Turning Recurring Decimals into Fractions

Line the tails up

Take the recurring decimal 0.474747…, in which the block 47 repeats forever, and call it x.

The repeating block has 2 digits. Multiplying by 100 moves every digit 2 places to the left, which is exactly one block, so 100x = 47.474747… with the same endless tail. After the point, 100x has the same digits as x, 474747… forever, and they sit in the same places.

Now subtract x from 100x. The endless tails after the point are identical, so they cancel completely: 100x − x = 47.474747… − 0.474747… = 47. So 99x = 47, and $x=\frac{47}{99}$x = 47/99.

After the point, 100x and x have the same digits in the same places, so the subtraction leaves exactly 47: 99x = 47.

Check by dividing

Divide 47 by 99 to check. 470 ÷ 99 = 4 remainder 74, and 740 ÷ 99 = 7 remainder 47. The remainder 47 has come back, so the digits 4 and 7 repeat, and $\frac{47}{99}=0.474747\dots$47/99 = 0.474747… as it should.

Dividing 47 by 99, the remainders are 47 and 74, and then 47 again, so the digits 47 repeat.

Multiply by the length of the block

The multiplier must move the digits by one whole repeating block. With a 2-digit block, 10x is no good: for x = 0.474747…, 10x = 4.747474…, whose tail starts with 7, not 4. The tails do not line up, so subtracting leaves an endless tail instead of a whole number.

With 1 repeating digit, multiply by 10. For x = 0.555…, 10x = 5.555…, so 10x − x = 5, 9x = 5, and $x=\frac{5}{9}$x = 5/9. It is the same move that showed 0.999… = 1, where 9x = 9.

With 3 repeating digits, multiply by 1000. For x = 0.123123…, 1000x = 123.123123…, so 999x = 123 and $x=\frac{123}{999}$x = 123/999, which simplifies to $\frac{41}{333}$41/333. A block of n digits over n nines is the pattern: $\frac{5}{9}$5/9, $\frac{47}{99}$47/99, $\frac{123}{999}$123/999.

A digit that does not repeat

In 0.8333…, only the 3 repeats: the 8 happens once. No multiple of x by a power of 10 has the same tail as x itself, because the tail of x starts with the 8. So make two multiples of x whose tails match instead.

Let x = 0.8333…, with the 3 repeating. Multiply by 10 to move past the 8, which gives 10x = 8.333… with threes forever. Multiply by 100 to move one more place, which is one block of the repeat: 100x = 83.333…, again with threes forever. After the point, the two are the same.

Subtract them: 100x − 10x = 83.333… − 8.333… = 75. So 90x = 75, and $x=\frac{75}{90}$x = 75/90. Divide the top and the bottom by 15: $x=\frac{5}{6}$x = 5/6.

100x and 10x both end in threes forever, so their difference is exactly 75: 90x = 75, and $x=\frac{75}{90}=\frac{5}{6}$x = 75/90 = 5/6.

Every recurring decimal is a fraction

The method never fails. Whatever the repeating block, two multiples of x by powers of 10 have the same tail (one of them may be x itself, which is 1 × x), so their difference is a whole number, and x is that whole number divided by a whole number. So every recurring decimal is a fraction, which means it is rational.

Every terminating decimal is a fraction as well: $0.35=\frac{35}{100}$0.35 = 35/100. Together with the fact that every fraction terminates or recurs, this settles the question both ways. A number is rational exactly when its decimal terminates or recurs.

Each of these recurring decimals has come out as a fraction, so each one is rational.

The usual mistakes

Cutting the decimal off. $0.47=\frac{47}{100}$0.47 = 47/100 is a terminating decimal. 0.474747… is a little more, and equals $\frac{47}{99}$47/99.

Taking the whole decimal as the block. 0.1666… is not $\frac{16}{99}$16/99, because $\frac{16}{99}=0.161616\dots$16/99 = 0.161616…, where the 1 repeats too. Only the 6 repeats, so the method for a digit that does not repeat is needed: 100x − 10x = 16.666… − 1.666… = 15, so 90x = 15 and $x=\frac{1}{6}$x = 1/6.

Stopping before lowest terms. $\frac{75}{90}$75/90 is correct, but it simplifies to $\frac{5}{6}$5/6.