Lesson 8.3.G. REPEATING DECIMALS (Pre-requisite for INFINITE GEOMETRIC SERIES). Example 1. Write decimal 0.2727272… as a fraction. We know that 0.2727272... can be written as the sum of 0.27 + 0.0027 + 0.000027 + 0.00000027 + … These series 0.27 + 0.0027 + 0.000027 + 0.00000027 + … can be written as the sum using summation notation n 1 27 . 100 n 1 To find an infinite sum, we use the formula for the sum (S) of an infinite geometric sequence: 27 27 a 27 3 S 1 100 100 . 1 99 1 r 99 11 1 100 100 Divide 3/11 on your calculator. 3/11= 0.2727272727…. Example 2. Use the same method even if your number begins as a non-repeating number. For instance, Write 5.13333333333… as a fraction. Writing this as the sum of its fractional parts yields 51/10 + 3/100 + 3/1000 + 3/10000 + …… You should recognize that the repeating part is a geometric series 3 3 a1 30 1 100 100 . Add this to the non-repeating part of the number and apply the formula S 1 9 1 r 900 30 1 10 10 (51/10) and you get 51/10 + 1/30 = 154/30 = 77/15. Another method This is made quite easy with the following observations: 5/9 = 0.5555 7/9 = 0.7777 12/99 = 0.121212 23/99 = 0.232323 456/999 = 0.456456456 78/999 = 078/999 = 0.078078078 So we can see that fractions with a denominator of 9, 99, 999, 9999, are very useful for making repeating decimals. All we have to do is to reduce the fraction to its lowest terms. For example, let's look at the repeating decimal 0.027027027... Clearly this is 27/999 = 1/37 (having divided top and bottom of the fraction by 27) Now let's try something more tricky. Take a look at 0.4588888888... This isn't simply something divided by 99, since the 45 bit doesn't repeat. What we need to do is move the decimal point over to the start of the repeating bit. In this case we multiply by 100 to get 45.888888... Now we know the fraction part is 8/9. In total we have 45 + 8/9 = 413/9 (changing into an improper fraction will make things easier for us). So 45.88888... = 413/9 Now just divide both sides by 100 to get: 0.458888... = 413/900
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