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Repeating decimals to fractions

Numbers · level 6, lesson 1 of 7

A question to start

Divide 1 by 7 on a calculator and you get \(0.142857142857\ldots\): the block 142857 over and over. Divide 1 by 3 and you get \(0.333\ldots\) forever. Divide 1 by 8 and you get \(0.125\), which simply stops.

Two puzzles. First, going the other way: if someone hands you \(0.272727\ldots\), which fraction is it? Second, a deeper one: is it just luck that every fraction we try either stops or repeats? Could some fraction, say with a huge bottom number, produce digits that wander forever with no pattern? This lesson answers both, and the answer to the second is a firm no, with a reason you can check yourself.

The idea

Words and notation

When you turn a fraction into a decimal by dividing the top by the bottom (level 4), one of two things happens:

  • the division stops: \(\tfrac18 = 0.125\). This is a terminating decimal;
  • the division never stops, and a block of digits repeats forever: \(\tfrac17 = 0.142857142857\ldots\) This is a repeating decimal (also called a recurring decimal).

The repeating block is called the period, and its number of digits is the period length. We mark it with a bar over the block:

\[\tfrac13 = 0.\overline{3} \qquad \tfrac17 = 0.\overline{142857} \qquad \tfrac{7}{55} = 0.1\overline{27}\]

In the last one, the 1 does not repeat; only the 27 does. It means \(0.1272727\ldots\)

From a repeating decimal to a fraction: one repeating digit

Take \(x = 0.777\ldots\) The trick is to make a second copy of \(x\) with the same endless tail, then subtract so that the tails cancel.

\[ \begin{array}{rl} 10x &= 7.777\ldots \\ x &= 0.777\ldots \\ \hline 10x - x &= 7 \end{array} \]

So \(9x = 7\), and \(x = \tfrac79\). Check: \(7 \div 9 = 0.777\ldots\) ✓

Why does the subtraction work? Multiplying by 10 moves every digit one place left. Because the 7s go on forever, after the shift there are still 7s in every decimal place. So \(10x\) and \(x\) have exactly the same digits after the point, and subtracting removes all of them, leaving a whole number. With a decimal that stopped, nothing so neat would happen; here the endlessness is exactly what makes it work.

Longer blocks: shift by the block length

For \(x = 0.272727\ldots\) the block "27" has 2 digits. Shifting by one place would line up 2s under 7s, which does not cancel. So shift by two places: multiply by \(10^2 = 100\).

\[ \begin{array}{rl} 100x &= 27.2727\ldots \\ x &= 0.2727\ldots \\ \hline 99x &= 27 \end{array} \]

So \(x = \tfrac{27}{99} = \tfrac{3}{11}\) (dividing top and bottom by 9).

The general rule: if the block has \(k\) digits, multiply by \(10^k\). A useful shortcut follows. A decimal that repeats from straight after the point is the block over that many 9s:

\[0.\overline{7} = \frac{7}{9} \qquad 0.\overline{27} = \frac{27}{99} \qquad 0.\overline{123} = \frac{123}{999} \qquad 0.\overline{142857} = \frac{142857}{999999} = \frac17\]

A delay before the repeat

\(x = 0.1272727\ldots = 0.1\overline{27}\). The 1 gets in the way. Use two shifts: one that moves the non-repeating part in front of the point, and one more that moves a whole block as well.

  1. \(10x = 1.272727\ldots\) (now the repeating part starts right after the point).
  2. \(1000x = 127.272727\ldots\) (two more places: one whole block).
  3. Subtract: \(1000x - 10x = 127.2727\ldots - 1.2727\ldots = 126\).
  4. So \(990x = 126\), and \(x = \tfrac{126}{990} = \tfrac{7}{55}\) (dividing by 18).

Check: \(7 \div 55 = 0.12727\ldots\) ✓

So every repeating decimal is a fraction

Look at what the method always does. However long the delay and however long the block, the two shifted copies have identical tails, and subtracting leaves

\[(\text{some power of 10} - \text{another power of 10}) \times x = \text{a whole number}\]

Dividing gives \(x\) as a whole number over a whole number. So every repeating decimal is a fraction. (Terminating decimals are fractions too: \(0.125 = \tfrac{125}{1000}\).)

Now the other way: why every fraction must stop or repeat

Here is the deeper question. Why can a fraction never give endless digits without a pattern? Watch the remainders in the long division of \(1 \div 7\):

step divide digit remainder
1 \(10 \div 7\) 1 3
2 \(30 \div 7\) 4 2
3 \(20 \div 7\) 2 6
4 \(60 \div 7\) 8 4
5 \(40 \div 7\) 5 5
6 \(50 \div 7\) 7 1
7 \(10 \div 7\) 1 3

At step 6 the remainder is 1 again, which is where we started. Two things are now certain.

Each step depends only on the remainder before it. Once the digits of the top number are used up, every step "brings down a 0": it takes the remainder, puts a 0 after it (multiplies by 10), and divides by 7. So the same remainder always leads to the same digit and the same next remainder. Once a remainder comes back, everything after it repeats, forever.

A remainder must come back. When you divide by 7, the remainder is always one of \(0, 1, 2, 3, 4, 5, 6\). If it is ever 0, the division stops: a terminating decimal. If not, there are only six possible remainders, 1 to 6. In seven steps you cannot have seven different remainders from only six choices, so some remainder must appear twice. (This "more steps than choices" idea is called the pigeonhole principle: put 7 letters into 6 pigeonholes and some hole gets two.)

132 645 142 857 circles: remainders bold: digits 1/7 = 0.142857...

The same argument works for any denominator \(q\). The remainders can only be \(0, 1, \ldots, q - 1\). Either a 0 turns up and the decimal stops, or within \(q\) steps some non-zero remainder repeats and the digits cycle. So:

Every fraction gives a terminating or a repeating decimal, and its repeating block is at most \(q - 1\) digits long.

For \(\tfrac17\) the block has the longest possible length, \(7 - 1 = 6\). For \(\tfrac{1}{13}\) it could have had 12 digits, but it has only 6: \(0.\overline{076923}\), because the remainders come back sooner.

The cycle picture gives a bonus: the decimals of \(\tfrac27\), \(\tfrac37, \ldots\) use the same cycle, just starting at a different remainder. \(\tfrac27\) starts at remainder 2, so it reads the cycle from there: \(0.\overline{285714}\).

Putting both halves together

\[\text{fractions} \quad \longleftrightarrow \quad \text{decimals that terminate or repeat}\]

The arrow goes both ways. This is exactly what you need for the next lesson: a decimal that never repeats cannot be a fraction.

And \(0.999\ldots\)?

Use the method on \(x = 0.\overline{9}\): \(10x = 9.999\ldots\), so \(9x = 9\), and \(x = 1\). So \(0.999\ldots = 1\). This surprises many people, but it is not a trick: they are two ways of writing the same number, as \(\tfrac12\) and \(\tfrac24\) are. Another way to see it: \(\tfrac13 = 0.333\ldots\), and three times both sides gives \(1 = 0.999\ldots\). If they were different numbers, there would have to be a number between them (lesson 2 of level 5), and you cannot squeeze anything between \(0.999\ldots\) and 1.

Worked examples

1. Write \(0.\overline{36}\) as a fraction in lowest terms.

  1. The block "36" has 2 digits, so multiply by 100: \(100x = 36.3636\ldots\)
  2. Subtract \(x\): \(99x = 36\), so \(x = \tfrac{36}{99}\).
  3. Divide top and bottom by 9: \(\tfrac{4}{11}\). Check: \(4 \div 11 = 0.3636\ldots\) ✓

2. Write \(2.\overline{3}\) as a fraction.

  1. \(x = 2.333\ldots\), \(10x = 23.333\ldots\), so \(9x = 21\) and \(x = \tfrac{21}{9} = \tfrac73\).
  2. Or: \(2.\overline{3} = 2 + 0.\overline{3} = 2 + \tfrac13 = \tfrac73\). Both ways agree. ✓

3. Write \(0.4\overline{6}\) (that is, \(0.4666\ldots\)) as a fraction.

  1. \(10x = 4.666\ldots\) (repeating part now starts right after the point).
  2. \(100x = 46.666\ldots\) (one more place, one whole block).
  3. \(100x - 10x = 42\), so \(90x = 42\) and \(x = \tfrac{42}{90} = \tfrac{7}{15}\).

4. Find the decimal for \(\tfrac37\) without dividing again.

\(\tfrac37\) starts with remainder 3. In the cycle 1 → 3 → 2 → 6 → 4 → 5 → 1, starting from 3 gives the digits 4, 2, 8, 5, 7, 1. So \(\tfrac37 = 0.\overline{428571}\). ✓

5. Without working it out, what is the longest the repeating block of \(\tfrac{1}{17}\) could be?

There are 16 possible non-zero remainders, so at most 16 digits. (In fact it is exactly 16: \(\tfrac{1}{17} = 0.\overline{0588235294117647}\).)

Try it

  1. Write \(0.\overline{4}\) as a fraction.
  2. Write \(0.\overline{81}\) as a fraction in lowest terms.
  3. Write \(0.\overline{123}\) as a fraction in lowest terms.
  4. Write \(0.1\overline{6}\) as a fraction. (You may recognise it.)
  5. Write \(3.\overline{12}\) as a fraction.
  6. Show that \(0.4\overline{9}\) (that is, \(0.4999\ldots\)) equals \(\tfrac12\).
  7. Use the remainder cycle for sevenths to write down \(\tfrac57\) as a decimal.
  8. Explain why the decimal for \(\tfrac{1}{23}\) must repeat with a block of at most 22 digits.
Answers
  1. \(10x = 4.444\ldots\), so \(9x = 4\) and \(x = \tfrac49\).
  2. \(100x - x = 81\), so \(x = \tfrac{81}{99} = \tfrac{9}{11}\).
  3. \(1000x - x = 123\), so \(x = \tfrac{123}{999} = \tfrac{41}{333}\) (dividing by 3).
  4. \(10x = 1.666\ldots\) and \(100x = 16.666\ldots\), so \(90x = 15\) and \(x = \tfrac{15}{90} = \tfrac16\).
  5. \(3.\overline{12} = 3 + \tfrac{12}{99} = 3 + \tfrac{4}{33} = \tfrac{99}{33} + \tfrac{4}{33} = \tfrac{103}{33}\).
  6. \(10x = 4.999\ldots\) and \(100x = 49.999\ldots\), so \(90x = 45\) and \(x = \tfrac{45}{90} = \tfrac12\). So \(0.4999\ldots\) and \(0.5\) are the same number.
  7. Start the cycle at remainder 5: the next digits are 7, 1, 4, 2, 8, 5. So \(\tfrac57 = 0.\overline{714285}\). (Check: \(5 \div 7 = 0.714\ldots\) ✓)
  8. When dividing by 23, the remainders can only be 0 to 22. 23 does not divide \(1 \times 10^n\), so the remainder is never 0 (it never stops). That leaves 22 possible remainders, so within 23 steps one must repeat, and the block has at most 22 digits. (In fact it has exactly 22.)

Common mistake

Multiplying by 10 when the block is longer than one digit. For \(0.\overline{27}\), \(10x = 2.7272\ldots\) does not line up with \(x = 0.2727\ldots\), and subtracting does not cancel the tails. Multiply by \(10^k\), where \(k\) is the length of the block.

Thinking \(0.999\ldots\) is "just less than 1". It is not a number creeping towards 1; it is a fixed number, and it equals 1. Writing it with a never-ending row of 9s is just an unusual name for 1.

Where it leads