Turning 0.272727… back into a fraction
Tonight's idea
Long division turns a fraction into a decimal that stops or repeats (grade 6, sheet 11). Tonight we go the other way. The bar in \(0.\overline{27}\) means the digits 27 repeat forever: \(0.272727\ldots\)
Call the number \(x\). The block is two digits long, so multiply by 100, shifting the point two places. The tail after the point is the same tail as before:
\(100x - x = 27.\overline{27} - 0.\overline{27}\), so \(99x = 27\) and \(x = \dfrac{27}{99} = \dfrac{3}{11}\).
Check by dividing 3 by 11. In general a block of \(n\) digits sits over \(n\) nines, because multiplying by \(10^n\) and subtracting leaves \((10^n - 1)x\). If the repeat starts later, multiply by two different powers of ten so the tails line up, then subtract.
Now \(x = 0.\overline{9}\): \(10x = 9.\overline{9}\), so \(9x = 9\) and \(x = 1\). Not "just under" 1: equal to 1. If you think there is a gap, name it. It would be smaller than \(0.1\), than \(0.01\), than every \(0.00\ldots01\). The only such number is 0. So \(0.\overline{9}\) and 1 are two names for one number.
Watch one: \(x = 2.1\overline{36}\)
The digits are \(2.1363636\ldots\) The repeat starts one place after the point, so shift once to reach it and then shift two more places for one full block:
\[10x = 21.\overline{36}, \qquad 1000x = 2136.\overline{36}.\]
Both have the tail \(.\overline{36}\). Subtract: \(990x = 2136 - 21 = 2115\), so
\[x = \frac{2115}{990} = \frac{47}{22}.\]
Check: \(47 \div 22 = 2.1363636\ldots\) Why 990? It is \(1000 - 10\), the gap between the two shifts. Choosing the two powers of ten was the whole skill; the subtraction did the rest.
Your turn
- Write each as a fraction in lowest terms: \(0.\overline{7}\), \(0.\overline{45}\), \(0.\overline{123}\). Check one of them by long division.
- Write \(0.41\overline{6}\) and \(0.1\overline{6}\) as fractions in lowest terms. Say which two powers of ten you multiplied by, and why.
- (a) Write \(0.\overline{142857}\) as a fraction and simplify it fully. (b) Show that \(0.\overline{09} = \tfrac{1}{11}\). Then write \(0.\overline{27}\) and \(0.\overline{63}\) as fractions with no further working.
- (a) Use the method on \(0.4\overline{9}\). (b) Use it on \(3.\overline{9}\). (c) What kind of number has two different decimal names? Describe them all.
Stretch: adding repeaters
Find \(0.\overline{1} + 0.\overline{2} + 0.\overline{3} + \cdots + 0.\overline{8}\) exactly. Then find \(0.\overline{1234} + 0.\overline{8765}\) without converting either to a fraction first. Explain the answer.
Answers
- \(0.\overline{7} = \tfrac{7}{9}\). \(0.\overline{45} = \tfrac{45}{99} = \tfrac{5}{11}\). \(0.\overline{123} = \tfrac{123}{999} = \tfrac{41}{333}\).
- \(0.41\overline{6}\): multiply by 100 and by 1000, since the repeat starts two places in and the block is one digit. \(1000x - 100x = 416.\overline{6} - 41.\overline{6} = 375\), so \(x = \tfrac{375}{900} = \tfrac{5}{12}\). \(0.1\overline{6}\): multiply by 10 and 100. \(90x = 16 - 1 = 15\), so \(x = \tfrac{15}{90} = \tfrac{1}{6}\).
- (a) \(\tfrac{142857}{999999} = \tfrac{1}{7}\), since \(142857 \times 7 = 999999\). (b) \(100x - x = 9\), so \(x = \tfrac{9}{99} = \tfrac{1}{11}\). Then \(0.\overline{27} = 3 \times 0.\overline{09} = \tfrac{3}{11}\) and \(0.\overline{63} = \tfrac{7}{11}\).
- (a) \(100x - 10x = 49.\overline{9} - 4.\overline{9} = 45\), so \(x = \tfrac{45}{90} = \tfrac{1}{2}\): \(0.4\overline{9} = 0.5\). (b) \(9x = 39.\overline{9} - 3.\overline{9} = 36\), so \(x = 4\). (c) Every decimal that stops (other than 0) has a second name: lower the last digit by one and follow it with nines forever. Numbers whose decimals never stop have only one name.
Stretch. Each \(0.\overline{d}\) is \(\tfrac{d}{9}\), so the sum is \(\tfrac{1 + 2 + \cdots + 8}{9} = \tfrac{36}{9} = 4\). Adding digit by digit, \(0.\overline{1234} + 0.\overline{8765} = 0.\overline{9999} = 0.\overline{9} = 1\). As fractions, \(\tfrac{1234 + 8765}{9999} = \tfrac{9999}{9999} = 1\). The two blocks add to 9999, which is why.