Turning a repeating decimal into a fraction

When you reach for it

Dividing turns any fraction into a decimal that either stops or repeats, as Terminating and repeating decimals showed. This lesson runs the other way: you are handed a repeating decimal, say 0.\overline{57} = 0.575757\ldots, and you want the fraction it equals.

A terminating decimal is easy to turn back, because place value already writes it as a fraction: 0.57 is \displaystyle \frac{57}{100}. A repeating decimal has no last place, so that reading fails. The earlier lesson gave a shortcut, the block over nines, which says \displaystyle 0.\overline{57} = \frac{57}{99}, and checked it by dividing. This lesson gives the reason the shortcut works, and a method that also handles the decimals the shortcut does not fit directly: blocks that start late, whole-number parts, negatives, and the puzzling 0.\overline{9}.

The whole method is one idea. Multiply the decimal by a power of ten so that its digits slide along until the repeating tail lines up with the tail of the original. Two numbers with the same tail differ by an ordinary number, so subtracting removes the part that never ends, and what is left is a short equation to solve.

It is taught over two days:

• Day 1: blocks that start straight after the decimal point, one digit long and longer.
• Day 2: blocks that start later, which need two multiplications; answers as mixed numbers; and why 0.\overline{9} = 1.

The procedure

Day 1: a block that starts straight after the point

One repeating digit. Take 2.\overline{3} = 2.333\ldots, and call the number x:

Multiply both sides by 10. Multiplying by ten moves every digit one place to the left, and multiplying both sides of an equation by the same number keeps it true (the multiplication property of equality):

Now compare 23.333\ldots with 2.333\ldots. After the point they are the same: every place, as far as you go, holds a 3. So the two numbers differ by exactly their whole-number parts, 23 - 2 = 21, and

Put that into the equation for 10x:

The never-ending tail has gone. What is left is an ordinary equation with the unknown on both sides, which is a look ahead to equations with variables on both sides; here it takes one step to bring the x terms together. Subtract x from both sides (the subtraction property of equality), and combine 10x - x into 9x:

Now it is a one-step equation. Divide both sides by 9, and write the fraction in lowest terms by dividing top and bottom by 3:

Check by dividing: 7 \div 3 = 2 remainder 1, and each step after that is 10 \div 3 = 3 remainder 1, so \displaystyle \frac{7}{3} = 2.333\ldots as it should.

The same work is often set out as a subtraction of one line from another:

Subtract the second line from the first, left side from left side and right side from right side, and 9x = 21 comes out directly. That is the same step as before. The second line says that x and 2.333\ldots are equal, so taking x from one side of the first line and 2.333\ldots from the other takes away equal amounts, and the first line stays true.

Why ten. The number 10 was chosen because the block is one digit long. Shifting by one place moves each 3 into the place where the next 3 was, and since every digit of the tail is the same, the shifted tail matches the old one exactly.

Longer blocks. Take 0.\overline{24} = 0.242424\ldots. The block 24 has two digits, so shift by two places, multiplying by 100:

After the point both read 2, 4, 2, 4, \ldots in the same places, so 100x = 24 + x. Subtract x from both sides and divide by 99:

The power must match the block. Watch what happens if you multiply by 10 instead: 10x = 2.424242\ldots. After the point that reads 4, 2, 4, 2, \ldots, while x reads 2, 4, 2, 4, \ldots. The tails are out of step, so subtracting does not remove them:

That gives 9x = 2.\overline{18}, which still has a repeating decimal in it, and nothing has been gained. Figure 1 sets the two attempts side by side. A block of k digits needs a shift of k places, so multiply by 10^k, a 1 followed by k zeros.

What this proves. Let x = 0.\overline{B}, where the block has k digits and reads as the whole number B. Multiplying by 10^k moves exactly one copy of the block in front of the point and leaves the same tail behind it, so

Subtract x from both sides to get (10^k - 1)x = B, and divide:

The number 10^k - 1 is written as k nines: 9, 99, 999 and so on. That is the block over nines from Terminating and repeating decimals, now with a reason behind it rather than a check. When there is a whole-number part, as in 2.\overline{3}, the same subtraction leaves the number in front of the tail in 10^k x minus the whole-number part of x: there it was 23 - 2 = 21.

Day 2: a block that starts late, mixed numbers, and repeating nines

A block that starts late. In 0.4\overline{6} = 0.4666\ldots the 4 happens once and only the 6 repeats. Multiplying by 10 once is not enough on its own: 10x = 4.666\ldots has a tail of sixes, but x has the tail 4, 6, 6, \ldots, so they still do not match.

The fix is to compare two shifted copies of x rather than a shifted copy with x itself. Multiply by 10 to move the digit that does not repeat to the left of the point, so that the copy repeats from the point on. Then shift that copy one block further, here one more place, which is multiplying x by 100:

Both tails are all sixes (Figure 2), so 46.666\ldots = 42 + 4.666\ldots, that is,

Subtract 10x from both sides, then divide by 90:

dividing top and bottom by 6. Check: 7 \div 15 gives 0.4 remainder 10, and from then on each step is 100 \div 15 = 6 remainder 10, so \displaystyle \frac{7}{15} = 0.4\overline{6}.

In general, if m digits come between the point and the start of the block, and the block has k digits, multiply by 10^m and by 10^{m + k}. The first moves the digits that do not repeat out of the way; the second shifts one block further. Subtracting leaves

and 10^{m + k} - 10^m is written as k nines followed by m zeros. For 0.4\overline{6}, with m = 1 and k = 1, that is 100 - 10 = 90. Why this is the right pair, and not, say, 1 and 10, is exercise 24.

Mixed numbers. A decimal with a whole-number part can go through the method whole, as 2.\overline{3} did. Often it is simpler to keep the whole number to one side: write 3.4\overline{6} as 3 + 0.4\overline{6}, change only the decimal part, and read off the mixed number \displaystyle 3\frac{7}{15}. Both routes give the same value, and Example 6 works one each way. If an improper fraction is wanted, convert at the end (Mixed numbers and improper fractions: rewriting and comparing): \displaystyle 3\frac{7}{15} = \frac{52}{15}.

Repeating nines. Run the method on a decimal whose block is 9 and the answer comes out as a terminating decimal, starting with 0.\overline{9} = 1. That surprises most people the first time, and Example 7 shows both why the method says so and why it is right.

Worked examples

Boundaries

The decimal has to repeat, and you have to know its block. The method lines up a tail with a shifted copy of itself, which needs the same block forever. A decimal such as 0.2022022202222\ldots, where each run of 2s is one longer than the last, never settles into a block, and no power of ten makes its tail match. Such a decimal is not a fraction at all; it is irrational. Nor does a calculator display settle the question: a screen showing 0.3157894737 has been cut off and rounded, and the screen alone cannot say what comes next.

The shift must be a whole number of blocks. Multiplying by 10^k for a block of k digits works. So does shifting by two blocks or three: 10\,000 \times 0.\overline{24} = 2424.\overline{24} gives 9999x = 2424, and \displaystyle \frac{2424}{9999} simplifies to the same \displaystyle \frac{8}{33}. Any other shift leaves the tails out of step, as Figure 1 shows.

The equation must be solved exactly. The coefficient of x is a string of nines, or nines followed by zeros, and it is never 0, so dividing by it is always allowed. Division leaves a fraction of two integers, which still has to be put in lowest terms.

A terminating decimal does not need the method. Place value gives its fraction directly: \displaystyle 0.375 = \frac{375}{1000} = \frac{3}{8}. Written with repeating nines, as 0.374\overline{9}, it goes through the method and comes out at the same \displaystyle \frac{3}{8}.

A negative sign stays outside. -2.\overline{06} means -(2.060606\ldots); the minus applies to the whole number, not to the tail alone.

Where this goes next

Together with dividing, this lesson closes a loop. Every fraction gives a decimal that stops or repeats, and now every decimal that stops or repeats gives back a fraction of two integers. So the rational numbers are exactly the decimals that stop or repeat, which is the full statement in What makes a number rational, and Which fractions give terminating decimals says which of the two you get. The decimals left over, those that neither stop nor repeat, are the irrational numbers, such as \displaystyle \sqrt{2} and \pi; together the two kinds fill the number line.

The step “bring the x terms to one side” here was a single subtraction. Equations with the unknown on both sides, and with more going on in each, are the subject of Equations with variables on both sides.

Later, 0.\overline{24} will be read in a third way, as an endless sum \displaystyle \frac{24}{100} + \frac{24}{10\,000} + \frac{24}{1\,000\,000} + \cdots. Adding up such a sum is Infinite geometric series, and the trick inside it, multiply the whole sum by a fixed number and subtract so the endless tail cancels, is the trick of this lesson.

Exercises

Day 1: one repeating digit

Write each as a fraction in lowest terms.

1. 7.\overline{7}
2. 3.\overline{4}
3. 8.\overline{3}

Day 1: longer blocks

Write each as a fraction in lowest terms.

4. 0.\overline{93}
5. 0.\overline{65}
6. 0.\overline{216}
7. 0.\overline{027}
8. 5.\overline{43}
9. -1.\overline{3}
10. 0.\overline{285714}

Day 1: reasoning

11. Aurelie is converting x = 0.\overline{61}. She multiplies by 10 and subtracts x. What does she get for 10x - x, why has it not removed the repeating part, and what should she do instead? Finish the conversion.
12. Show that for any block B of k digits, \displaystyle 0.\overline{B} = \frac{B}{10^k - 1}, and explain in words why multiplying by 10^k is what makes it work.

Day 2: blocks that start late

Write each as a fraction in lowest terms.

13. 0.1\overline{3}
14. 0.9\overline{4}
15. 0.0\overline{7}
16. 0.5\overline{21}
17. 0.02\overline{3}

Day 2: mixed numbers

Write each as a mixed number in lowest terms.

18. 2.8\overline{3}
19. 1.2\overline{3}
20. 6.\overline{03}

Day 2: repeating nines

Write each as a fraction in lowest terms, or as an integer.

21. 0.7\overline{9}
22. 4.\overline{9}

Day 2: reasoning

23. Benedek says that 0.\overline{9} is the largest number less than 1. Explain why he is wrong in two ways: that 0.\overline{9} is not less than 1, and that no largest number less than 1 exists.
24. A decimal has m digits between the point and the start of a block of k digits. Explain why the method multiplies by 10^m and 10^{m + k}, and not by 1 and 10^k, and say what the number multiplying x looks like after the subtraction.
25. Explain why every repeating decimal is a rational number.