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Decimal Expansion of Fractions Brent Murphy

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.

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Decimal Expansion of Fractions Brent Murphy

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  1. Decimal Expansion of FractionsBrent Murphy

  2. PQ 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).

  3. 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.

  4. 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.

  5. 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.

  6. 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.

  7. 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.

  8. 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

  9. 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.

  10. 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

  11. 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

  12. Repeating decimalsStep 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.

  13. Repeating DecimalsStep 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!!!

  14. 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

  15. 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

  16. Problem 3 • Express as rationals: • A) 13.201… • B) .27… • C) .23… • D)4.163333… • Show that .99…. Represents 1.0

  17. 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

  18. B • N = .27… • 100N = 27.27 • 100N – N = 27 • 99N = 27 • N = 27/99 • N = 3/11

  19. C • N = .23 • 100N = 23.23 • 100N – N = 23 • 99N = 23 • 23/99 = N

  20. 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

  21. Show that .9999 represents 1.0 • N = .999… • 10N = 9.999 • 10N – N = 9 • 9N = 9 • N = 1.0

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