Converting Repeating Decimals to Fractions | Grade 8 Math | LevelKite

You already know how to turn a fraction into a decimal with long division. When you divide $1$ by $3$ , the digit $3$ shows up in every decimal place and never stops. In this lesson you go the other way: you start with a decimal like that and find the fraction it equals.

A repeating decimal is a decimal in which one digit, or a group of digits, after the decimal point repeats forever. The part that repeats is called the repeating block. You mark it with a bar over the block:

$0.\overline{3}$ means $0.333\ldots$ , with the $3$ repeating forever.
$0.\overline{45}$ means $0.454545\ldots$ , with the block $45$ repeating forever.
$0.8\overline{3}$ means $0.8333\ldots$ , where only the $3$ is under the bar, so only the $3$ repeats forever.

A rational number is a number you can write as $\frac{p}{q}$ , where $p$ and $q$ are integers (whole numbers and their opposites) and $q \neq 0$ . The decimal form of a rational number either terminates, which means its digits stop, or repeats. A decimal that neither terminates nor repeats is the decimal form of an irrational number, a number that cannot be written as such a fraction.

The reverse is also true: every repeating decimal is a rational number, and this lesson shows you how to find its fraction. You cannot simply read it off. A terminating decimal has a last digit, so you can read its place value: $0.25$ is $25$ hundredths, or $\frac{25}{100}$ . A repeating decimal has no last digit, so you need a different idea.

Here is the idea. Compare $0.\overline{3}$ with ten times that number. Multiplying by $10$ moves each digit one place to the left, which looks the same as moving the decimal point one place to the right, so $10 \times 0.\overline{3} = 3.\overline{3}$ .

Because the $3$ repeats forever, both numbers have a $3$ in every decimal place. The table lines them up by place value. When you subtract, every decimal place gives $3 - 3 = 0$ , and only the ones place is left. The difference between these two never-ending decimals is exactly $3$ . Subtracting has canceled the endless part.

Subtracting 0.333... from ten times 0.333... place by place (the 3 repeats forever in both)

Number	Ones	Tenths	Hundredths	Thousandths	Every later place
$3.\overline{3}$	$3$	$3$	$3$	$3$	$3$
$0.\overline{3}$	$0$	$3$	$3$	$3$	$3$
Difference	$3 - 0 = 3$	$3 - 3 = 0$	$3 - 3 = 0$	$3 - 3 = 0$	$3 - 3 = 0$

Worked example

A one-digit repeating block

Write the repeating decimal $0.\overline{3}$ as a fraction in simplest form, which means the numerator and the denominator share no factor greater than $1$ .

Name the number so you can write equations about it. Let x stand for the value of the repeating decimal. Its repeating block is the single digit 3, and the block starts right after the decimal point.
$x = 0.\overline{3}$
Multiply both sides by 10. Multiplying both sides of an equation by the same number keeps it true. The decimal point moves one place to the right, past exactly one block, so the digits after it are still all 3s.
$10x = 3.\overline{3}$
Subtract the first equation from the second. On the left, ten copies of x minus one copy leaves 9x. On the right, the digits after the decimal point match in every place, so they cancel and leave 3. The two sides stay equal because you subtracted x on the left and the decimal equal to x on the right.
$10x - x = 3.\overline{3} - 0.\overline{3} \quad \text{so} \quad 9x = 3$
Divide both sides by 9. This is allowed because 9 is not zero.
$x = \frac{3}{9}$
Write the fraction in simplest form. Dividing the numerator and the denominator by the same nonzero number, here 3, gives an equal fraction. Now 1 and 3 share no factor greater than 1, so you are done.
$\frac{3}{9} = \frac{3 \div 3}{9 \div 3} = \frac{1}{3}$

Answer

The repeating decimal $0.\overline{3}$ is equal to $\frac{1}{3}$ . This is the fraction that long division turns into $0.\overline{3}$ , so the method agrees with what you already know.

Key concept

Line up the repeating digits, then subtract

Let $x$ stand for the repeating decimal. Multiply by $10$ , $100$ , $1000$ , and so on to write two equations whose right sides have the same digits after the decimal point:

one with the decimal point just before the first repeating block (this is $x$ itself if the block starts right after the decimal point);
one with the decimal point just after that block.

Subtract the first equation from the second. The matching digits cancel and leave integers. Divide both sides by the number in front of $x$ ; it is not zero because the two equations multiply $x$ by different numbers. Then write the fraction in simplest form.

The result is $\frac{p}{q}$ with integers $p$ and $q$ , $q \neq 0$ , so the repeating decimal is rational.

Worked example

A two-digit repeating block

Write the repeating decimal $0.\overline{45}$ as a fraction in simplest form.

Name the number. Let x stand for the value of the repeating decimal. The repeating block is 45. It has two digits and starts right after the decimal point.
$x = 0.\overline{45}$
Multiply both sides by 100, which moves the decimal point two places to the right. The block has two digits, so two places is exactly one full block, and the digits after the decimal point are still 45 repeating.
$100x = 45.\overline{45}$
Subtract the first equation from the second. On the left, 100x minus x is 99x. On the right, the matching digits after the decimal point cancel and leave 45.
$100x - x = 45.\overline{45} - 0.\overline{45} \quad \text{so} \quad 99x = 45$
Divide both sides by 99, which is not zero.
$x = \frac{45}{99}$
Write the fraction in simplest form. Both 45 and 99 are multiples of 9, so divide the numerator and the denominator by 9. Now 5 and 11 share no factor greater than 1.
$\frac{45}{99} = \frac{45 \div 9}{99 \div 9} = \frac{5}{11}$
Check with long division. Dividing 5 by 11 gives the digits 4 and 5 over and over, which matches the decimal you started with.
$5 \div 11 = 0.\overline{45}$

Answer

The repeating decimal $0.\overline{45}$ is equal to $\frac{5}{11}$ .

Watch the size of the shift. A common mistake is to multiply by $10$ every time. For $x = 0.\overline{45}$ that gives $10x = 4.\overline{54}$ . If you then treat the repeating parts as if they cancel, you get $9x = 4$ and $x = \frac{4}{9}$ . But long division shows $\frac{4}{9} = 0.\overline{4}$ , not $0.\overline{45}$ .

Here is why it fails. After the decimal point, $10x$ has the digits $5, 4, 5, 4, \ldots$ while $x$ has $4, 5, 4, 5, \ldots$ The digits in each place are different, so they do not cancel: $10x - x = 4.\overline{09}$ , which still repeats. Moving the decimal point one full block lines the digits up again: two places (multiplying by $100$ ) for a two-digit block, three places (multiplying by $1000$ ) for a three-digit block.

Worked example

A digit before the repeating block

Write the repeating decimal $0.8\overline{3}$ as a fraction in simplest form.

Name the number. Let x stand for the value of the repeating decimal. The 8 does not repeat. The repeating block is the single digit 3, which starts after the 8 instead of right after the decimal point.
$x = 0.8\overline{3}$
Multiply both sides by 10 to move the decimal point past the 8. Now the decimal point sits just before the first repeating block, and only 3s come after it.
$10x = 8.\overline{3}$
Go back to the first equation and multiply both sides by 100. This moves the decimal point two places, past the 8 and past one block of 3. Again only 3s come after the decimal point.
$100x = 83.\overline{3}$
Subtract the equation for 10x from the equation for 100x. On the left, 100x minus 10x is 90x. On the right, the matching 3s after the decimal point cancel, and 83 minus 8 is 75.
$100x - 10x = 83.\overline{3} - 8.\overline{3} \quad \text{so} \quad 90x = 75$
Divide both sides by 90, which is not zero. Then write the fraction in simplest form: 75 and 90 are both multiples of 15, so divide the numerator and the denominator by 15. Now 5 and 6 share no factor greater than 1.
$x = \frac{75}{90} = \frac{75 \div 15}{90 \div 15} = \frac{5}{6}$
Check with long division. Dividing 5 by 6 gives 8 in the tenths place and then 3 in every later place, which matches the decimal you started with.
$5 \div 6 = 0.8\overline{3}$

Answer

The repeating decimal $0.8\overline{3}$ is equal to $\frac{5}{6}$ .

A repeating decimal never ends, but its repeating block lets you pin it down exactly. Write two equations whose digits after the decimal point match, subtract so that the endless part cancels, and solve what is left. You get a fraction of two integers with a nonzero denominator, which shows that the repeating decimal is a rational number. To check, divide the numerator by the denominator: if you get the decimal you started with, your fraction is correct.