N10 - Recurring decimals and fractions

A recurring decimal can be an exact fraction even though its digits never stop. Learn why division repeats, how multiplication and subtraction recover the fraction, and how to handle a non-repeating start before the repeating digits.

Why decimal digits repeat (Higher)

Is $0.9999\ldots$ , with nines continuing forever, a little less than one or exactly one? Finite and infinite decimals behave differently. Start by seeing how an infinite pattern arises.

You already know

A fraction means numerator divided by denominator. A terminating decimal has finitely many decimal places, and a simplified fraction terminates when its denominator has only prime factors two and five.

In division, bring down a zero at each decimal place. The remainder determines the next digit. If the remainder is zero, division ends; if an earlier non-zero remainder returns, everything from there repeats.

Recurring decimal

A decimal with a digit or block of digits repeating forever. One dot marks a single repeating digit: $0.1\dot6=0.1666\ldots$ . Dots above the first and last digits mark a whole block: $0.\dot2\dot7=0.272727\ldots$ , whereas $0.2\dot7=0.277777\ldots$ .

The dots describe exact values. A finite calculator display cannot show infinitely many digits, so its last visible digit does not prove termination.

Worked example 1: division that ends

Question. Convert $\frac18$ to a decimal using division. Explain why it ends.

Plan. Track the digit and remainder at each decimal place.

\begin{aligned} 10&=8\times1+2 && \text{tenths digit }1 \\ 20&=8\times2+4 && \text{hundredths digit }2 \\ 40&=8\times5+0 && \text{thousandths digit }5 \end{aligned}

Why? A zero remainder leaves nothing to divide, so there are no further non-zero digits.

Answer. $\frac18=0.125$ , a terminating decimal.

Check. $0.125\times8=1$ . Similarly, $\frac38=3\times0.125=0.375$ .

Change only the denominator, from eight to six. Watch what happens to the remainder.

Worked example 2: division that repeats

Question. Convert $\frac16$ to a decimal and explain which digit repeats.

Plan. Continue the division until the remainder is zero or repeats.

\begin{aligned} 10&=6\times1+4 && \text{first digit }1 \\ 40&=6\times6+4 && \text{next digit }6 \\ 40&=6\times6+4 && \text{same remainder returns} \end{aligned}

Why? Remainder four always leads to digit six and remainder four again.

Answer. $\frac16=0.1\dot6$ .

Check. Six copies of $0.1666\ldots$ give one; a finite value such as $0.1666$ is only approximate.

For denominator $b$ , only the remainders from zero to $b-1$ are possible. Division must therefore stop or eventually revisit a remainder. That is why every fraction has a terminating or eventually recurring decimal. Simplify before using denominator factors: $\frac3{12}=\frac14$ terminates even though twelve contains a factor three.

Watch out

$0.\dot2\dot7$ repeats both digits, while $0.2\dot7$ repeats only seven. Expand several digits before starting algebra.

Change only the denominator again, from six to twelve.

Converting a repeated block into a fraction (Higher)

Multiplying by a power of ten can put identical infinite tails directly underneath each other. Subtraction then removes the entire tail exactly.

Worked example 3: a two-digit repeated block

Question. Express $0.\dot2\dot7$ as a fraction in simplest form. Show your algebraic working.

Plan. Name the value $x$ , multiply by a hundred, then subtract the original equation.

\begin{aligned} x&=0.272727\ldots && \text{define the value} \\ 100x&=27.272727\ldots && \text{shift one full block} \\ 99x&=27 && \text{subtract }x\text{ from }100x \\ x&=\frac{27}{99}=\frac3{11} && \text{divide and simplify} \end{aligned}

Why? Both decimal tails are exactly $0.272727\ldots$ ; subtracting aligned tails leaves zero.

Answer. $\frac3{11}$ .

Check. Division gives $3\div11=0.272727\ldots$ .

Change only the repeated block, from twenty-seven to zero-three. Its leading zero still counts as a digit.

Worked example 4: a repeated block containing zero

Question. Express $0.\dot0\dot3$ as a fraction in simplest form. Show your algebraic working.

Plan. Shift the two-digit block using a hundred again.

\begin{aligned} x&=0.030303\ldots && \text{write the whole block} \\ 100x&=3.030303\ldots && \text{shift two places} \\ 99x&=3 && \text{subtract aligned tails} \\ x&=\frac3{99}=\frac1{33} && \text{simplify} \end{aligned}

Why? Multiplying by ten would misalign the zero and three, so the tails would not cancel.

Answer. $\frac1{33}$ .

Check. $1\div33=0.030303\ldots$ , not $0.3333\ldots$ .

A one-digit block needs ten: for $y=0.\dot6$ , $10y-y=6$ , so $y=\frac69=\frac23$ . A three-digit block needs a thousand: $0.\dot1 2\dot5=\frac{125}{999}$ . An integer part works too: if $z=1.\dot2$ , then $10z-z=12.222\ldots-1.222\ldots=11$ , so $z=\frac{11}{9}$ .

Now the opening surprise follows from exactly the same reasoning. Let $u=0.\dot9$ . Then $10u-u=9$ , hence $9u=9$ and $u=1$ . Every finite string of nines is below one, but the infinite recurring decimal is another name for one. There is no final decimal place hiding a gap.

Watch out

Wrong: $0.\dot0\dot3=\frac39$ . The zero belongs to the two-digit block, so use $100x-x$ , giving $\frac3{99}$ . For “Show algebraically”, display that alignment and subtraction; the final fraction alone is not a demonstration.

Change only the repeated block in Example 4.

Decimals with an initial non-repeating section (Higher)

When some digits do not repeat, shift past them before lining up the recurring tails. For $0.1\dot6$ , the one is fixed and only the six repeats.

Worked example 5: one fixed digit before recurrence

Question. Express $0.1\dot6$ as a fraction in simplest form using algebra.

Plan. Multiply by ten to move past the fixed digit, then by a hundred to move one repeating digit as well.

\begin{aligned} x&=0.1666\ldots && \text{name the value} \\ 100x&=16.666\ldots && \text{shift two places} \\ 10x&=1.666\ldots && \text{shift one place} \\ 90x&=15 && \text{subtract aligned tails} \\ x&=\frac{15}{90}=\frac16 && \text{simplify} \end{aligned}

Why? Subtract $10x$ , rather than $x$ , because the recurring tails then begin at the same decimal place.

Answer. $\frac16$ .

Check. The division in Part 1 gives $0.1\dot6$ .

Change only the repeating digit, from six to seven. The alignment stays the same.

Worked example 6: keeping the alignment

Question. Express $0.1\dot7$ as a fraction in simplest form using algebra.

Plan. Use the same two multipliers, a hundred and ten.

\begin{aligned} x&=0.1777\ldots && \text{expand the notation} \\ 100x&=17.777\ldots && \text{move a repeated digit} \\ 10x&=1.777\ldots && \text{move the fixed digit} \\ 90x&=16 && \text{subtract} \\ x&=\frac{16}{90}=\frac8{45} && \text{simplify} \end{aligned}

Why? The denominator is $100-10=90$ , not ninety-nine; the initial digit does not repeat.

Answer. $\frac8{45}$ .

Check. $8\div45=0.1777\ldots$ , a little greater than one sixth.

The same alignment handles a longer block and an integer part. For $y=2.3\dot1\dot8$ , one digit is fixed and two recur:

\begin{aligned} 1000y&=2318.181818\ldots && \text{shift past fixed and repeated digits} \\ 10y&=23.181818\ldots && \text{shift past the fixed digit} \\ 990y&=2295 && \text{subtract} \\ y&=\frac{2295}{990}=\frac{51}{22} && \text{simplify} \end{aligned}

Division checks $51\div22=2.3181818\ldots$ . For a negative recurring decimal, convert its positive magnitude and restore the minus sign, or retain the sign throughout every equation.

Watch out

Wrong: $0.1\dot7=\frac{17}{99}$ . That fraction repeats both digits. Align the tails first; the correct fraction is $\frac8{45}$ .

Change only the fixed digit from one to two in your matched question.

Quick Recap

Division stops at remainder zero; a repeated remainder makes the later digits repeat.
Pearson's dots identify exactly which digits recur, including zeros inside a block.
Multiply to align infinite tails, subtract them, then simplify the resulting fraction.
For a fixed initial section, shift past it before aligning the repeated block.

The opening decimal $0.9999\ldots$ equals one exactly; a finite string of nines does not.

Last Accessed

Papers 1, 2 and 3

N10 - Recurring decimals and fractions

Why decimal digits repeat (Higher)

Converting a repeated block into a fraction (Higher)

Decimals with an initial non-repeating section (Higher)

Quick Recap

Learn faster with Nova