2.02a - Recurring decimals as fractions

2.02a - Recurring decimals as fractions

A recurring decimal has an infinite tail, but it can still represent an exact fraction. You will use division to identify repeating blocks, then align and subtract shifted copies to convert recurring decimals back into fractions. The method works when the repeating block starts immediately or follows a non-recurring section.

Why decimals recur

How can 0.181818…0.181818\ldots, which never stops, be an exact number rather than an unfinished calculation?

Higher only. It is exactly 2/112/11. Long division explains both its digits and why the same block keeps returning. When you bring down the next zero after the decimal point, the current remainder determines the next digit and the next remainder.

Recurring decimal

A recurring decimal repeats a fixed block of digits indefinitely. Dots over the first and last digits of the repeating block show its extent: 0.1˙8˙=0.181818…0.\dot1\dot8=0.181818\ldots. A one-digit block needs one dot, as in 0.3˙=0.3333…0.\dot3=0.3333\ldots.

For a block longer than two digits, mark its two ends, not every digit: 0.1˙42857˙0.\dot1 4285\dot7 means 0.142857142857…0.142857142857\ldots. Digits before the first dotted digit do not repeat.

Worked example 1: a returning remainder

Question. Write 2/112/11 as a recurring decimal.

Plan. Divide and watch for a remainder that has already appeared.

20=1×11+9first digit 1, remainder 990=8×11+2next digit 8, remainder 220=1×11+9same state as before\begin{aligned} 20&=1\times11+9 && \text{first digit 1, remainder 9} \\ 90&=8\times11+2 && \text{next digit 8, remainder 2} \\ 20&=1\times11+9 && \text{same state as before} \end{aligned}

Answer. 2/11=0.1˙8˙2/11=0.\dot1\dot8.

Check. The repeating number dd satisfies 100d=18+d100d=18+d, so 99d=1899d=18 and d=18/99=2/11d=18/99=2/11. The next part explains this subtraction method fully.

Change only the numerator from 22 to 33. The division follows the same pattern with a different remainder cycle.

Worked example 2: a different repeating block

Question. Write 3/113/11 as a recurring decimal.

Plan. Divide until the original remainder returns.

30=2×11+8first digit 2, remainder 880=7×11+3next digit 7, remainder 330=2×11+8the cycle restarts\begin{aligned} 30&=2\times11+8 && \text{first digit 2, remainder 8} \\ 80&=7\times11+3 && \text{next digit 7, remainder 3} \\ 30&=2\times11+8 && \text{the cycle restarts} \end{aligned}

Answer. 3/11=0.2˙7˙3/11=0.\dot2\dot7.

Check. Eleven times 0.272727…0.272727\ldots is 2.727272…+0.272727…=32.727272\ldots+0.272727\ldots=3. The repeating tails combine exactly, not just to a rounded value.

For a positive integer denominator qq, the remainders can only be 0,1,…,q−10,1,\ldots,q-1. A zero remainder ends the decimal. Otherwise a remainder must eventually repeat, and the subsequent digits repeat with it. Thus fractions have terminating or recurring decimal expansions. An infinite decimal that never settles into a repeating block, such as the decimal expansion of 2\sqrt2, is irrational.

Watch out

0.1˙8˙0.\dot1\dot8 does not mean 0.18888…0.18888\ldots. Both 11 and 88 belong to the repeating block. The latter number is 0.18˙0.1\dot8.

Use division rather than a rounded calculator display to establish the recurring block.

Subtracting shifted copies

Multiplying a recurring decimal by a suitable power of ten lines up identical infinite tails, which subtraction removes exactly.

Let x=0.4˙x=0.\dot4. Then 10x=4.4˙10x=4.\dot4. The fractional tails are the same number, so 10x−x=410x-x=4. This is not subtraction of two finite rounded versions; it uses the repeating pattern itself.

Worked example 3: a one-digit repeating block

Question. Convert 0.4˙0.\dot4 to an exact fraction.

Plan. Shift one place because the repeating block has one digit.

x=0.4444…name the exact value10x=4.4444…shift one place10x−x=4identical tails cancel9x=4collectx=49divide by 9\begin{aligned} x&=0.4444\ldots && \text{name the exact value} \\ 10x&=4.4444\ldots && \text{shift one place} \\ 10x-x&=4 && \text{identical tails cancel} \\ 9x&=4 && \text{collect} \\ x&=\frac49 && \text{divide by 9} \end{aligned}

Answer. 4/94/9.

Check. Dividing 44 by 99 produces 40=4×9+440=4\times9+4 repeatedly, so every decimal digit is 44.

Lengthen the repeating block by appending the digit 55. You now need a two-place shift so that a whole block moves in front of the decimal point.

Worked example 4: a two-digit repeating block

Question. Convert 0.4˙5˙0.\dot4\dot5 to an exact fraction in its simplest form.

Plan. Multiply by 100100, subtract the original, then simplify.

x=0.454545…block 45 repeats100x=45.454545…shift one whole block99x=45subtract xx=4599=511divide and cancel 9\begin{aligned} x&=0.454545\ldots && \text{block 45 repeats} \\ 100x&=45.454545\ldots && \text{shift one whole block} \\ 99x&=45 && \text{subtract }x \\ x&=\frac{45}{99}=\frac5{11} && \text{divide and cancel 9} \end{aligned}

Answer. 5/115/11.

Check. Independent division gives remainder steps 50=4×11+650=4\times11+6 and 60=5×11+560=5\times11+5. They generate the repeating digits 4545.

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