GCSE · Number

Recurring Decimals and Fractions

Type 1 ÷ 8 into a calculator and you get 0.125. Type 1 ÷ 3 and you get 0.3333333, and the threes would go on for ever if the screen were big enough. Why do some fractions stop and others repeat? Which ones stop? And how do you turn 0.272727… back into a fraction? This page answers all three, and finishes by proving something that surprises everyone: 0.999… is exactly 1.

DfE GCSE subject content (Number): work interchangeably with terminating decimals and their corresponding fractions (such as 3.5 and 7/2 or 0.375 and 3/8); Higher tier: change recurring decimals into their corresponding fractions and vice versa

What you need from KS3

1 · Terminating, recurring and dot notation

A terminating decimal stops: 0.5, 0.375, 2.04. A recurring decimal has a block of digits that repeats for ever: 0.333…, 0.181818…, 0.1666…. The repeating block can start straight after the point, or after a few non-repeating digits.

Writing "…" is vague, so we use dot notation: put a dot over the first and last digit of the repeating block (one dot if the block is a single digit). So 0.181818… = 0.18 and 0.1666… = 0.16. Build your own: type the digits that don't repeat and the block that does.

Dot notation traps: for 0.142857142857… the dots go over the 1 and the 7 only (0.142857), not over every digit. And 0.16 (= 0.1666…) is different from 0.16 (= 0.161616…).

2 · Fraction to decimal: short division

A fraction is a division: 38 means 3 ÷ 8. Write 3 as 3.000… and divide with the bus stop method. Each step, the remainder is carried to the next 0, making 10 times the remainder. Step through it.

For 38 the remainder reaches 0 and the division stops: 0.375, a terminating decimal. Now try 17 or 56: the remainder never reaches 0.

3 · Why some decimals recur: remainders repeat Higher

When you divide by 7, the remainder at each step must be one of 0, 1, 2, 3, 4, 5 or 6. If it ever hits 0, the decimal stops. If it never hits 0, there are only six other possible remainders, so within six steps a remainder must come round again. And once a remainder repeats, every step after it repeats too, because each step depends only on the remainder before it. That's why the digits cycle. The "remainder clock" shows the journey.

So dividing by d, the repeating block can be at most d − 1 digits long. 1/7 uses all six possible non-zero remainders, so its block is six digits: 0.142857. 1/13 has a six-digit block too, but uses only six of its twelve possible remainders (try 2/13 to meet the other six).

4 · Which fractions terminate?

A terminating decimal is a whole number over 10, 100, 1000 …: 0.375 = 3751000. And 10 = 2 × 5, 100 = 2² × 5², 1000 = 2³ × 5³. These powers of 10 have no prime factors except 2 and 5. So:

A fraction in its simplest form terminates exactly when its denominator has no prime factors other than 2 and 5.

If the denominator is only 2s and 5s, you can always top it up to a power of 10: 38 = 32³ = 3 × 5³2³ × 5³ = 3751000. A factor of 3, 7, 11 … can never be turned into a power of 10, so those decimals recur. Tap any denominator on the grid, or type a fraction.

Exam trap: simplify first. 312 has a 3 in its denominator (12 = 2² × 3), but 312 = 14, which terminates (0.25). The test only works once the fraction is in its simplest form. Try 3/12 and 9/75 above.

5 · Recurring decimal to fraction Higher

Here's the clever part. Call the decimal x. Multiply by 10, 100 or 1000 so the repeating block shifts exactly one block to the left. Then x and the new number have identical tails, and subtracting makes the endless tail vanish. Rule of thumb: block length 1 → × 10, block length 2 → × 100, block length 3 → × 1000.

Notice the shortcut hiding in the answers: a block of 1 digit goes over 9, 2 digits over 99, 3 digits over 999. So 0.27 = 2799 = 311. But in an exam that says "prove" or "show that", you must write out the algebra.

6 · When the repeat starts later Higher

For a decimal like 0.23 = 0.2333…, multiplying x by 10 doesn't line the tails up: x = 0.2333… has a 2 before the threes. The trick is to pick two multiples of x whose tails match. Choose any two multipliers and see whether the parts after the point line up exactly.

pick two:

The reliable recipe: multiply by 10 for each non-repeating digit to bring the repeat to just after the point; then multiply by a further 10, 100, … (one 0 per digit in the block) to shift one whole block. For 0.2333…: 10x = 2.333… and 100x = 23.333…, so 90x = 21. Common mistakes: subtracting x from 10x when the tails don't match (you get 9x = 2.1, which isn't a whole number: a sure sign something's wrong), and forgetting to simplify the final fraction.

7 · Proof: 0.999… = 1 Higher

Use exactly the same method on 0.9:

x = 0.9999…
10x = 9.9999…
10x − x = 9.9999… − 0.9999…
9x = 9
x = 1

Most people's instinct says 0.999… must be "just below" 1. But think about the gap. Slide to add nines. The gap shrinks by a factor of 10 every time, and with infinitely many nines there is no gap left at all: no positive number is smaller than every 0.000…01. Two numbers with nothing between them are the same number.

One more way to see it: 13 = 0.333…, and 3 × 13 = 1. But 3 × 0.333… = 0.999…. So 0.999… = 1.

8 · Practice

New questions each time. Higher-tier questions are tagged.

Worked examples

Write 58 as a decimal
Set up: 5 ÷ 8, written as 5.000 ÷ 8.
Divide: 8 into 5 goes 0 times, remainder 5. 8 into 50 = 6 remainder 2. 8 into 20 = 2 remainder 4. 8 into 40 = 5 remainder 0.
Remainder 0, so it stops: 58 = 0.625. (Check: 8 = 2³, only 2s, so it had to terminate.)
Write 211 as a recurring decimal Higher
Divide 2.0000… by 11: 11 into 20 = 1 remainder 9. 11 into 90 = 8 remainder 2.
Remainder 2 is where we started (the 2 in 2.000…), so the digits 1, 8 repeat for ever: 0.181818…
Dot notation: 0.18
Convert 0.45 to a fraction in its simplest form Higher
Let x = 0.454545…
Block length 2, so × 100: 100x = 45.454545…
Subtract: 100x − x = 45.4545… − 0.4545…, so 99x = 45.
Divide and simplify: x = 4599 = 511 (÷ 9 top and bottom).
Prove algebraically that 0.25 = 2390 Higher
Let x = 0.2555…
Move the repeat to just after the point: 10x = 2.555…
Shift one more block (length 1): 100x = 25.555…
Subtract: 100x − 10x = 25.555… − 2.555…, so 90x = 23.
Divide: x = 2390 as required. (23 is prime and doesn't divide 90, so it's already simplest.)
Convert 0.123 to a fraction Higher
Let x = 0.123123123…; the block has 3 digits, so × 1000: 1000x = 123.123123…
Subtract: 999x = 123, so x = 123999.
Simplify (both divisible by 3): 41333.
Which of 740, 512 and 975 are terminating decimals?
740: already simplest; 40 = 2³ × 5. Only 2s and 5s, so it terminates: 0.175.
512: simplest; 12 = 2² × 3. The 3 means it recurs: 0.41666… = 0.416.
975: 75 = 3 × 5², which has a 3, but simplify first: 975 = 325, and 25 = 5². It terminates: 0.12.

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