# Converting Repeating Decimal Numbers to Fractions

Is it possible to write any decimal number, with a repeating decimal part, and be able to convert it into the form n/d (where both n and d are natural numbers)?

I know rational numbers that are expressed in decimal notation will either terminate exactly (such as 1.25, which is the value 5/4), or repeat forever (such as 0.333..., which is the value 1/3).

So if I just come up with any random repeating decimal, like 2.175175175..., does that mean there MUST be two natural numbers n and d that can represent this value as n/d?

I'm just trying to get a better feel for rational numbers and decimals.

Yes, as long at the repeating decimal is a positive number. Here's how: Let x = .175175175... Then 1000x - x = 175. This implies 999x = 175 and we have .175175175... = 175/999.

Finally, 2.175175175... = 2 + 175/999 = 2173/999

Let $x=y.a_1a_2\ldots a_m b_1b_2\ldots b_p b_1b_2\ldots b_p \ldots$, where $y\in \mathbb N$.

Then $10^m x=t+f$, where $t\in \mathbb N$ and $f=0.b_1b_2\ldots b_p b_1b_2\ldots b_p \ldots$ .

Now, $10^p f=b+f$, where $b=(b_1b_2\ldots b_p)_{10}\in \mathbb N$. So, $f=\dfrac{b}{10^p-1}$

Thus $x=\dfrac{t+\dfrac{b}{10^p-1}}{10^m}=\dfrac{t(10^p-1)+b}{10^m(10^p-1)}$ is a quotient of two natural numbers.

Both answers given so far are good, but it seems you might be looking for a general rule to follow. Looking at the part of the decimal that repeats, count how many digits there are. Place the repeating part over the same number of nines. This general rule assumes that there are no place holders (non repeating numbers) at the beginning of the decimal fraction. AsdrubalBeltran's second example shows how to handle these.

For example, .444... can be written as 4/9.

.141414... can be written as 14/99

.235235235... can be written as 235/999

.076307630763... can be written as 763/9999

(Please take note of that last one, as it has a repeating zero and is not to be confused with the example of a non-repeating placeholder.)

This general rule is based on the proof that .9999... equals 1.

Suppose x = .999...

Then 10x = 9.999...

10x-x = 9.999... - .999...

9x = 9

x=1

But as x was set to equal .999... and has been shown to equal 1, then .999... = 1.

Two examples: First write in the numerator: the number but not decimal point, less the part not periodic:

$$2.757575...=\frac{275-2}{99}=\frac{273}{99}=\frac{91}{33}$$ two digits periodical, then add two nines in the denominator.

$$3.0412412412...=\frac{30412-30}{9990}=\frac{15191}{4995}$$

3 digits periodic then 3 nines, and one digit decimal no periodic then add a zero.

Some of these answer responses tell you how to do it, but only one of them explains why, and it probably looks convoluted to you, so I'll try explain it as simply as possible.

When you take any repeating decimal, the first step is to change the repeating bit to a bunch of zeros followed by a one, like this: $$x = 0.123123123.... \\ \\ \frac{x}{123} = 0.001001001...$$

Now we can write this value as a sum: $$\begin{eqnarray} 0.001001001... &=& 0.001 + 0.000001 + 0.000000001 ... \\ \\ &=& \frac{1}{1000} + \frac{1}{1000000} + \frac{1}{1000000000}... \\ \\ &=& \frac{1}{1000} + \frac{1}{1000^2} + \frac{1}{1000^3} + ... \end{eqnarray}$$

This now becomes an infinite geometric series: $$\sum\limits_{k=0}^\infty r^k = \frac{1}{1-r} \\ \\ \sum\limits_{k=0}^\infty \left(\frac{1}{1000}\right)^k = \frac{1}{1-\frac{1}{1000}} = \frac{1}{\frac{999}{1000}} = \frac{1000}{999} \\ \\ \sum\limits_{k=1}^\infty \left(\frac{1}{1000}\right)^k = \sum\limits_{k=0}^\infty \left(\frac{1}{1000}\right)^k - 1 = \frac{1000}{999}-1 = \frac{1}{999} \\$$

Substituting our new calculations, we now know that $$\frac{x}{123} = \frac{1}{999} \\ \\ x = \frac{123}{999}$$

That's where the all of the $9$ digits in the denominator comes from.

You can also adapt this method to decimals of any base. For example, $0.252525...$ in base $8$ would be: $$0.252525... = \frac{25}{77}$$ $25$ in base $8$ is $2*8 + 5 = 21$ in base $10$.

$77$ is $8^2-1$ or $7*8 + 7 = 63$ in base $10$.

The decimal is $\frac{21}{63}$ in base $10$.

• When you take any repeating decimal, the first step is to change the repeating bit to a bunch of zeros followed by a one - but why? – Max Aug 28 '17 at 11:09
• You can do the step where you split it up into fractions first, but either way you end up factoring out the digit sequence (123 in the first example). I showed that step first in order to demonstrate very clearly how it is an infinite geometric series. – CosmoVibe Nov 20 '17 at 22:22