HN10 - Recurring decimals as fractions

HN10 - Recurring decimals as fractions

Learn to read recurring-decimal notation, recognise repetition in division, and convert recurring decimals into exact fractions. This develops HN10 in both directions, including decimals with a non-recurring start before their repeating block.

Reading recurring decimals

Is 0.6˙0.\dot6 exactly 0.6660.666, or does the notation say more? A dot above a digit means that the indicated block continues forever; writing only three sixes gives a different, terminating value.

Recurring decimal

A recurring decimal has a block of digits that repeats indefinitely. A dot above one digit marks a one-digit block. For a longer block, dots above its first and last digits mark the whole block between them.

0.6˙=0.6666…,0.2˙7˙=0.272727…,0.27˙=0.27777…0.\dot6=0.6666\ldots,\qquad 0.\dot2\dot7=0.272727\ldots,\qquad 0.2\dot7=0.27777\ldots

In 0.12˙7˙=0.1272727…0.1\dot2\dot7=0.1272727\ldots, the initial 11 occurs once; only 2727 repeats. The dots mark the boundaries of a block, not two separate digits to repeat individually. An ellipsis alone needs a stated pattern: 0.12345…0.12345\ldots does not tell you which digits, if any, repeat.

A rational number can be written p/qp/q, where p,qp,q are integers and q≠0q\ne0. Long division explains why a rational decimal terminates or eventually repeats. With positive denominator qq, there are only qq possible remainders, from 00 to q−1q-1. Remainder zero ends the division; a repeated non-zero remainder restarts the same sequence of digits.

Worked example 1: a repeated remainder

Question. Write 2/112/11 as a recurring decimal, using dot notation.

Plan. Divide, recording the remainder after each decimal digit until one repeats.

The whole-number part is zero. Each next step multiplies the remainder by 1010.

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

Remainder 22 is the starting remainder again, so the digits 1818 recur.

Answer. 2/11=0.181818…=0.1˙8˙2/11=0.181818\ldots=0.\dot1\dot8.

Check. The truncated decimal gives 0.181818×11=1.9999980.181818\times11=1.999998, just below 22, as expected. The repeating remainder establishes the exact expansion.

Keep denominator 1111 but change the numerator. The available remainders are the same, but the repeating digits change.

Worked example 2: the same denominator

Question. Write 5/115/11 as a recurring decimal, using dot notation.

Plan. Follow the remainders until the starting one returns.

50=4×11+6first digit 4, remainder 660=5×11+5next digit 5, remainder 5\begin{aligned} 50&=4\times11+6 &&\text{first digit }4,\ \text{remainder }6\\ 60&=5\times11+5 &&\text{next digit }5,\ \text{remainder }5 \end{aligned}

The return to remainder 55 repeats the digits 4545.

Answer. 5/11=0.454545…=0.4˙5˙5/11=0.454545\ldots=0.\dot4\dot5.

Check. 0.454545×11=4.9999950.454545\times11=4.999995, consistent with truncating an exact quotient of 55.

After a fraction is simplified, a positive denominator containing only prime factors 22 and 55 gives a terminating decimal: it can be scaled to a power of 1010. A different prime factor prevents this, so long division must eventually repeat a non-zero remainder. For example, 3/8=0.3753/8=0.375, whereas denominator 1111 produces recurrence. Apply this test after cancellation: 3/6=1/23/6=1/2 terminates.

Watch out

0.27˙0.2\dot7 and 0.2˙7˙0.\dot2\dot7 are different: the first begins 0.27777…0.27777\ldots, the second 0.272727…0.272727\ldots. Write enough digits to show the block before calculating.

Use the same division process with a fresh numerator.

One repeating block

Multiplying by a power of 1010 moves decimal digits to the left without changing their repeating order. Choose a shift of one whole repeating block, then subtract so the infinite tails cancel exactly.

If a block has one digit, compare 10x10x with xx. For two digits, compare 100x100x with xx. The resulting multipliers are 10−1=910-1=9 and 100−1=99100-1=99, explaining the nines in the denominator.

Worked example 3: a one-digit block

Question. Express 0.6˙0.\dot6 as a fraction in its simplest form.

Plan. Call the decimal xx and subtract it from 10x10x.

x=0.6666…10x=6.6666…shift one digit10x−x=6identical decimal tails cancel9x=6x=69=23divide and simplify\begin{aligned} x&=0.6666\ldots\\ 10x&=6.6666\ldots &&\text{shift one digit}\\ 10x-x&=6 &&\text{identical decimal tails cancel}\\ 9x&=6\\ x&=\frac69=\frac23 &&\text{divide and simplify} \end{aligned}

Answer. 0.6˙=2/30.\dot6=2/3.

Check. Dividing 22 by 33 gives digit 66 and remainder 22 repeatedly, so the original recurring decimal returns.

Lengthen the repeating block from 66 to 6464. A one-digit shift no longer aligns the tails; use two digits.

Worked example 4: a two-digit block

Question. Express 0.6˙4˙0.\dot6\dot4 as a fraction in its simplest form.

Plan. Subtract xx from 100x100x, because the block has two digits.

x=0.646464…100x=64.646464…shift one complete block100x−x=64matching tails cancel99x=64x=6499no common factor remains\begin{aligned} x&=0.646464\ldots\\ 100x&=64.646464\ldots &&\text{shift one complete block}\\ 100x-x&=64 &&\text{matching tails cancel}\\ 99x&=64\\ x&=\frac{64}{99} &&\text{no common factor remains} \end{aligned}

Answer. 0.6˙4˙=64/990.\dot6\dot4=64/99.

Check. Division gives 640=6×99+46640=6\times99+46, then 460=4×99+64460=4\times99+64. Remainder 6464 returns, generating 6464 repeatedly.

This cancellation concerns infinite decimals, not finite approximations. For instance, 100(0.6464)−0.6464=63.9936100(0.6464)-0.6464=63.9936, which is not 6464. Replacing the recurrence by four digits before subtracting changes the number.

Key idea

The same method gives 0.9˙=9/9=10.\dot9=9/9=1 exactly. A decimal ending in infinitely many nines is another representation of a terminating value; there is no positive gap between 0.999…0.999\ldots and 11.

The opening decimal is therefore exactly 2/32/3, whereas 0.666=333/5000.666=333/500 is only a finite approximation.

Mixed recurring decimals

If some digits occur once before repetition begins, first move that non-recurring start to the left of the decimal point. Then shift by a whole repeating block, so both decimal tails match before subtraction.

For 0.27˙0.2\dot7, comparing 10x10x with xx does not align the tails: one starts with repeated copies of 77, the other starts with 22. Comparing 100x100x with 10x10x aligns them.

Worked example 5: one non-recurring digit

Question. Express 0.27˙0.2\dot7 as a fraction in its simplest form.

Plan. Compare 100x100x and 10x10x to leave the same 0.777…0.777\ldots tail.

x=0.27777…10x=2.7777…move the initial 2100x=27.7777…shift one repeating digit100x−10x=25cancel equal tails90x=25x=2590=518simplify\begin{aligned} x&=0.27777\ldots\\ 10x&=2.7777\ldots &&\text{move the initial }2\\ 100x&=27.7777\ldots &&\text{shift one repeating digit}\\ 100x-10x&=25 &&\text{cancel equal tails}\\ 90x&=25\\ x&=\frac{25}{90}=\frac5{18} &&\text{simplify} \end{aligned}

Answer. 5/185/18.

Check. 50=2×18+1450=2\times18+14, then 140=7×18+14140=7\times18+14. The remainder 1414 repeats, giving 0.27777…0.27777\ldots.

Keep the non-recurring digit 22, but lengthen the repeating block from 77 to 7373. This needs a two-digit shift after 10x10x.

Worked example 6: a longer recurring tail

Question. Express 0.27˙3˙0.2\dot7\dot3 as a fraction in its simplest form.

Plan. Compare 1000x1000x and 10x10x, which both end in repeated 7373.

x=0.2737373…10x=2.737373…move the initial digit1000x=273.737373…move one further complete block1000x−10x=271identical tails cancel990x=271x=271990no common factor remains\begin{aligned} x&=0.2737373\ldots\\ 10x&=2.737373\ldots &&\text{move the initial digit}\\ 1000x&=273.737373\ldots &&\text{move one further complete block}\\ 1000x-10x&=271 &&\text{identical tails cancel}\\ 990x&=271\\ x&=\frac{271}{990} &&\text{no common factor remains} \end{aligned}

Answer. 271/990271/990.

Check. Division gives first digit 22 with remainder 730730. Next, 7300=7×990+3707300=7\times990+370; then 3700=3×990+7303700=3\times990+730. The return to remainder 730730 repeats the block 7373.

An integer part does not change the principle: 1.27˙=1+5/18=23/181.2\dot7=1+5/18=23/18. For a negative recurring decimal, convert its magnitude and keep the minus sign. In every case, the shifted multipliers must be different, so their difference is non-zero and division is valid.

Watch out

Do not put all displayed digits over nines when there is a non-recurring start. For 0.27˙0.2\dot7, 27/99=0.272727…27/99=0.272727\ldots is the wrong decimal. Align the tails first; the correct denominator comes from 100−10=90100-10=90.

Use equal tails and finish in simplest form.

Quick Recap

  • Dots mark the first and last digits of the recurring block.
  • A repeated division remainder generates repeated decimal digits.
  • Shift by powers of 1010 until decimal tails match, then subtract exactly.
  • Include any non-recurring start in your choice of shifts and simplify the resulting fraction.

The opening value 0.6˙0.\dot6 is exactly 2/32/3; 0.6660.666 has only three decimal digits and is smaller.