thanks Plato. Care to break that down a little further? :)
• Dec 21st 2010, 05:46 PM
Soroban
Hello, BIOS!

Quote:

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

I'll show you what skeeter did . . .

Let $\displaystyle x \,=\,0.0315315\hots$

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

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

• Dec 21st 2010, 05:58 PM
BIOS
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
• Dec 21st 2010, 08:07 PM
Prove It
Quote:

Originally Posted by BIOS
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...

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

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

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

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

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

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

$\displaystyle \displaystyle = \frac{7}{222}$.
• Dec 22nd 2010, 01:48 PM
BIOS
Thanks Prove it! Makes it alot clearer :D