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Decimal Expansion of Fractions Brent Murphy P Q Problem: Under What conditions will the decimal expansion of p/q terminate? Under what conditions will it repeat? p/q can be investigated as p*(1/q). Terminating • When placing 1 over q as a fraction the decimal expansion will terminate if it can be factored down to a prime number < 5, unless q is also a multiple of 3 and a prime number > 5 is not a factor. Examples of Terminating decimal expansion • ½ and 1/5 terminate. • 1/10 will terminate because 10 can be factored down to 2 and 5 and it is not a multiple of 3. • 1/16 terminates. 16 can be factored down to four 2’s, and is not a multiple of 3. • 1/25 terminates. 25 can be factored down to 5 and 5 and is not a multiple of 3. Repeating Decimal Expansion • If q is a multiple of 3 it will repeat. • If q can be factored down to a prime number > 5 then it will repeat. More examples of repeating Decimals • 1/29 repeats. 29 is a prime number > 5. • 1/35 repeats. 35 can be factored to 5 and 7. 7 is a prime number greater than 5, so it must repeat the decimal when placed under a number as a fraction. • 1/14 repeats. It can be factored to a 2 and 7. 7 is prime and > 5 so 1/14 repeats. Examples of repeating Decimals • 1/3 repeats. • 1/9 repeats. 9 is a multiple of 3. • 1/15 repeats. 15 can be factored to 3 and 5. The fact that it is a factor of 5 (one of the first 3 prime numbers) would cause it to terminate if it were not a multiple of 3. 15 is a multiple of 3 so it repeats. Problem 2 • Suppose you are given a decimal expansion of a fraction. How can you represent that as p/q? • For Terminating decimals • For Repeating decimals Terminating Decimals • Terminating decimals are a little easier than repeating decimals. If you make the decimal a whole number by multiply it by 10^x, where x is the number of decimal places needed to move to make the decimal a whole number. Put that number over 10^x and reduce. Examples of terminating decimals to fractions Example 1: • N= 3.74 • N*10^2 = 374 • 374/10^2= 374/100 = 187/50 Example 2: N = 4.169 N*10^3 = 4,169 N/10^3= 4,169/1000 Repeating Decimals • Repeating decimals are a little harder and require a little more logic. N= a repeating decimal number. Declare a variable (M) that M= N*10^x, where x is the number of decimal places needed to move the decimal to where it begins to repeat. • Note: This step may not be needed if the decimal begins to repeat after immediately. In this case, assume M=N Repeating decimals Step 2 • Next, multiple M by 10^y, where y is the number terms in the geometric sequence before it begins to repeat once more. Then, setup an equation where 10^yM = the number M multiplied by 10^y. Repeating Decimals Step 3 • Subtract M from both sides leaves the right side of the equation as a whole number. Divide that whole number by 10^yM – M and it will create a fraction. Remember, that M = N*10^y. You are looking for N, not M so you must setup and equation for N*10^y and then divide the fraction found for M by N*10^y and it will give you the fraction for N. Reduce and Enjoy!!! Examples of repeating decimals • • • • • • • N = 3.135135 N=M M*10^3 = 3,135.135135 1000M – M = 3,132 999M = 3,132 M = 3,132/999 M = 116/37 M= N so N = 116/37 Repeating Decimal Example • • • • • • • • • N = 5.163333 M = 516.3333 = 100N 10M = 5,163.3333 10M – M = 4,647 9M = 4,647 M = 4647/9 M = 1549/3 100N = 1549/3 N = 1549/300 Problem 3 • • • • • Express as rationals: A) 13.201… B) .27… C) .23… D)4.163333… • Show that .99…. Represents 1.0 A • • • • • • N = 13.201… 1000N = 13,201.201 1000N – N = 13,201.201 – N 999N = 13,188 N = 13,188/999 N = 4,396/333 B • • • • • • N = .27… 100N = 27.27 100N – N = 27 99N = 27 N = 27/99 N = 3/11 C • • • • • N = .23 100N = 23.23 100N – N = 23 99N = 23 23/99 = N D • • • • • • • • • N = 4.16333… M= 100N M = 416.33 10M = 4,163.33 10M – M = 3,747 9M = 3,747 M = 3,747/9 = 1249/3 100N = 1249/3 N = 1249/300 Show that .9999 represents 1.0 • • • • • N = .999… 10N = 9.999 10N – N = 9 9N = 9 N = 1.0