E1.4 - Recurring decimals as fractions

E1.4 - Recurring decimals as fractions

A decimal can continue forever and still represent an exact fraction. You will read Cambridge recurring-dot notation, use division to find recurring decimals and convert them back to fractions with algebra. The repeating digits provide the structure that makes the algebra work.

Recurring-dot notation

A calculator shows only a finite row of digits. How could you record 0.142424242…0.142424242\ldots exactly, and how could infinitely many decimal places ever give a fraction?

The block 4242 repeats forever after the initial 11. Cambridge uses dots above the first and last digits of the repeating block:

0.14˙2˙=0.142424242…0.1\dot4\dot2=0.142424242\ldots

The undotted 11 occurs before the repeating part starts. The dots tell you which digits to keep copying; they are not decimal points or rounding instructions.

Recurring-dot notation

For a one-digit repeating block, put a dot over that digit: 0.7˙=0.7777…0.\dot7=0.7777\ldots. For a longer block, put dots over its first and last digits. For example, 0.1˙06˙=0.106106106…0.\dot1 0\dot6=0.106106106\ldots, with the zero included in each block.

The location of the first dot matters as much as the last. These two expansions begin with the same three digits but then separate.

Worked example 1: a two-digit block after a prefix

Question. Write the first six digits after the decimal point of 0.12˙7˙0.1\dot2\dot7.

Plan. Keep the undotted prefix once, then repeat the block between the dots.

prefix=1before the first dotrepeating block=27from first dot to last dot0.12˙7˙=0.1 27 27 27…copy the whole block\begin{aligned} \text{prefix}&=1 && \text{before the first dot}\\ \text{repeating block}&=27 && \text{from first dot to last dot}\\ 0.1\dot2\dot7&=0.1\,27\,27\,27\ldots && \text{copy the whole block} \end{aligned}

Answer. The first six digits are 127272127272, giving 0.127272…0.127272\ldots. This is an expansion, not a rounded value.

Check. After removing the first digit, the remaining digits occur in pairs 27,27,27,…27,27,27,\ldots.

Move only the first dot from the 22 to the 11. This makes the repeating block three digits long and removes the non-recurring prefix.

Worked example 2: a three-digit block

Question. Write the first six digits after the decimal point of 0.1˙27˙0.\dot1 2\dot7.

Plan. Repeat every digit from the first dotted digit through the last dotted digit.

repeating block=127include the middle digit0.1˙27˙=0.127 127 127…repeat three-digit blocksfirst six digits=127127take two complete blocks\begin{aligned} \text{repeating block}&=127 && \text{include the middle digit}\\ 0.\dot1 2\dot7&=0.127\,127\,127\ldots && \text{repeat three-digit blocks}\\ \text{first six digits}&=127127 && \text{take two complete blocks} \end{aligned}

Answer. 0.127127…0.127127\ldots, with first six digits 127127127127.

Check. Digits one to three and four to six are identical. That was not true in the previous example.

Watch out

Dots over the endpoints include the digits between them. Writing 0.1˙27˙0.\dot1 2\dot7 does not mean that only 11 and 77 repeat while the 22 disappears.

Fractions that produce recurring decimals

Long division tells you both the digits and when they start repeating. At each decimal place, multiply the current remainder by 1010, divide by the denominator and keep the new remainder.

For 5/65/6, start with remainder 55. Then 50=6(8)+250=6(8)+2, giving the first digit 88. Next 20=6(3)+220=6(3)+2, giving digit 33 and remainder 22 again. The same remainder creates the same next step, so the 33 repeats:

56=0.83˙.\frac56=0.8\dot3.

[DIAGRAM: asset_name: A repeated remainder produces a repeating digit; asset_slug: repeated_remainder; caption: Once remainder two returns, the same division step produces another three.; description: A flow diagram for five divided by six. Initial remainder five gives fifty divided by six, digit eight and remainder two. Remainder two gives twenty divided by six, digit three and remainder two again. A loop returns to that same remainder-two state and the decimal reads zero point eight followed by repeating threes.]
Once remainder two returns, the same division step produces another three.

Why must rational decimals eventually terminate or recur? Dividing by a positive integer denominator qq can leave only remainders 0,1,…,q−10,1,\ldots,q-1. Remainder zero ends the division. If it never reaches zero, eventually a non-zero remainder returns, and all the subsequent digits repeat from that point. A repeated digit alone is not the test: watch the remainder.

Worked example 3: a two-step remainder cycle

Question. Write 3/113/11 as a recurring decimal, using dot notation.

Plan. Continue division until a remainder returns to one already used.

30=11(2)+8first digit two, remainder eight80=11(7)+3next digit seven, remainder three311=0.272727…=0.2˙7˙initial remainder three returns\begin{aligned} 30&=11(2)+8 && \text{first digit two, remainder eight}\\ 80&=11(7)+3 && \text{next digit seven, remainder three}\\ \frac3{11}&=0.272727\ldots=0.\dot2\dot7 && \text{initial remainder three returns} \end{aligned}

Answer. 0.2˙7˙0.\dot2\dot7.

Check. Multiplying the expansion by 11=10+111=10+1 gives 2.72727…+0.27272…=32.72727\ldots+0.27272\ldots=3.

Change only the numerator. The denominator is still 1111, but the digits and remainder cycle change.

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