Converting Between Fractions and Decimals

Learning goals

Convert a decimal that ends to a fraction by reading its last place as the denominator
Divide numerator by denominator to turn any fraction into a decimal
Scale to a denominator of ten or a hundred instead of dividing
Predict whether a fraction terminates from the prime factors of its denominator
Explain why a fraction's decimal, when it does not end, must repeat, since a remainder must come back

A decimal that ends is already a fraction

You saw in the place-value lesson that the digits after the decimal point are tenths, hundredths, and thousandths. That single fact is the whole method for turning a decimal that ends into a fraction.

Take $0.49$ . The $4$ is four tenths and the $9$ is nine hundredths, so written out it is $\tfrac{4}{10} + \tfrac{9}{100}$ . Put both pieces over a hundred and add:

0.49 = \frac{40}{100} + \frac{9}{100} = \frac{49}{100}.

The two-digit string " $49$ " lands directly on top of $100$ , which is the shortcut. A decimal that ends always works this way, because its last place names a denominator every digit can be written over. The rule is short: the number of digits after the point is the number of zeros in the denominator.

0.7 = \frac{7}{10}, \qquad 0.49 = \frac{49}{100}, \qquad 0.013 = \frac{13}{1000}.

A fraction read straight off the digits may not yet be in lowest terms, so the conversion has a second half. A fraction is in lowest terms when its top and bottom share no factor larger than $1$ . Reduce by dividing the top and bottom by their greatest common factor, the same simplifying you did in the fractions chapter.

Worked example 1 Write $0.35$ as a fraction in lowest terms

There are two digits after the point, so the denominator is a hundred and the digits sit on top:

0.35 = \frac{35}{100}.

Now reduce. The greatest common factor of $35$ and $100$ is $5$ , so divide the top and bottom by $5$ :

\frac{35}{100} = \frac{35 \div 5}{100 \div 5} = \frac{7}{20}.

Since $7$ and $20$ share no common factor larger than $1$ , the fraction $\tfrac{7}{20}$ is in lowest terms, and that is the answer.

A number with a whole part converts the same way: leave the whole part alone and convert only the decimal piece. The result is a mixed number you can read off directly. For instance $2.6$ is $2\tfrac{6}{10} = 2\tfrac{3}{5}$ .

Three of four equal parts have the same shaded length as 0.75 of the whole: the decimal 0.75 and the fraction 3/4 mark the same amount.

Check your understanding

Write $0.16$ as a fraction in lowest terms.

\tfrac{4}{25}

\tfrac{16}{100}

\tfrac{8}{50}

\tfrac{1}{6}

A fraction is a division waiting to happen

Going the other direction, from a fraction to a decimal, rests on a fact you have known since the fractions chapter. The fraction bar means divide.

Take $\tfrac{3}{8}$ . The bar says divide $3$ by $8$ . Since $8$ does not go into $3$ , write $0$ in the ones place, bring the point up, and keep dividing into $3.000$ , exactly as you did with whole-number long division, appending a zero each time there is still a remainder to carry:

\frac{3}{8} = 3 \div 8 = 0.375.

Every fraction goes the same way: it is the numerator divided by the denominator, so carry out that division and keep the decimal point in the quotient lined up above the point in the dividend. The remainder does not always clear, and the last section of this lesson says which fractions it clears for.

Dividing 7 by 8: the quotient's decimal point sits directly above the dividend's, so each digit of the answer lands in its own place, ones over ones and tenths over tenths.

Why dividing the numerator by the denominator gives the decimal#

Run the argument on $\tfrac{9}{4}$ . Split $9$ wholes equally among $4$ groups. Collecting all four of those shares has to rebuild the $9$ you started with. So the value of $\tfrac{9}{4}$ is the number that gives $9$ when you take $4$ of it, which is exactly what $9 \div 4$ asks for. Long division answers that a place at a time. Four goes into $9$ twice, leaving $1$ over. That $1$ becomes $10$ tenths, and four goes into $10$ twice, leaving $2$ tenths. Those $2$ tenths become $20$ hundredths, and four goes into $20$ exactly five times. The quotient is $2.25$ , and four copies of $2.25$ do rebuild the $9$ .

Nothing in that depended on the digits $9$ and $4$ : the same reasoning turns any fraction into the division that produces its decimal.

Worked example 2 Write $\tfrac{7}{8}$ as a decimal

The bar means divide, so compute $7 \div 8$ . Eight does not fit into $7$ , so the quotient starts with $0$ and a point, and the division runs into $7.000$ .

Eight into $70$ tenths goes $8$ times ( $8 \times 8 = 64$ ), leaving $6$ tenths, which become $60$ hundredths. Eight into $60$ goes $7$ times ( $56$ ), leaving $4$ hundredths, which become $40$ thousandths. Eight into $40$ goes exactly $5$ times with no remainder:

\frac{7}{8} = 7 \div 8 = 0.875.

The remainder reached $0$ , so the decimal terminates. A quick sanity check: $\tfrac{7}{8}$ is just under $1$ , and $0.875$ is just under $1$ , so the size is right.

Check your understanding

Write $\tfrac{5}{8}$ as a decimal by dividing.

0.625

1.6

0.58

0.85

A shortcut when the denominator scales cleanly

Long division always works, but you can often skip it. The fraction $\tfrac{3}{4}$ is the classic case. A hundred is $4 \times 25$ , so multiply the top and bottom by $25$ :

\frac{3}{4} = \frac{3 \times 25}{4 \times 25} = \frac{75}{100} = 0.75.

That step uses equivalent fractions: multiplying the top and bottom by the same number does not change the value. So $\tfrac{75}{100}$ is still three quarters, now written with a bottom you can read as hundredths. Once the bottom is a hundred, the top is simply the hundredths, and the decimal is read off directly. This is the same conversion as the long division $3 \div 4 = 0.75$ , just reached by scaling instead of dividing.

The general rule follows. If the denominator divides evenly into ten, a hundred, a thousand, or any longer string of zeros, then scaling the fraction to that denominator turns it into a decimal. You choose the multiplier that makes the denominator into that power of ten. You can then read the decimal straight off, with no division of the numerator at all. Ten, a hundred, and a thousand are simply the first three worth trying by hand; if none of them works, a larger one sometimes still does; the last section of this lesson gives the test that settles the question for good.

Choosing the multiplier is the whole skill here. Take $\tfrac{3}{5}$ : multiplying top and bottom by $2$ gives $\tfrac{6}{10}$ , which reads straight off as $0.6$ . Multiplying by $3$ or by $4$ instead gives $\tfrac{9}{15}$ and $\tfrac{12}{20}$ , equally true names for the same amount. But $15$ and $20$ are not ten, a hundred, or a thousand, so neither can be read off as a decimal without more scaling.

Worked example 3 Write $\tfrac{7}{20}$ as a decimal by scaling

Look at the denominator $20$ . It does not divide $10$ , but it does divide $100$ , because $100 = 20 \times 5$ . So multiply the top and bottom by $5$ to make the denominator a hundred:

\frac{7}{20} = \frac{7 \times 5}{20 \times 5} = \frac{35}{100}.

A denominator of a hundred means two decimal places, and the top " $35$ " gives those two digits:

\frac{7}{20} = \frac{35}{100} = 0.35.

This is the reverse of Worked Example 1, where $0.35$ reduced to $\tfrac{7}{20}$ , a useful check that the two directions agree.

The picture below shows why scaling does not change the amount: cutting each of the five fifths into two makes ten tenths. The shaded portion is unchanged, so $\tfrac{2}{5}$ and $\tfrac{4}{10} = 0.4$ are the same number.

Splitting each fifth into two equal parts turns 2/5 into 4/10 without changing the shaded amount, and 4/10 is the decimal 0.4.

Check your understanding

Which is the quickest correct way to convert $\tfrac{9}{25}$ to a decimal, and what is the result?

A Scale to a denominator of a hundred to get

\tfrac{36}{100} = 0.36

B Scale to a denominator of ten to get

0.036

C Divide

25 \div 9

to get

0.36

D It cannot be written as a terminating decimal

Terminating or repeating: it depends on the denominator

Some fractions, like $\tfrac{3}{8} = 0.375$ , give a decimal that ends. Others do not. Try $\tfrac{1}{3}$ with long division: $1 \div 3$ gives $0.333\ldots$ , where the remainder is $1$ at every step, so the $3$ s never stop. We write a repeating decimal with a bar over the block of digits that repeats forever:

\frac{1}{3} = 0.333\ldots = 0.\overline{3}, \qquad \frac{1}{6} = 0.1666\ldots = 0.1\overline{6}.

You can predict which happens before doing any division by looking at the prime factors of the denominator. The dividing line is clean: a fraction in lowest terms terminates exactly when its denominator has no prime factor other than $2$ and $5$ . If any other prime, such as $3$ or $7$ , survives in the denominator, the decimal repeats.

Why only the primes 2 and 5 give a terminating decimal#

First see why $2$ and $5$ are the special primes. Start with the denominators a terminating decimal can have. A decimal with one place is a count of tenths. A decimal with two places is a count of hundredths. Three places count thousandths. So the denominators that can occur are ten, a hundred, a thousand, and so on. They are precisely the ” $1$ followed by zeros” numbers. A decimal therefore terminates exactly when the fraction can be rewritten with one of those denominators. Now factor those denominators: ten is $2 \times 5$ , a hundred is $2 \times 2 \times 5 \times 5$ , a thousand is three $2$ s times three $5$ s. Every one of them is built from nothing but $2$ s and $5$ s. Now think about a fraction in lowest terms. Such a fraction can be scaled up to one of these denominators only if its denominator already divides one of them. And a denominator divides one of them only when it is built from nothing but $2$ s and $5$ s. For example $\tfrac{7}{20}$ has denominator $20 = 2 \times 2 \times 5$ , all $2$ s and $5$ s, and indeed $\tfrac{7}{20} = \tfrac{35}{100}$ terminates. A denominator does not need both primes, and $\tfrac{1}{8}$ shows why. Its denominator is $8 = 2 \times 2 \times 2$ , with no $5$ at all, yet $1000 \div 8 = 125$ , so $8$ divides a thousand and $\tfrac{1}{8} = \tfrac{125}{1000} = 0.125$ . But $\tfrac{1}{6}$ has denominator $6 = 2 \times 3$ . That stubborn factor of $3$ is the obstacle. It can never be turned into a string of $2$ s and $5$ s. So no ” $1$ followed by zeros” number is a multiple of $6$ . The decimal for $\tfrac{1}{6}$ therefore cannot terminate.

Why a fraction's decimal, when it does not end, must repeat#

When dividing one whole number by another does not stop, the digits do not wander on forever without a pattern; they settle into a repeating block. Watch that happen on $\tfrac{4}{11}$ , where the starting remainder comes back after two steps. Dividing $40$ by $11$ gives the digit $3$ and leaves a remainder of $7$ . Dividing $70$ by $11$ gives the digit $6$ and leaves a remainder of $4$ . That $4$ is the remainder the division started with, so the next two digits have to be $3$ and $6$ again. Only the remainders $4$ and $7$ ever appeared, and the return of the $4$ forces that same block of digits again. The decimal is $0.\overline{36}$ , and it cycles because a remainder came back.

Nothing about $11$ was essential to that. In any long division, every remainder is a whole number smaller than the divisor, so there are only finitely many remainders the division can ever produce. If a remainder of $0$ never turns up, the division runs forever while drawing from that same short, fixed list, so eventually some remainder has to repeat. The instant it does, the digits that follow it must match the digits that followed it the first time, because each digit is decided entirely by the current remainder. From there the same block of digits cycles forever. That is why dividing one whole number by another gives only one of two kinds of decimal: it ends, or it settles into a repeating block. Nothing else can happen.

Check your understanding

Dividing $5$ by $6$ , the remainder $2$ shows up twice in a row. What does that guarantee about the decimal?

A The digit that followed the first

2

must keep following every later

2

, forever B The division is about to terminate C The decimal must be

0.5\overline{2}

specifically D Nothing extra; each later digit still has to be checked on its own

To apply the test, reduce the fraction to lowest terms first, then prime factorize the denominator. Reducing first matters: $\tfrac{6}{12}$ looks like it has the bad factor $3$ . But it reduces to $\tfrac{1}{2}$ , whose denominator is just $2$ , so it terminates ( $0.5$ ).

Worked example 4 Without dividing, decide whether $\tfrac{9}{40}$ and $\tfrac{5}{12}$ terminate

Apply the test to each: reduce to lowest terms, then check the prime factors of the denominator.

For $\tfrac{9}{40}$ , the numerator $9$ and denominator $40$ share no common factor, so it is already in lowest terms. Prime factorize the denominator:

40 = 2 \times 2 \times 2 \times 5.

No prime other than $2$ and $5$ appears, so $\tfrac{9}{40}$ terminates. Scaling confirms it, since $40 \times 25 = 1000$ , giving $\tfrac{9}{40} = \tfrac{225}{1000} = 0.225$ .

For $\tfrac{5}{12}$ , the fraction is already in lowest terms, and

12 = 2 \times 2 \times 3.

The factor of $3$ is neither $2$ nor $5$ , so $\tfrac{5}{12}$ repeats. Dividing confirms it: $5 \div 12 = 0.41666\ldots = 0.41\overline{6}$ .

Check your understanding

Which of these fractions gives a repeating decimal?

\tfrac{7}{30}

\tfrac{7}{8}

\tfrac{7}{50}

\tfrac{7}{16}

Common mistakes

Dividing the wrong way. The fraction $\tfrac{a}{b}$ is $a \div b$ , top divided by bottom, no matter which number is bigger. For $\tfrac{3}{4}$ that is $3 \div 4 = 0.75$ , not $4 \div 3$ ; for $\tfrac{9}{4}$ it is still numerator over denominator, $9 \div 4$ , even though $9$ is the larger number. Reversing the order is the most common slip.
Forgetting to reduce a decimal-to-fraction answer. $0.35 = \tfrac{35}{100}$ is correct but not finished. Divide by the greatest common factor to reach $\tfrac{7}{20}$ .
Miscounting decimal places. $0.013$ has three digits after the point, so the denominator is a thousand, giving $\tfrac{13}{1000}$ , not $\tfrac{13}{100}$ . The number of digits is the number of zeros.
Testing for termination before reducing. $\tfrac{6}{12}$ seems to have a factor of $3$ in the bottom, but it reduces to $\tfrac{1}{2}$ and terminates. Always reduce to lowest terms before prime factorizing the denominator.
Writing a repeating decimal as if it ends. Rounding $\tfrac{1}{3}$ to $0.33$ is an approximation, not an equality. The exact value is $0.\overline{3}$ , with the $3$ s continuing forever.

Key takeaways

A decimal that ends is already a fraction: the digits after the point sit over ten, a hundred, or a thousand (one zero per digit). So $0.49 = \tfrac{49}{100}$ ; then reduce with the GCF, for example $0.35 = \tfrac{7}{20}$ .
A fraction is a division: the bar means $a \div b$ , so dividing the numerator by the denominator gives the decimal, for example $\tfrac{3}{8} = 3 \div 8 = 0.375$ .
When the denominator divides ten, a hundred, a thousand, or any longer string of zeros, scale to that denominator and read the decimal off instead of dividing, for example $\tfrac{3}{4} = \tfrac{75}{100} = 0.75$ .
A fraction in lowest terms terminates exactly when its denominator has no prime factor other than $2$ and $5$ . Any other prime in the reduced denominator makes the decimal repeat, written with an overline like $\tfrac{1}{3} = 0.\overline{3}$ .
A fraction’s decimal, when it does not end, must repeat, because long division has only finitely many possible remainders. So one remainder is forced to recur, and the digits cycle from there.

Practice

Multiple Choice Questions (MCQ)

Progressively harder sets of questions. Each opens on its own page.

Level 1 Foundational 12 questions Start → Level 2 Intermediate 12 questions Start → Level 3 Challenge 12 questions Start →

Core practice

Practice problems at the level of the course, to be worked out on paper. Hints one at a time, then the answer or the full worked solution, with your progress kept in this browser.

More practice (optional)

Extra sets, as hard as the Challenge set. Each one opens on its own page.

More resources (optional)

Other explanations of this lesson, if you want a second take.

Converting decimals to fractions (Khan Academy)
Converting fractions to decimals (Khan Academy)
Converting fractions to decimals (Math is Fun)
Converting decimals to fractions (Math is Fun)
Prealgebra: decimals and fractions (free textbook) (OpenStax)

Go deeper (optional)

You can skip this and keep going. Read it if you want to know more.

Why the fraction bar always means divide

The lesson showed the reasoning on $\tfrac{9}{4}$ : sharing $9$ wholes into $4$ equal groups has to rebuild the $9$ , which is exactly what $9 \div 4$ means. Here is that same argument written with letters, so it covers every fraction at once, together with the reason long division actually builds the decimal one place at a time.

Why dividing the numerator by the denominator gives the decimal, for any fraction#

Call the decimal you are looking for $d$ . By the meaning of the fraction bar, $\tfrac{a}{b}$ is the number $d$ for which $b \times d = a$ : split $a$ into $b$ equal shares, and collecting all $b$ of those shares must rebuild $a$ . But “the number that gives $a$ when multiplied by $b$ ” is the definition of $a \div b$ . So the value of the fraction and the result of the division $a \div b$ are the same number, and computing one computes the other.

Long division produces that number one decimal place at a time. How many times does the divisor fit into the current amount? You record that digit in the quotient. Then you carry the leftover remainder to the next place by attaching a zero, which makes the remainder ten times as many of the next smaller place. Every digit you write is the correct count for its place value. Digit by digit, that quotient is the decimal expansion of $a \div b$ , so it is the decimal expansion of $\tfrac{a}{b}$ as well.

Counting exactly how soon a remainder must repeat

The lesson gives the reason a repeat is forced: only finitely many remainders exist, so one has to recur. Here that count is made exact, for any denominator $b$ .

A guaranteed bound on how long a repeating block can be#

Take the long division of $a$ by $b$ . Every step produces a remainder, and each remainder is a whole number from $0$ up to $b - 1$ , so the division has only $b$ possible remainders to offer. If a remainder of $0$ ever appears, the division stops and the decimal terminates. Now suppose $0$ never appears. That leaves only the $b - 1$ values from $1$ up to $b - 1$ , so within at most $b$ steps some earlier remainder has to show up again: there are more steps than there are available remainders, so there is nowhere else for the remainders to go. The moment a remainder repeats, the digits that follow it must match the digits that followed it the first time, because each digit is decided entirely by the current remainder. So the repeating block can never be longer than $b - 1$ digits. For $\tfrac{4}{11}$ , $b - 1 = 10$ , and the actual block, $36$ , is well inside that bound.

A bit of history (optional)

Divide $1$ by $7$ and the digits keep coming. Nothing on the page promises that they will ever settle into a loop. You could grind out a hundred places and still be guessing.

The proof that they must comes from an idea small enough to hold in one hand, one that mathematicians eventually named the pigeonhole principle, or the drawer principle: put more socks into drawers than there are drawers, and some drawer ends up holding two. The name is often linked to the German mathematician Peter Gustav Lejeune Dirichlet, who used the idea in the early 1800s. It sounds too obvious to be worth naming, yet it decides questions that nothing else can reach.

Now look again at your long division. Dividing by $7$ , every remainder has to be one of $0, 1, 2, 3, 4, 5, 6$ . That is seven drawers, and the division keeps handing you socks. Some remainder must come back a second time. The moment it does, the same digits that followed it the first time must follow it again.

That is the whole reason a fraction’s decimal, when it does not end, must repeat. It is the argument you worked through in this lesson.