How to Turn a Repeating Decimal into a Fraction

$\frac{1}{3}$ can never stop, because $3$ is neither $2$ nor $5$ . Its digits run forever, but they are not random. They fall into a loop, and the remainders in long division are what force that loop.

Problem

Long-divide

1 \div 3

. At each step the remainder decides the next digit. What number is the remainder at the end of every step?

Show a hint

Don't watch the quotient digit, watch what is left over after you subtract. Each step you have some number, you take away the biggest multiple of $3$ you can, and a small leftover stays behind. Write down that leftover for the first two or three steps and see if it changes.
On step one you compute $10 - 9 = 1$ . On step two you pull down a zero to make $10$ again, and $10 - 9 = 1$ . The leftover after subtracting is the same single digit every single time. Type that digit.

Show the full solution

Bring down a zero and divide.

10 \div 3

gives digit

3

with

10 - 9 = 1

left over. Bring down another zero and

10 \div 3

gives

3

again with

1

left over. The remainder at the end of every step is

1

. Since that leftover is the number we started with, the step copies itself forever, which is why

\frac{1}{3} = 0. \overline{3}

Problem

Long-dividing

1 \div 7

cycles through remainders

1, 3, 2, 6, 4, 5

before

1

returns, giving

\frac{1}{7} = 0. \overline{142857}

. How many digits are in the repeating block?

Show a hint

Only $0$ through $6$ can ever be a remainder, and $0$ would stop the decimal. So while it keeps going, every remainder is one of $1, 2, 3, 4, 5, 6$ . The block repeats once a remainder returns, so the longest a block can run before looping is exactly the number of different remainders that show up.
List the remainders in order until one comes back. They are $1, 3, 2, 6, 4, 5$ , and then $1$ appears again. Count how many remainders you went through before that first repeat. That count is the length of the repeating block.

Show the full solution

Follow the remainders.

10 \div 7

gives

1

remainder

3

, then

30 \div 7

gives

4

remainder

2

, then

20 \div 7

gives

2

remainder

6

, then

60 \div 7

gives

8

remainder

4

, then

40 \div 7

gives

5

remainder

5

, then

50 \div 7

gives

7

remainder

1

. The starting remainder is back after six steps, so the block is

6

digits long, namely

142857

. A block can never run longer than the count of available nonzero remainders, and here all six of

1

through

6

get used.

Dividing

1

7

, the leftover at each step rides around this ring in the order

1, 3, 2, 6, 4, 5

, and the quotient digit stamped on each arrow spells

0. \overline{142857}

. The closing gold arrow from

5

back to

1

is the whole point. Every leftover is smaller than the divisor, so only finitely many leftovers are possible, and the moment one returns the digits are forced to repeat. The size of the loop is the length of the repeating block, the period. Here that loop holds six remainders, so the period is

6

. Beside it,

\frac{1}{3}

loops straight back onto remainder

1

at once, a loop of one, giving the single digit block

0. \overline{3}

Problem

\frac{1}{6} = 0.1 \overline{6}

, only the

6

repeats. How many digits come before the repeating block begins, meaning how many are not under the bar?

Show a hint

Write out the decimal with the bar in the right place. In $0.1 \overline{6}$ , the bar covers the looping part. Everything to the left of where the bar starts is the non-repeating head. Just count those digits.
The repeating block is the $6$ , and the bar starts right at it. The only digit sitting before the bar is the $1$ . So count the digits before the loop begins.

Show the full solution

Divide

1

6

10 \div 6 = 1

remainder

4

, so the first digit is

1

. Then

40 \div 6 = 6

remainder

4

, the same remainder, so the

6

repeats from there. That gives

\frac{1}{6} = 0.1 \overline{6}

, and only

1

digit sits outside the bar. The head comes from the

2

6 = 2 \times 3

. One factor of

2

buys one settling digit, and the

3

makes the loop.

Problem

The long division

1 \div 11

gives

\frac{1}{11} = 0. \overline{09}

. The bar covers both the

0

and the

9

. How many digits are in the repeating block? Give the period.

Show a hint

The period is just a count. Look at the block sitting under the bar in $0. \overline{09}$ and count its digits, including the leading zero.
The block is $09$ . Count its digits one at a time. There is a $0$ , and there is a $9$ . That is two digits.

Show the full solution

10 \div 11 = 0

remainder

10

, so the first digit after the point is

0

. Then

100 \div 11 = 9

remainder

1

, which is the remainder we began with, so the digits cycle as

0, 9, 0, 9, \dots

and

\frac{1}{11} = 0. \overline{09}

. The block

09

has

2

digits. Keep that leading zero, since

0. \overline{9}

is a completely different number.

Problem

Long division gives

\frac{1}{12} = 0.08 \overline{3}

. What is the repeating block? Type the digit or digits under the bar.

Show a hint

The repeating block is the part that keeps coming back once the digits settle into their loop. The $0$ and the $8$ at the front each happen only once, so neither of them is part of the loop. Watch for the moment a remainder shows up a second time, because that is exactly when the repetition kicks in.
Carry the division out to $0.0833333 \dots$ . The remainder $4$ appears, gives the digit $3$ , and then comes back as the remainder again, so the same step repeats and you keep writing $3$ . Only the $3$ sits under the bar, so $\frac{1}{12} = 0.08 \overline{3}$ .

Show the full solution

Divide

1

12

10 \div 12 = 0

remainder

10

, then

100 \div 12 = 8

remainder

4

, then

40 \div 12 = 3

remainder

4

. Remainder

4

has come back, so that step repeats forever and

\frac{1}{12} = 0.08 \overline{3}

. The block under the bar is

3

. The head

08

is two digits because

12 = 2^{2} \times 3

carries two factors of

2

, and the

3

is what loops.

Problem

Which of

\frac{1}{16}

\frac{1}{20}

\frac{1}{25}

\frac{1}{15}

repeats? Factor each denominator and apply the

2

s-and-

5

s test. Give the denominator.

Show a hint

Break each denominator into primes. $16 = 2^{4}$ , $20 = 2^{2} \times 5$ , $25 = 5^{2}$ , and $15 = 3 \times 5$ . The terminating ones use only the "friendly" primes $2$ and $5$ . Look for the denominator that has a different prime.
Three of the denominators are made purely of $2$ s and $5$ s, so those three terminate. Only $15 = 3 \times 5$ carries a prime that is neither $2$ nor $5$ . That stray factor of $3$ is what forces the digits to repeat, so the answer is its denominator.

Show the full solution

Factor the four denominators.

16 = 2^{4}

20 = 2^{2} \times 5

25 = 5^{2}

, and

15 = 3 \times 5

. The first three are built only from

2

s and

5

s, so they terminate at

0.0625

0.05

, and

0.04

. Only

15

carries a

3

, and

\frac{1}{15} = 0.0 \overline{6}

, so the denominator is

15

. Only

2

and

5

divide a power of ten, so any other prime can never be scaled away and the remainders cycle instead of reaching

0

The bar sits only over the block that cycles.

\frac{1}{7} = 0. \overline{142857}

repeats from the first digit (period 6), while

\frac{1}{12} = 0.08 \overline{3}

settles at

08

first, then repeats the

3

So far we have gone from a fraction to its repeating decimal. The trip back works too. Multiply by a power of ten to make a second copy of the same endless tail, subtract so the two tails cancel, and the infinite decimal collapses into an ordinary fraction.

Problem

Let

x = 0. \overline{7}

. Multiply by

10

to get

10 x = 7. \overline{7}

. The tails of

7

s cancel when you subtract:

9 x = 7

. Write

0. \overline{7}

as a fraction in lowest terms.

Show a hint

You built two equations, $x = 0.7777 \dots$ and $10 x = 7.7777 \dots$ . Stack the second over the first and subtract straight down. On the left you get $10 x - x$ . On the right, the endless $7$ s line up exactly and wipe each other out, so what whole number is left?
Subtracting gives $9 x = 7$ . To get $x$ by itself, divide both sides by $9$ . That leaves $x = \frac{7}{9}$ . Check whether $7$ and $9$ share any common factor before you call it done.

Show the full solution

Let

x = 0.7777 \dots

, so

10 x = 7.7777 \dots

. Past the decimal point both numbers carry the identical endless string of

7

s, so subtracting wipes it out.

10 x - x = 7.7777 \dots - 0.7777 \dots

9 x = 7

Divide by

9

to get

x = \frac{7}{9}

, and

7

is prime so nothing reduces.

7 / 9

Multiplying by

10

shifts past exactly one block, which is what lines the two tails up.

Shifting by exactly one block length lines the endless tails up perfectly, so when you subtract they erase each other and an ordinary equation is left behind. Here

10 x - x = 7. \overline{7} - 0. \overline{7}

gives

9 x = 7

, so

0. \overline{7} = \frac{7}{9}

. The same trick scales to any pure repeater. A block of length

k

becomes the block written over

k

nines, for example

0. \overline{12} = \frac{12}{99}

Problem

Let

x = 0. \overline{12}

. Multiply by

100

to get

100 x = 12. \overline{12}

. Subtracting gives

99 x = 12

, so

x = \frac{12}{99}

. Write

0. \overline{12}

as a fraction in lowest terms.

Show a hint

You already know it begins as $\frac{12}{99}$ . Now ask the lowest-terms question. What whole number divides both $12$ and $99$ evenly?
Both $12$ and $99$ are divisible by $3$ . Divide the top and the bottom by $3$ , which gives $12 \div 3 = 4$ and $99 \div 3 = 33$ . Check that $4$ and $33$ share no common factor, then you are finished.

Show the full solution

Let

x = 0.121212 \dots

. The block is two digits, so multiply by

100

100 x = 12.121212 \dots

The tails match, so subtracting leaves

99 x = 12

and

x = \frac{12}{99}

. Both parts are divisible by

3

\frac{12}{99} = \frac{4}{33}

Since

4 = 2 \times 2

and

33 = 3 \times 11

, this is fully reduced, so

0. \overline{12} = 4 / 33

. The block-over-nines shortcut hands you a correct fraction but rarely a reduced one, so always check.

Problem

Let

x = 0. \overline{9}

. Multiply by

10

10 x = 9. \overline{9}

. Subtract:

9 x = 9

. Divide by

9

. What does

0. \overline{9}

equal?

Show a hint

You already did the hard part for $0. \overline{7}$ . Write $x = 0. \overline{9}$ and $10 x = 9. \overline{9}$ , then subtract. Trust the cancellation even though the result feels surprising.
Subtracting gives $10 x - x = 9. \overline{9} - 0. \overline{9}$ . The infinite tails of nines are identical, so they vanish and leave $9 x = 9$ . Now divide both sides by 9 and read off the single whole number you get.

Show the full solution

Let

x = 0.9999 \dots

, so

10 x = 9.9999 \dots

. After the decimal point both numbers carry the same endless string of nines, so subtracting cancels it.

10 x - x = 9.9999 \dots - 0.9999 \dots

9 x = 9

Divide by

9

and

x = 1

. Nothing was rounded or dropped anywhere in that work, so

0. \overline{9}

is not creeping up toward

1

. It is another name for

1

Problem

Write

0.41 \overline{6}

as a fraction in lowest terms. Multiply by

1, 000

and by

100

, subtract to cancel the

\overline{6}

tails, then solve and reduce.

Show a hint

You want two shifted copies of $x$ whose decimal tails are identical, so the subtraction wipes the tail out. The repeating part is $0.00 \overline{6}$ , sitting two places in. Try sliding the point so one copy stops just before the loop and the other stops just after one block of it.
Let $x = 0.41 \overline{6}$ . Multiply by $100$ to land just before the loop, $100 x = 41. \overline{6} = 41.666 \dots$ . Multiply by $1, 000$ to land one block past it, $1, 000 x = 416. \overline{6} = 416.666 \dots$ . Both end in the same $. 666 \dots$ tail. Subtract the smaller from the larger and solve for $x$ , then reduce.

Show the full solution

Let

x = 0.41666 \dots

. Shift past the head with

100 x = 41.666 \dots

, then one block further with

1, 000 x = 416.666 \dots

. Both end in the same

. 666 \dots

tail, so subtracting cancels it.

1, 000 x - 100 x = 416.666 \dots - 41.666 \dots

900 x = 375

Then

x = \frac{375}{900}

, and both share a factor of

75

, since

375 = 75 \cdot 5

and

900 = 75 \cdot 12

x = 5 / 12

When there is a non-repeating head you need two shifts, one landing just before the loop and one a full block later.

Problem

\frac{1}{7} = 0. \overline{142857}

, a block of

6

digits. Divide

50 \div 6

to find the position within the block. What is the 50th digit after the decimal point?

Show a hint

The block $142857$ is six digits long, so the pattern resets every $6$ positions. Every time you pass a full group of $6$ , you are back at the start of the block. So really you just need to know how far past the last full group position $50$ sits.
Divide $50$ by $6$ . You get $8$ full blocks with $2$ left over. That remainder of $2$ tells you the $50$ th digit is the same as the $2$ nd digit of the block $142857$ . Read off the second digit.

Show the full solution

The block

142857

is six digits long, so divide.

50 = 6 \times 8 + 2

, which puts position

50

two digits into a fresh block. The second digit of

142857

4

. Only the remainder picks the digit. The quotient

8

just counts how many full blocks you skipped past.

Problem

Find

0. \overline{72} - 0. \overline{27}

as a single fraction in lowest terms. Convert each to a fraction over

99

, subtract, then reduce.

Show a hint

Each of these has a two-digit repeating block, and you found a shortcut for exactly this. A repeating two-digit block sits over $99$ . So turn each decimal into a fraction with denominator $99$ before you do anything else.
Once both are written over $99$ , they share a denominator, so you can subtract the top numbers directly and keep the $99$ underneath. That leaves one fraction. Now reduce it by dividing the top and bottom by their greatest common factor.

Show the full solution

A two-digit block sits over

99

, so

0. \overline{72} = \frac{72}{99}

and

0. \overline{27} = \frac{27}{99}

. The denominators already match, so subtract the numerators.

\frac{72}{99} - \frac{27}{99} = \frac{45}{99}

Both

45

and

99

are divisible by

9

, which gives

\frac{5}{11}

, and

5

shares no factor with

11

5 / 11

Checking,

\frac{5}{11} = 0. \overline{45}

, and

0.7272 \dots

minus

0.2727 \dots

does leave

0.4545 \dots

Practice these ideas

Practice

Long-divide

2 \div 3

. Each step ends with the same remainder. What single digit repeats forever in the decimal

0. \overline{6}

Show the solution

Since

3

is bigger than

2

, the whole-number part is

0

and the remainder is

2

. Then

20 \div 3 = 6

with remainder

20 - 18 = 2

, the remainder we started with, so every step after that prints another

6

. That makes

2 \div 3 = 0. \overline{6}

, so the repeating digit is

6

. A returning remainder is exactly what locks a decimal into a loop.

Practice

Long-divide

1 \div 9

. The remainder returns immediately, giving

\frac{1}{9} = 0. \overline{1}

. How many digits are in the repeating block? Give the period.

Show the solution

Bring down a zero to get

10

. Then

10 \div 9 = 1

with remainder

1

, the number we started with, so every step writes another

1

and

\frac{1}{9} = 0. \overline{1}

. The bar covers a single digit, so the period is

1

Practice

Long-divide

1 \div 11

. The remainder returns after two digits, giving

\frac{1}{11} = 0. \overline{09}

. Write the repeating block as a digit string, including the leading zero.

Show the solution

Start with remainder

1

. Attach a zero to get

10

, and

11

fits into

10

zero times, so the first digit is

0

and the remainder is

10

. Attach another zero to get

100

, and

11

fits

9

times since

9 \times 11 = 99

, leaving remainder

1

again. The digits cycle as

0, 9, 0, 9, \dots

, so the block is

09

. The leading zero is part of the block, not decoration.

Practice

5 \div 6 = 0.8 \overline{3}

. The digit

8

settles before the loop. How many digits come before the repeating block begins?

Show the solution

50 \div 6 = 8

with remainder

2

, so the first decimal digit is

8

. Then

20 \div 6 = 3

with remainder

2

again, so the

3

loops from there and

\frac{5}{6} = 0.8 \overline{3}

. Exactly

1

digit sits outside the bar. A decimal like this is called eventually repeating, since it settles into its loop after a short run-up instead of right away.

Practice

Long-divide

7 \div 30

to find

\frac{7}{30} = 0.2 \overline{3}

. The

2

settles first. What single digit is the repeating block?

Show the solution

70 \div 30 = 2

with remainder

10

, so the first decimal digit is

2

. Then

100 \div 30 = 3

with remainder

10

, the same remainder, so every later step writes another

3

. That gives

\frac{7}{30} = 0.2 \overline{3}

, so the repeating block is

3

. The

2

and the

5

30

produce the one settling digit, and the

3

produces the loop.

Practice

Let

x = 0. \overline{5}

. Multiply by

10

10 x = 5. \overline{5}

. Subtract:

9 x = 5

. Write

0. \overline{5}

as a fraction in lowest terms.

Show the solution

Let

x = 0.5555 \dots

, so

10 x = 5.5555 \dots

. Both numbers end in the identical endless string of fives, so subtracting cancels the tails and leaves

9 x = 5

. Divide by

9

to get

x = \frac{5}{9}

, and since

9 = 3 \times 3

shares nothing with

5

, it is already in lowest terms.

5 / 9

Practice

Using the block-over-nines shortcut,

0. \overline{2}

has a one-digit block, so it equals

\frac{2}{9}

. Write

0. \overline{2}

as a fraction in lowest terms.

Show the solution

The block is one digit, so it goes over a single nine, and

0. \overline{2} = \frac{2}{9}

. The factors of

2

are

1

and

2

, the factors of

9

are

1

3

, and

9

, so they share only

1

and the fraction cannot shrink.

2 / 9

That nine is really

10 - 1

. Setting

x = 0. \overline{2}

and subtracting gives

10 x - x = 9 x = 2

Practice

The decimal

0. \overline{81}

equals

\frac{81}{99}

. Reduce

\frac{81}{99}

to lowest terms. What fraction do you get?

Show the solution

A two-digit block sits over

99

, so

0. \overline{81} = \frac{81}{99}

. The greatest common factor of

81

and

99

9

, since

81 = 9 \times 9

and

99 = 9 \times 11

. Dividing top and bottom by

9

gives

9 / 11

. Checking,

9 \div 11 = 0.818181 \dots

. The block-over-nines shortcut gives a correct value but rarely a reduced one.

Practice

Write

0.0 \overline{45}

as a fraction in lowest terms. Use

1, 000 x

and

10 x

to cancel the repeating tails, giving

990 x = 45

, then reduce.

Show the solution

Let

x = 0.0454545 \dots

. Shift three places to land just before a fresh block,

1, 000 x = 45.4545 \dots

, and one place to clear the head zero,

10 x = 0.4545 \dots

. Both end in the same

. 4545 \dots

tail, so subtracting gives

990 x = 45

and

x = \frac{45}{990}

. Both share

45

, since

990 = 45 \cdot 22

, so

x = 1 / 22

. The leading zero appears once and never repeats, which is why a single shift will not line the tails up.

Practice

Let

x = 0. \overline{9}

. Multiply by

10

10 x = 9. \overline{9}

. Subtract and solve. What single whole number does

0. \overline{9}

equal?

Show the solution

Let

x = 0. \overline{9}

, so

10 x = 9. \overline{9}

. After the decimal point both carry the identical endless string of nines, so subtracting cancels it and leaves

9 x = 9

. Divide by

9

and

x = 1

. A second check agrees.

0. \overline{1} = \frac{1}{9}

, and nine copies give

\frac{9}{9} = 1

while

9 \cdot 0. \overline{1} = 0. \overline{9}

Practice

\frac{1}{13} = 0. \overline{076923}

, a block of

6

digits. Divide

100 \div 6

to place the position within the block. What is the 100th digit after the decimal point?

Show the solution

The block

076923

is six digits long, so divide.

100 = 6 \times 16 + 4

, since

6 \times 16 = 96

, which puts position

100

four digits into a fresh block. Reading

076923

, the 4th digit is

9

. The quotient

16

only counts the full blocks that went by. The remainder is what picks the digit.

Practice

\frac{4}{33} = 0. \overline{12}

, a block of

2

digits. Divide

99 \div 2

to find the position in the block. What is the 99th digit after the decimal point?

Show the solution

The block

12

is two digits long, so divide.

99 = 2 \times 49 + 1

, so position

99

lands on the first digit of the block, which is

1

. With a block of length

2

, odd positions all hold the

1

and even positions all hold the

2

, and

99

is odd.

Practice

Factor each denominator:

8 = 2^{3}

40 = 2^{3} \times 5

50 = 2 \times 5^{2}

24 = 2^{3} \times 3

. Which denominator contains a prime other than

2

5

, forcing its decimal to repeat?

Show the solution

Read the factorizations.

8 = 2^{3}

40 = 2^{3} \times 5

, and

50 = 2 \times 5^{2}

use only

2

s and

5

s, so all three terminate.

24 = 2^{3} \times 3

carries a

3

, which is neither

2

nor

5

, so the repeating denominator is

24

. Its decimal is

\frac{1}{24} = 0.041 \overline{6}

, and that stray

3

is the reason it never ends.

Practice

Find

0. \overline{81} - 0. \overline{36}

as a single fraction in lowest terms. Both two-digit repeaters sit over

99

. Subtract the numerators and reduce.

Show the solution

A two-digit block sits over

99

, so

0. \overline{81} = \frac{81}{99}

and

0. \overline{36} = \frac{36}{99}

. The denominators match, so subtract the numerators.

\frac{81}{99} - \frac{36}{99} = \frac{45}{99}

Both

45

and

99

are divisible by

9

, since

45 = 9 \times 5

and

99 = 9 \times 11

, so the answer is

5 / 11

. Checking,

5 \div 11 = 0. \overline{45}

, exactly the gap between the two decimals.

Repeating Decimals

Practice these ideas