MathsCBSE Class 9 · The World of Numbers

Decimal Expansions and Recurring Decimals as p/q: Class 9

Updated 17 October 2026 · Fact-checked

A fraction in lowest terms has a terminating decimal exactly when its denominator has no prime factor other than 2 and 5; otherwise the decimal repeats. To convert a recurring decimal to $\frac{p}{q}$, multiply by a power of 10 that moves one full block of repeating digits to the left of the point, subtract the original, and divide: for example $0.\overline{36} = \frac{36}{99} = \frac{4}{11}$. $0.\overline{9} = 1$.

The block 36 has 2 digits, so multiply by 100; subtracting cancels the endless tail, leaving 99x = 36 and x = 4/11.

Illustration drawn for QuestionBuddy.

Understand Decimal Expansions and Converting Recurring Decimals to p/q

Decimal expansion of a fraction. Dividing the numerator by the denominator gives either a terminating decimal ( $\frac{7}{8} = 0.875$ ) or a non-terminating repeating decimal ( $\frac{5}{12} = 0.41666 \dots = 0.41 \overline{6}$ ). The repeating block is the period.

Which fractions terminate? Write the fraction in lowest terms. The decimal terminates if and only if the denominator's prime factors are only $2$ and $5$ , because then the fraction can be written with denominator $1 0^{k}$ : $\frac{7}{8} = \frac{7 \times 125}{1000}$ and $\frac{13}{20} = \frac{65}{100}$ . So $\frac{3}{16}$ ( $16 = 2^{4}$ ) gives $0.1875$ and $\frac{11}{40}$ ( $40 = 2^{3} \times 5$ ) gives $0.275$ and both terminate. $\frac{7}{15}$ ( $15 = 3 \times 5$ ) gives $0.4666 \dots$ and $\frac{13}{28}$ ( $28 = 2^{2} \times 7$ ) gives $0.464285714285 \dots$ , which repeat. Reduce first: $\frac{6}{15} = \frac{2}{5}$ terminates even though $15$ has the factor $3$ .

**Converting a recurring decimal to $\frac{p}{q}$ .** Let $x$ be the number. Multiply by $1 0^{k}$ where $k$ is the length of the repeating block (and, if there is a non-repeating part, first multiply by a power of ten to put the block just after the point), then subtract so that the repeating part cancels.

Pure repeating: $x = 0. \overline{36}$ ; $100 x = 36. \overline{36}$ ; $100 x - x = 36$ , so $x = \frac{36}{99} = \frac{4}{11}$ . In general $0. \overline{a_{1} a_{2} \dots a_{k}} = \frac{a _{1} a _{2} \dots a _{k}}{1 0 ^{k} - 1}$ , so $0. \overline{47} = \frac{47}{99}$ and $0. \overline{123} = \frac{123}{999} = \frac{41}{333}$ .
Mixed: $x = 0.2 \overline{3}$ ; $10 x = 2. \overline{3}$ and $100 x = 23. \overline{3}$ ; $100 x - 10 x = 21$ , so $90 x = 21$ and $x = \frac{21}{90} = \frac{7}{30}$ .

A longer mixed example. For $x = 2.1 \overline{56}$ (that is $2.1565656 \dots$ ): $10 x = 21. \overline{56}$ and $1000 x = 2156. \overline{56}$ ; $1000 x - 10 x = 2135$ , so $990 x = 2135$ and $x = \frac{2135}{990} = \frac{427}{198}$ . Check: $\frac{427}{198} = 2.1565656 \dots$

**Is $0. \overline{9} = 1$ ?** Let $x = 0.999 \dots$ . Then $10 x = 9.999 \dots$ and $10 x - x = 9$ , so $9 x = 9$ and $x = 1$ . This is consistent with $\frac{1}{3} + \frac{2}{3} = 1$ and $0. \overline{3} + 0. \overline{6} = 0. \overline{9}$ . Similarly $0. \overline{4} + 0. \overline{5} = \frac{4}{9} + \frac{5}{9} = 1$ .

Cyclic patterns. $\frac{1}{7} = 0. \overline{142857}$ : the period is $6$ , and $\frac{2}{7}, \frac{3}{7}, \dots, \frac{6}{7}$ use the same digits in a shifted order: $\frac{3}{7} = 0. \overline{428571}$ and $\frac{6}{7} = 0. \overline{857142}$ , because $3 \times 142857 = 428571$ and $6 \times 142857 = 857142$ . Also $142857 \times 7 = 999999$ , which is why the block divided by $999999$ gives $\frac{1}{7}$ . The decimal of $\frac{22}{7}$ is $3. \overline{142857}$ , so the 100th digit after the point is the 4th digit of the block (since $100 = 16 \times 6 + 4$ ), which is $8$ .

Irrational decimals. A decimal like $0.101001000100001 \dots$ , whose pattern never repeats, is not a fraction and is irrational. A calculator's finite display (such as $0.14285714$ for $\frac{1}{7}$ ) is a truncation of the recurring decimal and does not mean the decimal terminates.

Method. To classify: reduce the fraction and examine the denominator. To convert: set $x$ , shift, subtract, solve, and reduce to lowest terms. Verify by division.

Decimal Expansions and Converting Recurring Decimals to p/q: formulas to remember

Terminating test: $\frac{p}{q} (lowest terms) terminates ⟺ q = 2^{a} 5^{b}$; No other prime factor.
Pure repeating decimal: $0. \overline{a_{1} \dots a_{k}} = \frac{a _{1} \dots a _{k}}{1 0 ^{k} - 1}$; Block over $9 \dots 9$ .
Method with $x$: $1 0^{k} x - x (or 1 0^{k + m} x - 1 0^{m} x)$; Subtract to cancel the repeating part.
0.999...: $0. \overline{9} = 1$; $9 x = 9$ .
Cycle of 1/7: $\frac{1}{7} = 0. \overline{142857}, 7 \times 142857 = 999999$; Period 6.
Irrational decimal: $no repeating block$; Not of the form $\frac{p}{q}$ .

How to solve Decimal Expansions and Converting Recurring Decimals to p/q questions

Classify by the denominator; convert by subtracting shifted copies.

1Reduce the fraction to lowest terms.
2Factorise the denominator: if only 2s and 5s, the decimal terminates; otherwise it repeats.
3To convert, call the decimal $x$ and find the length $k$ of the repeating block.
4Multiply by $1 0^{k}$ (after clearing any non-repeating digits with a power of ten) so that the repeating parts line up.
5Subtract the smaller multiple from the larger, solve for $x$ , and reduce.
6Check by dividing the numerator by the denominator.

Decimal Expansions and Converting Recurring Decimals to p/q in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Repeating decimal period

Suppose a calculator shows 22 divided by 7 as 3.142857142857 and a student wonders what the 100th digit after the decimal point is.

Find the 100th digit after the decimal point of $\frac{22}{7}$ .

Try it first, then show the solutionSolution

$\frac{22}{7} = 3. \overline{142857}$ with a repeating block of 6 digits.
$100 = 16 \times 6 + 4$ , so the 100th digit is the 4th digit of the block $142857$ , which is $8$ .

Answer: The 100th digit after the decimal point is 8.

Memory hook: Divide the position by the period; the remainder picks the digit.

Tricks to remember Decimal Expansions and Converting Recurring Decimals to p/q

Short memory aids. Say each one aloud once, then use it on a question the same day.

Only 2s and 5s end
A reduced fraction terminates only when the denominator is made of 2s and 5s.
Use it when: Terminating or repeating?
Block over nines
A pure repeating block of $k$ digits is the block divided by $k$ nines.
Use it when: Quick conversion.
Line up and subtract
Multiply so the repeating tails match, subtract and the tails vanish.
Use it when: Mixed recurring decimals.

Decimal Expansions and Converting Recurring Decimals to p/q: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Terminating or not · 3 marks · Short answerBoard level

Without dividing, decide whether each fraction has a terminating or a repeating decimal, and write the decimal: $\frac{7}{8}$ , $\frac{5}{12}$ , $\frac{13}{20}$ , $\frac{4}{7}$ .

Try it first, then show the solutionSolution

$8 = 2^{3}$ : terminating, $\frac{7}{8} = 0.875$ . (1)
$12 = 2^{2} \times 3$ has the factor $3$ : repeating, $\frac{5}{12} = 0.41 \overline{6}$ . (1)
$20 = 2^{2} \times 5$ : terminating, $\frac{13}{20} = 0.65$ . $7$ is a prime other than 2 and 5: $\frac{4}{7}$ repeats, $0. \overline{571428}$ . (1)

Answer: $\frac{7}{8}$ and $\frac{13}{20}$ terminate ( $0.875$ , $0.65$ ); $\frac{5}{12}$ and $\frac{4}{7}$ repeat.

Example 2: Pure recurring decimal · 3 marks · Short answerBoard level

Express $0. \overline{36}$ and $3. \overline{6}$ in the form $\frac{p}{q}$ .

Try it first, then show the solutionSolution

$x = 0. \overline{36}$ ; $100 x = 36. \overline{36}$ ; $100 x - x = 36$ , so $99 x = 36$ and $x = \frac{36}{99} = \frac{4}{11}$ . (1½)
$y = 3. \overline{6}$ ; $10 y = 36. \overline{6}$ ; $10 y - y = 33$ , so $9 y = 33$ and $y = \frac{33}{9} = \frac{11}{3}$ . (1½)

Answer: $0. \overline{36} = \frac{4}{11}$ and $3. \overline{6} = \frac{11}{3}$ .

Example 3: Mixed recurring decimals · 5 marks · Long answerHarder board level

Express $0.2 \overline{3}$ and $2.1 \overline{56}$ as fractions in lowest terms, and check one by division.

Try it first, then show the solutionSolution

$x = 0.2 \overline{3}$ : $10 x = 2. \overline{3}$ and $100 x = 23. \overline{3}$ . (1)
$100 x - 10 x = 21$ , so $90 x = 21$ and $x = \frac{21}{90} = \frac{7}{30}$ . (1)
$y = 2.1 \overline{56}$ : $10 y = 21. \overline{56}$ and $1000 y = 2156. \overline{56}$ . (1)
$1000 y - 10 y = 2135$ , so $990 y = 2135$ and $y = \frac{2135}{990} = \frac{427}{198}$ . (1)
Check: $\frac{7}{30} = 0.2333 \dots$ and $\frac{427}{198} = 2.1565656 \dots$ (since $198 \times 2 = 396$ and $\frac{31}{198} = 0.15656 \dots$ ). (1)

Answer: $0.2 \overline{3} = \frac{7}{30}$ and $2.1 \overline{56} = \frac{427}{198}$ .

Example 4: Cycle of 1/7 · 3 marks · Short answerHarder board level

(i) Show that $0.4 \overline{9} = 0.5$ . (ii) Given $\frac{1}{7} = 0. \overline{142857}$ , write $\frac{2}{7}$ and $\frac{5}{7}$ as decimals without dividing.

Try it first, then show the solutionSolution

(i) $x = 0.4 \overline{9}$ ; $10 x = 4. \overline{9}$ , and $4. \overline{9} = 5$ (since $0. \overline{9} = 1$ ), so $10 x = 5$ and $x = 0.5$ . (1)
(ii) $\frac{2}{7} = 2 \times 0. \overline{142857}$ and $2 \times 142857 = 285714$ , so $\frac{2}{7} = 0. \overline{285714}$ . (1)
$5 \times 142857 = 714285$ , so $\frac{5}{7} = 0. \overline{714285}$ . (1)

Answer: $0.4 \overline{9} = 0.5$ ; $\frac{2}{7} = 0. \overline{285714}$ ; $\frac{5}{7} = 0. \overline{714285}$ .

Example 5: A calculator display · 4 marks · Case-basedHOTS / competency

Case: Priya's calculator shows $\frac{1}{7}$ as $0.14285714$ and she says "it ends, so it is a terminating decimal". (i) Is she right? (ii) What is the repeating block, and how long is it? (iii) Show that the block $142857$ gives back $\frac{1}{7}$ . (iv) Find the 100th digit after the point of $\frac{22}{7}$ .

Try it first, then show the solutionSolution

(i) No. The display is cut off at eight digits; since $7$ is a prime factor other than 2 and 5, $\frac{1}{7}$ is a non-terminating repeating decimal. (1)
(ii) The block is $142857$ of length $6$ . (½)
(iii) $0. \overline{142857} = \frac{142857}{999999}$ and $999999 = 7 \times 142857$ , so it equals $\frac{1}{7}$ . (1½)
(iv) $\frac{22}{7} = 3. \overline{142857}$ ; $100 = 16 \times 6 + 4$ , so the 100th digit is the 4th digit of the block: $8$ . (1)

Answer: (i) no; (ii) 142857, length 6; (iii) verified; (iv) 8.

Where students lose marks in Decimal Expansions and Converting Recurring Decimals to p/q

The slips examiners see most often, and what to write instead.

Mistake: Testing an unreduced fraction
Reading the factor 3 in an unreduced denominator.
Wrong working
$\frac{6}{15}$ has $3$ in the denominator, so it repeats.
Right working
$\frac{6}{15} = \frac{2}{5} = 0.4$ terminates.
Do this: Reduce first.
Usually costs: 1 mark.
Mistake: Subtracting without lining up the repeating parts
Wrong multiples of 10.
Wrong working
$0.2 \overline{3}$ : $100 x - x = 23.1$ .
Right working
$100 x - 10 x = 21$ , so $x = \frac{21}{90}$ .
Do this: Multiply so the repeating tails are identical.
Usually costs: The conversion.
Mistake: Leaving the fraction unreduced
Stopping at $\frac{36}{99}$ .
Wrong working
$0. \overline{36} = \frac{36}{99}$ .
Right working
$\frac{36}{99} = \frac{4}{11}$ .
Do this: Divide by the common factor.
Usually costs: ½ mark.

Exam tips

Show the two equations (for example $10 x$ and $100 x$ ) in the conversion.
Reduce fractions to lowest terms before testing the denominator.
Use the period to find far digits: divide the position by the period length.
State that a finite calculator display does not prove a decimal terminates.

Practice questions on Decimal Expansions and Converting Recurring Decimals to p/q

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which of the following has a terminating decimal expansion?
(a) $\frac{1}{3}$
(b) $\frac{7}{16}$
(c) $\frac{5}{6}$
(d) $\frac{2}{7}$
Hint
Look for denominators with only 2s and 5s.
Answer
(b) $\frac{7}{16}$ . $16 = 2^{4}$ , so $\frac{7}{16} = 0.4375$ . The denominators $3$ , $6 = 2 \times 3$ and $7$ contain other primes.
Q2 · 3 marks · Short answerBoard level
Express $0. \overline{47}$ and $0. \overline{123}$ in the form $\frac{p}{q}$ in lowest terms.
Hint
Block over 99 and 999.
Answer
$0. \overline{47} = \frac{47}{99}$ (already in lowest terms). $0. \overline{123} = \frac{123}{999} = \frac{41}{333}$ (dividing by 3).
Q3 · 3 marks · Short answerHarder board level
Express $0.1 \overline{5}$ (that is $0.1555 \dots$ ) as a fraction in lowest terms.
Hint
Use $10 x$ and $100 x$ .
Answer
$x = 0.1555 \dots$ ; $10 x = 1.555 \dots$ ; $100 x = 15.555 \dots$ . $100 x - 10 x = 14$ , so $90 x = 14$ and $x = \frac{14}{90} = \frac{7}{45}$ . Check: $\frac{7}{45} = 0.1555 \dots$
Q4 · 5 marks · Long answerHarder board level
Which of $\frac{3}{16}$ , $\frac{7}{15}$ , $\frac{11}{40}$ , $\frac{13}{28}$ have terminating decimals? Give the reason for each and write the decimal expansion (the repeating block marked) for each.
Hint
Factorise the denominators and divide to get the decimals.
Answer
$\frac{3}{16}$ : $16 = 2^{4}$ , terminates, $0.1875$ . $\frac{7}{15}$ : $15 = 3 \times 5$ , has the factor $3$ , repeats, $0.4 \overline{6}$ . $\frac{11}{40}$ : $40 = 2^{3} \times 5$ , terminates, $0.275$ . $\frac{13}{28}$ : $28 = 2^{2} \times 7$ , has the factor $7$ , repeats, $0.46 \overline{428571}$ (digits $0.4642857142857 \dots$ ). So $\frac{3}{16}$ and $\frac{11}{40}$ terminate.
Q5 · 4 marks · Case-basedHOTS / competency
Case: a student adds recurring decimals. (i) Show that $0. \overline{3} + 0. \overline{6} = 1$ using fractions. (ii) Show that $0. \overline{4} + 0. \overline{5} = 1$ . (iii) Express $0. \overline{123}$ in lowest terms. (iv) Is $0.101001000100001 \dots$ rational? Give a reason.
Hint
$0. \overline{3} = \frac{1}{3}$ and $0. \overline{9} = 1$ .
Answer
(i) $\frac{1}{3} + \frac{2}{3} = 1$ . (ii) $\frac{4}{9} + \frac{5}{9} = 1$ . (iii) $\frac{123}{999} = \frac{41}{333}$ . (iv) No. The number of zeros between the ones keeps increasing, so there is no repeating block; it is a non-terminating, non-repeating decimal, so it is irrational.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.When does a fraction terminate?Show answer
When the reduced denominator has only the primes 2 and 5.
2. $0. \overline{9}$ equals?Show answer
1.
3. $0. \overline{45}$ as a fraction?Show answer
$\frac{45}{99} = \frac{5}{11}$ .
4.Period of $\frac{1}{7}$ ?Show answer
6.
5.Can an irrational number be a fraction?Show answer
No.

Decimal Expansions and Converting Recurring Decimals to p/q: frequently asked questions

Why does 0.999... equal 1?

Letting x = 0.999\ldots, 10x - x = 9 gives x = 1. There is no number between 0.\overline{9} and 1, so they are the same number.

How do I find the repeating block length?

Divide until a remainder repeats; the digits between the repeated remainders form the block. For denominators like 7 the period divides 6.

Do all fractions have an expansion that repeats or terminates?

Yes, because there are only finitely many possible remainders when dividing by q.