Converting Decimals (Non-ending + Repeating) to Fractions

**BIOS** · December 21st 2010, 04:29 PM

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

**skeeter** · December 21st 2010, 04:49 PM

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

**BIOS** · December 21st 2010, 04:56 PM

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.

**Plato** · December 21st 2010, 05:25 PM

$0.0\overline {315} = \dfrac{1}{{10}} + \sum\limits_{k = 0}^\infty {\dfrac{{315}}{{10^{4 + 3k} }}}$

**BIOS** · December 21st 2010, 05:30 PM

Err thanks Plato. Care to break that down a little further?

**Soroban** · December 21st 2010, 05:46 PM

Hello, BIOS!

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

I'll show you what skeeter did . . .

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

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

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

**BIOS** · December 21st 2010, 05:58 PM

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

**Prove It** · December 21st 2010, 08:07 PM

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...

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

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

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

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

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

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

$\displaystyle = \frac{7}{222}$ .

**BIOS** · December 22nd 2010, 01:48 PM

Thanks Prove it! Makes it alot clearer

Thread: Converting Decimals (Non-ending + Repeating) to Fractions

LinkBack

Thread Tools

Display

Converting Decimals (Non-ending + Repeating) to Fractions

Similar Math Help Forum Discussions

Writing repeating decimals as fractions (complex repeating decimals)

Converting Decimals to Fractions

Converting fractions to decimals--without a calculator

Change repeating decimals to fractions

Converting decimals to fractions and a word problem on interest rate

Search Tags