1.3 - Recurring decimals as fractions

An endless decimal can still represent an exact fraction. You will use place value and algebra to cancel repeating tails, including decimals with nonrecurring digits or a whole-number part. The key is to align the tails before subtracting.

Converting a repeating block (Higher)

Can $0.\dot9$ , with nines continuing for ever, equal $1$ exactly? It feels as if there should be a tiny gap. We will use the same algebra that converts any repeating block to a fraction to settle this.

You already know

Multiplying a decimal by $10$ shifts each digit one place to the left. You can subtract equal quantities from both sides of an equation, then divide to solve it. Equivalent fractions can be simplified by cancelling common factors.

A finite decimal display stops. A recurring decimal does not, so its notation must tell the reader what continues.

Recurring decimal

A recurring decimal repeats a digit or block of digits for ever. Pearson dot notation puts a dot over a single repeating digit, or over the first and last digits of a repeating block. Thus $0.\dot3=0.333333\ldots$ and $0.\dot2\dot7=0.272727\ldots$ . The terminating decimal $0.27$ is different from $0.\dot2\dot7$ .

Let the exact recurring number be $x$ . Move it left by one whole repeating block. Its fractional tail then matches the original tail. Subtracting leaves a whole number, so you can solve for $x$ . These are exact infinite decimals, not rounded approximations. The method is absent from the Higher formulae sheet: learn it.

Worked example 1: cancel one repeating digit

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

Plan. Call the number $x$ , multiply by $10$ , then subtract $x$ .

\begin{aligned} x&=0.333333\ldots &&\text{define the exact value}\\ 10x&=3.333333\ldots &&\text{move one repeating digit}\\ 10x-x&=3 &&\text{identical tails cancel}\\ 9x&=3 &&\text{collect the copies of }x\\ x&=\frac39=\frac13 &&\text{divide, then simplify} \end{aligned}

Why multiply by $10$ ? There is one digit in each repeating block, so a one-place shift matches the tails.

Answer. $\frac13$ .

Check. Dividing $1$ by $3$ produces digit $3$ and remainder $1$ repeatedly.

Now change only the repeating block from one digit to two. You must shift two places, not one.

Worked example 2: cancel a two-digit block

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

Plan. Use $100x-x$ , because the block $27$ has two digits.

\begin{aligned} x&=0.272727\ldots &&\text{define the recurring value}\\ 100x&=27.272727\ldots &&\text{move one whole block}\\ 100x-x&=27 &&\text{cancel matching tails}\\ 99x&=27 &&\text{subtract the coefficients}\\ x&=\frac{27}{99}=\frac3{11} &&\text{divide and simplify} \end{aligned}

Why subtract $x$ , rather than $10x$ ? The tails of $100x$ and $x$ both start $272727\ldots$ ; the tail of $10x$ starts $727272\ldots$ .

Answer. $\frac3{11}$ .

Check. $3\div11=0.272727\ldots$ , and $99\times\frac3{11}=27$ .

The denominator before simplification is $9$ for one repeating digit, $99$ for two and $999$ for three, when the repetition begins immediately after the point. A zero inside the block still counts: $0.\dot0 2\dot7=\frac{27}{999}=\frac1{37}$ .

Watch out

Wrong: $0.\dot2\dot7=\frac{27}{100}$ . That fraction stops at $0.27$ . The correct subtraction gives $99x=27$ , so the recurring value is $\frac{27}{99}$ .

Use dots or an unmistakable repeating expansion in each line. “Show algebraic working” requires the aligned equations and subtraction; a calculator display does not show this argument.

Nonrecurring digits before the block (Higher)

A digit before the repeating block occurs once, so it needs different treatment. In $0.1\dot6$ , the $1$ is not part of the block. First shift past it, then shift one whole block farther.

Nonrecurring prefix

The nonrecurring prefix is the finite set of decimal digits before repetition starts. In $0.3\dot1\dot2=0.3121212\ldots$ , the prefix is $3$ , and the repeating block is $12$ . Repeating $312$ instead would give a different number.

Choose two shifted equations with identical fractional tails. You may need to subtract $10x$ , rather than $x$ . The multiplier is chosen from the decimal's structure, not from a remembered string of nines and zeros.

Worked example 3: pass one nonrecurring digit

Question. Express $0.1\dot6$ as a fraction in simplest form. Show algebraic working.

Plan. Compare $100x$ and $10x$ , which both have a tail of sixes.

\begin{aligned} x&=0.166666\ldots &&\text{only the six recurs}\\ 10x&=1.666666\ldots &&\text{pass the prefix}\\ 100x&=16.666666\ldots &&\text{pass one more block}\\ 100x-10x&=16-1 &&\text{matching tails cancel}\\ 90x&=15 &&\text{subtract both sides}\\ x&=\frac{15}{90}=\frac16 &&\text{divide and simplify} \end{aligned}

Why not $100x-x$ ? Its tails start $666\ldots$ and $1666\ldots$ , so they do not cancel completely.

Answer. $\frac16$ .

Check. $1\div6=0.166666\ldots$ ; the first digit is different from the repeating digit.

Now keep the one-digit prefix but change the repeating block to two digits. The second shift must travel two places beyond the first.

Worked example 4: a two-digit block after the prefix

Question. Express $0.1\dot1\dot6$ as a fraction in simplest form. Show algebraic working.

Plan. Compare $1000x$ and $10x$ , leaving the same tail $161616\ldots$ .

\begin{aligned} x&=0.1161616\ldots &&\text{one prefix digit, then }16\\ 10x&=1.161616\ldots &&\text{pass the prefix}\\ 1000x&=116.161616\ldots &&\text{pass a two-digit block}\\ 1000x-10x&=116-1 &&\text{cancel identical tails}\\ 990x&=115 &&\text{subtract the coefficients}\\ x&=\frac{115}{990}=\frac{23}{198} &&\text{divide and simplify} \end{aligned}

Why is the coefficient $990$ ? We subtract $10x$ from $1000x$ , not $x$ .

Answer. $\frac{23}{198}$ .

Check. $990\times\frac{23}{198}=115$ , and division gives $0.1161616\ldots$ .

With two prefix digits, first use $100x$ ; then move another whole repeating block. The same reasoning works for any finite prefix. For example, $0.3\dot1\dot2$ uses $1000x-10x=312-3$ , giving $x=\frac{309}{990}=\frac{103}{330}$ .

Watch out

Wrong: $0.1\dot6=\frac{16}{99}$ . That fraction repeats the entire block $16$ . Here the $1$ occurs once, so use $\frac{16-1}{100-10}=\frac16$ .

The tails, not merely the whole-number parts, decide whether your two equations can be subtracted usefully.

Whole-number parts and exact checks (Higher)

A whole-number part stays in both equations. Keep it during subtraction, or convert the fractional part separately and add the whole number afterwards. Either route must give a fraction with the right size.

Key idea

An exact check reverses the equation or reproduces the recurring pattern. Agreement with a few digits on a calculator is useful evidence, but it does not prove an infinite decimal identity.

The full-value method is often the shortest route. It uses exactly the same tail alignment as before.

Worked example 5: keep the whole-number part

Question. Express $2.\dot1\dot8$ as a fraction in simplest form. Show algebraic working.

Plan. Move one two-digit block, then subtract the full original number.

\begin{aligned} x&=2.181818\ldots &&\text{define the full value}\\ 100x&=218.181818\ldots &&\text{move two places}\\ 100x-x&=218-2 &&\text{subtract both whole parts}\\ 99x&=216 &&\text{cancel the common tails}\\ x&=\frac{216}{99}=\frac{24}{11} &&\text{divide and simplify} \end{aligned}

Why subtract $2$ from $218$ ? The original $x$ includes that whole part as well as its recurring tail.

Answer. $\frac{24}{11}$ .

Check. $\frac{24}{11}=2+\frac2{11}=2.181818\ldots$ , so the size and block match.

Now change only the whole-number part from $2$ to $3$ . The denominator stays the same, because the repeating block has not changed.

Worked example 6: change the whole-number part

Question. Express $3.\dot1\dot8$ as a fraction in simplest form. Show algebraic working.

Plan. Again subtract $x$ from $100x$ , including the whole parts.

\begin{aligned} x&=3.181818\ldots &&\text{define the new full value}\\ 100x&=318.181818\ldots &&\text{same two-place shift}\\ 99x&=318-3=315 &&\text{subtract aligned equations}\\ x&=\frac{315}{99}=\frac{35}{11} &&\text{divide and simplify} \end{aligned}

Why has the answer increased by $1$ ? Only the whole part changed, so the recurring fractional part is unchanged.

Answer. $\frac{35}{11}$ .

Check. $\frac{35}{11}-\frac{24}{11}=1$ , and $\frac{35}{11}=3+\frac2{11}$ .

We can now settle the opening puzzle. Put $y=0.\dot9$ . Then $10y=9.999999\ldots$ , so $9y=9$ and $y=1$ . Thus $0.\dot9=1$ exactly: two decimal representations of the same number. Every finite truncation, such as $0.999$ , is smaller than $1$ ; the infinite recurring value is different from each truncation.

Watch out

Wrong: $100x-x=218$ for $x=2.\dot1\dot8$ . This forgets the whole part in $x$ . The correct right-hand side is $218-2=216$ , giving $\frac{24}{11}$ , a value between $2$ and $3$ .

A fraction outside that interval would reveal the error before you reached for a calculator. Use size and exact algebra together.

Quick Recap

Dots identify exactly which digits repeat.
Choose two shifts with identical fractional tails, then subtract the whole equations.
A prefix changes which multiples of $x$ you subtract; a whole-number part must stay in the calculation.
Simplify the exact fraction and check its size and recurring pattern.

The opening puzzle has an exact answer: $0.\dot9=1$ , because subtracting the aligned equations gives $9y=9$ .

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1.3 - Recurring decimals as fractions

Converting a repeating block (Higher)

Nonrecurring digits before the block (Higher)

Whole-number parts and exact checks (Higher)

Quick Recap

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