Why 1/7 goes round in circles
Tonight's idea
Divide 3 by 8 by long division and it stops: \(3/8 = 0.375\). Divide 1 by 3 and it never stops: \(0.3333\ldots\) Divide 1 by 7 and you get \(0.142857\,142857\,142857\ldots\), the same six digits forever. A decimal that stops is called terminating. One that falls into a loop is called repeating, and we write a bar over the loop: \(1/7 = 0.\overline{142857}\).
The rule: a fraction in lowest terms terminates exactly when its denominator has no prime factors other than 2 and 5. That is because \(10 = 2 \times 5\), so a denominator like \(8 = 2^3\) or \(40 = 2^3 \times 5\) can be scaled up to a power of ten. A 3 or a 7 in the denominator can never be scaled away, so the division never finishes.
The deeper fact: every fraction's decimal either stops or repeats. Why? Each step of long division by 7 leaves a remainder from 0 to 6. Within seven steps a remainder must come back, and from then on the digits replay. Here is 1 ÷ 7:
| remainder | divide | digit | next remainder |
|---|---|---|---|
| 1 | 10 ÷ 7 | 1 | 3 |
| 3 | 30 ÷ 7 | 4 | 2 |
| 2 | 20 ÷ 7 | 2 | 6 |
| 6 | 60 ÷ 7 | 8 | 4 |
| 4 | 40 ÷ 7 | 5 | 5 |
| 5 | 50 ÷ 7 | 7 | 1 again, so the digits replay |
The same argument works for any denominator \(d\): the loop is at most \(d - 1\) digits long.
The six digits of 1/7 are strange. \(142857 \times 2 = 285714\). \(142857 \times 3 = 428571\). Multiply by 4, 5 or 6 and you get 571428, 714285, 857142. The same six digits, rotated. It is called a cyclic number, and it comes from 1/7.
Watch one: turn \(0.272727\ldots\) back into a fraction
Call it \(x\). The loop is two digits long, so multiply by 100 to slide one full loop past the point:
\(100x = 27.272727\ldots\) and \(x = 0.272727\ldots\)
Subtract. The endless tails are identical and cancel: \(99x = 27\), so \(x = 27/99 = 3/11\).
Check by dividing 3 by 11: remainders go 3, 8, 3, 8, … and the digits go 2, 7, 2, 7. Two remainders, loop of two.
Your turn
- Without dividing, say which of these terminate and which repeat: \(\tfrac{1}{40}\), \(\tfrac{1}{30}\), \(\tfrac{7}{20}\), \(\tfrac{1}{14}\), \(\tfrac{9}{12}\). Careful with the last one.
- Do the long division for \(\tfrac{5}{12}\) and \(\tfrac{2}{11}\). Write each with a bar over the repeating part. For \(\tfrac{5}{12}\), which digits come before the loop?
- Multiply 142857 by 2, 3, 4, 5 and 6 and check that each answer is a rotation of 142857. Then multiply by 7. What happens, and why does that make sense for the decimal of 1/7?
- Turn these into fractions in lowest terms using the subtraction trick: \(0.\overline{8}\), \(0.\overline{45}\), \(0.8\overline{3}\). For the last one, multiply by 10 first to get the non-repeating digit out of the way.
- Use the remainder argument. If you divide by 13, what is the longest the repeating block could possibly be? Now actually divide 1 by 13 and find the block.
- What is \(0.\overline{9}\) as a fraction? Do the subtraction trick honestly and accept what it tells you.
Stretch: how long is the loop?
For 1/7 the loop has 6 digits, the maximum possible. For 1/13 you found a loop shorter than the maximum 12. Divide 1 by 17 (this takes patience, keep a neat column of remainders) and find the length of its loop. Is it the maximum?
Primes whose loop is the full maximum, like 7 and 17, are called full reptend primes; their blocks are cyclic numbers like 142857. Nobody knows whether there are infinitely many.
Answers
- \(40 = 2^3 \times 5\) terminates (0.025). \(30 = 2 \times 3 \times 5\) repeats (\(0.0\overline{3}\)). \(20 = 2^2 \times 5\) terminates (0.35). \(14 = 2 \times 7\) repeats (\(0.0\overline{714285}\)). \(9/12 = 3/4\) in lowest terms, and \(4 = 2^2\), so it terminates (0.75). You must reduce first.
- \(5/12 = 0.41\overline{6}\): the 4 and 1 are a lead-in, then 6 forever (because \(12 = 4 \times 3\), the 4 part settles and the 3 part loops). \(2/11 = 0.\overline{18}\).
- 285714, 428571, 571428, 714285, 857142: all rotations. \(142857 \times 7 = 999999\). That makes sense because \(7 \times \tfrac{1}{7} = 1\) and \(7 \times 0.142857142857\ldots = 0.999999\ldots\), which is 1 (see question 6).
- \(0.\overline{8}\): \(10x - x = 8\), \(x = 8/9\). \(0.\overline{45}\): \(99x = 45\), \(x = 45/99 = 5/11\). \(0.8\overline{3}\): \(10x = 8.333\ldots\), \(100x = 83.333\ldots\), subtract: \(90x = 75\), \(x = 75/90 = 5/6\).
- At most 12 digits (remainders 1 to 12). In fact \(1/13 = 0.\overline{076923}\), a loop of 6. For a prime denominator \(p\) other than 2 or 5, the loop length always divides \(p - 1\). That stronger rule does not apply to composite denominators: \(1/21 = 0.\overline{047619}\) has a loop of 6, which does not divide 20.
- \(10x - x = 9\), so \(9x = 9\) and \(x = 1\). \(0.\overline{9}\) is exactly 1, not slightly less. Two different decimal names for the same number.
Stretch. \(1/17 = 0.\overline{0588235294117647}\), a loop of 16 digits, which is the maximum. So 17 is a full reptend prime and 0588235294117647 is cyclic: multiply it by 2 and you get 1176470588235294, a rotation. (Keep the leading zero when you rotate.)