Converting Repeating Decimal Numbers to Fractions

Question

Is it possible to write any decimal number, with a repeating decimal part, and be able to convert it into the form $\frac nd$ (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 $\frac54$), or repeat forever (such as $0.333\cdots$, which is the value $\frac13$).

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

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

Answer 1

Yes, as long at the repeating decimal is a positive number. Here's how: Let $x = 0.175175175\cdots$ Then $1000x - x = 175$. This implies $999x = 175$ and we have $0.175175175\cdots = \frac{175}{999}$.

Finally, $2.175175175\cdots = 2 + \frac{175}{999} =\frac{2173}{999}$

Answer 2

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.

Answer 3

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, $0.444\cdots$ can be written as $\frac49$.

$0.141414\cdots$ can be written as $\frac{14}{99}$

$0.235235235\cdots$ can be written as $\frac{235}{999}$

$0.076307630763\cdots$ can be written as $\frac{763}{9999}$

(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 $0.9999\cdots$ equals $1$.

Suppose $x = 0.999\cdots$

Then $10x = 9.999\cdots$

$10x-x = 9.999\cdots - 0.999\cdots$

$9x = 9$

$x=1$

But as $x$ was set to equal $0.999\cdots$ and has been shown to equal $1$, then $0.999\cdots = 1$.

Answer 4

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.