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# Methods for Converting Recurring Decimals into Fractions

## Methods for Converting Recurring Decimals into Fractions

A decimal form of a number is said to become recurring if at any point it becomes repeated indefinitely.

Keying **1 / 3** into a calculator will give **0.3333333333...** which is a recurring decimal.

Keying **1 / 12** into a calculator will give **0.0833333....** which is also a recurring decimal.

The **3**s on the end of both of these numbers actually continue forever and are limited only by the number of digits on the calculator. These examples only involve 1 repeating digit, but as shown below, they may have more than 1 repeating digit.

**0.9** recurring **(0.9999..)** is actually **equal **to **1**, and a simple proof of this is shown in** Section 3** of this article.

__Section 1: The sum of an infinite geometric series__

We will start by representing our recurring (or repeating) decimal as a (geometric) series. A geometric series is defined as a series with a constant ratio between successive terms.

For example:

**X = 0.111111... **

**X = 0.1 + 0.01 + 0.001 + 0.0001 + 0.00001 + ...** **= 0.111...**

**X = 1 / 10 + 1 / 100 + 1/ 1000 + 1/10000 + ... = 0.111....**

We have now created a series to represent **0.111.... **We can write this as

**X = a _{1 }+ a_{2 }+ a_{3 }+ a_{4 }+ a_{5 }+ ...** +

**a**

_{k }In the example above, the ratio between the terms is **1 / 10, **which we will call **r.** How this is derived is shown below.

*Setting term a_{1 }= 0.1, term a_{2 }= 0.01 as above, *

*it is clear that*

**1 / 10 *** *a _{1 }= a*

_{2 }

**1 / 10 * 0.1 = 0.01**

We can see this continues with any successive terms we choose:

**a**_{5 }= 0.00001, **a _{6 }= 0.000001**

** 1 / 10 * a_{5 }**=

**a**_{6 }**1 / 10 * 0.00001 **= **0.000001**

This shows that the recurrence relation looks like this:

**r * a _{n }**

*=*

**a**_{n+1}And for this example:

**1 / 10 *** **a _{n }**=

**a**

_{n+1 }

Using our ratio** r = 1/10,** we can simplify much further to:

**a _{1 }+ a_{1}r + a_{1}r^{2}+ a_{1}r^{3}+ a_{1}r^{4 }+ ... + a_{1}r**

^{n }=

**0.11111...**

where

**a**_{1 }=** 0.1**

**a**_{2 }= **a _{1}r** =

**0.1*0.1**=

**0.01**

**a**_{3 }=** a _{1}r^{2 }**=

**0.1*0.01**=

**0.001**

**a**_{k }= **a _{1}r**

^{n }where

**n**=

**k - 1.**

This result can be proven for when the recurrence relationship has this form, but it is not detailed here as it is rather more complicated.**Summing the geometric series: Derivation of the formula**

Now, using we can derive a formula which will allow us to calculate the value of this series as a fraction:

Let **S(n) **= **a _{1 }+ a_{1}r + a_{1}r^{2}+ a_{1}r^{3}+ a_{1}r^{4 }+ ... + a_{1}r^{n }**

Then** r.S(n) **=** a _{1}r + a_{1}r^{2}+ a_{1}r^{3}+ a_{1}r^{4 }+ a_{1}r^{5 }... + a_{1}r**

^{n+1 }(using

**.**for simple multiplication here instead of * or x)

Then **S(n) - r.S(n)** =** ( 1 - r ).S(n) **= **a _{1 }- a_{1}r + a_{1}r - a_{1}r^{2} + a_{1}r^{2 }- ... - a_{1}r^{n }+ a_{1}r^{n }- a_{1}r^{n+1}**

All terms cancel apart from **a _{1 }**and

**- a**giving:

_{1}r^{n+1}**(1-r).S(n) **= **a _{1}- a_{1}r^{n+1}**

**S(n)** =** ( a _{1}- a_{1}r^{n+1 }) / ( 1 - r )**

As **n → ∞ **(as **n **approaches infinity) to include the infinite terms in the series we constructed with our recurring decimal, the term **a _{1}r**

^{n+1 }will approach

**0**as

**-1 < r < 1**and therefore

**a**will become more and more negligible compared to other terms when raised to a larger and larger powers,

_{1}r^{n+1}**n**, and can therefore be ignored giving:

**S(n)** = **a _{1 }/ ( 1 - r )**

*Some information on limits can be found on another page: **http://hubpages.com/hub/Limits-in-Mathematics*

**-1 < r < 1 **or equivalently **|r| < 1** (absolute value) is necessary to give convergence of this series as values of **r** with absolute value larger than **1** will result in larger and larger terms for **n > 1, **and therefore will sum to infinity.

If we now go back to the simple example of **X = 1 / 9 = 0.111... **using the formula with the first term in the series we constructed from the decimal expansion of **X, a _{1} = 0.1, **and

**r = 0.1.**

S(n)=

S(n)

**0.1 / (1 - 0.1)**=

**0.1 / 0.9**=

**1 / 9**

And a slightly more difficult example:

**X** =** 0.3181818181818...**

We must take the **0.3 **out of the series to start with as it does not repeat:

**X** = **0.3 + 0.018181818......** = **0.3 + 18 / 1000 + 18 / 100000 + 18 / 10000000 + ....X** =

**0.3 + a**where

_{1 }+ a_{1}r + a_{1}r^{2}+ a_{1}r^{3}+ a_{1}r^{4 }+ ... + a_{1}r^{n}**a**

_{1 }=

**18 / 1000,**and

**r**=

**1/100**

We apply the formula **S(n)** = **a _{1 }/ ( 1 - r )** to the last part of

**X**

**X** = **0.3 + [S(n)] = 0.3 + [ ( 18 / 1000 ) / ( 1 - ( 1 / 100 ) ) ]** = **0.3 + [ ( 18 / 1000 ) / ( 99 / 100 ) ]**

Simplyfying of the **S(n)** term gives:

**X** = **0.3 + [ 18 / 990 ]**

**X **= **0.3 + [ 1 / 55 ]X** =

**3 / 10 + 1 / 55**=

**33 / 110 + 2 / 110**=

**35 / 110**

**X** = **7 / 22**

__Section 2: Simultaneous equations__

There is a also a way to obtain fractions from recurring decimals using algebra and simultaneous equations.

*Example 1*

**X** = **0.111111... (1)**

We wish to set up a second equation such that the repeating part after the decimal is 'aligned' with the first equation, so that when we subtract them from each other it is eliminated.

We however need to shift this second in some direction in order for **(2) - (1) **to be non-zero. The easiest way to do this is multiply the first equation by **10 **to give the second.

**10X** = **1.1111... (2)**Now calculating

**(2) - (1)**

10X - X=

10X - X

**1.1111... - 0.1111..**

**9X** = **1**

**X** = **1/9**

*Example 2*

**X**=

**0.3181818181818...**

We need to use different equations this time as the number does not start repeating until after the

**0.3**which requires a shift of

**1**digit, therefore multiplication of

**X**by

**10**needs to be done to give equation

^{1 }**(1)**. This can be extended to a general rule where equation

**(1) =**

**10**to give a left shift of

^{a }X**a**digits:

**10X **=** 3.1818181818... (1)**

*(In the earlier example this multiplication of X to set up equation (1) was not necessary as it started repeating immediately after the decimal point.)*

It then starts to repeat every **2** digits, so a second 'aligned' equation can be easily set up by multiplying the first equation by **10 ^{2}. **This can again be extended to a general rule where if it starts repeating every

**b**digits

**,**then you will multiply the first equation by

**10**Or equation

^{b}.**(2) = 10**

^{a}10^{b}X**1000X** =** 318.18181818.... (2)**

*Calculating (2) - (1):*

**1000X - 10X** =** 318.1818181... - 3.181818181...**

**990X **= **315**

**X **= **315 / 990 **

**X **= **7 / 22**

__Section 3: Showing that 0.9 recurring is equal to 1 (0.99999... = 1)__

The recurring decimal **0.99999...** can be shown to equal **1** in a number of ways, two of these are listed below. A proof can be constructed using the epsilon-delta method or the completeness theorem, although this is more closely related to the idea of Limits.

This may come as a surprise to some, who may believe **0.99999...** is less than ** 1**. A simplified example for why this is not the case, is as follows:

You are asked to draw a mark showing **0.99999...cm **on a ruler. You may initially imagine drawing the mark as close to left of **1cm** as possible, in order for it to be just less than **1cm**, but this would mean there would be gap between your mark and the **1cm** mark. This gap can then be measured (albeit with very accurate equipment and a microscope) to give a measurement we shall call *x*.

This means that we require **1 - x = 0.99999...**

*,*so suppose our original gap was

**0.1cm,**we would get

**1**

**-**

**0.1**=

**0.9cm**, which is less than

*0.99999..*. cmWe could then decrease our gap to **0.01cm **to attempt to get closer to* 0.99999..*.cm**, **this would give us **1 - 0.01 = 0.99cm, **which is still less than *0.99999...cm*

We can then imagine decreasing our gap even further to **0.001cm**, this would give us **1- 0.001 = 0.999cm**, which is still less than **0.99999...cm ***etc.*

It quickly becomes clear that no matter how small we choose the gap between the mark for **1cm **and the mark for** 0.99999...cm, **it will always be too big. Therefore we can conclude the only correct place to put the line is **exactly **on the mark for **1cm.**

*Method 1: Using the formula for an infinite geometric series with |r| < 1*

**S(n)** = **a _{1 }/ ( 1 - r )**

Recall that **a _{1 }**is our first term, and

**r**is the ratio between the successive terms as in previous examples.

**S(n) **= **0.99999...**

This can be broken down into the below geometric series, where **a _{1 }**=

**0.9,**...etc

**a**=_{2 }**0.9r,****a**=_{3 }**0.9r**^{2},**a**=_{4 }**0.9r**^{4}_{}

giving:_{}

**S(n) = 0.9 + 0.09 + 0.009 + 0.0009 + 0.00009 + ...**

So we see that **a _{1 }**=

**0.9**and

**r = 0.1,**and put these values into our formula:

**S(n)** =** 0.9 / ( 1 - 0.1)**

**S(n) **=** 0.9 / 0.9**

**S(n)** =** 1**.

We can see the result of our sum is **1**, and as there are infinity successive terms in our series **S(n), **we can conclude **0.99999...** is exactly equal to **1 **in the same way** **as we concluded the recurring decimals in the previous examples were exactly equal to their respective fraction forms.

*Method 2: Using simultaneous equations*

**X** =** 0.99999... (1)**

**10X** =** 9.99999... (2)**

*Then take (2) - (1) as in previous examples*

**10X - X**=

**9**

**9X**=

**9**

**X **= **1.**

Related articles by the same author:

Absolute Value

Irrational Numbers