1. Decimal Expansion of Fractions Brent Murphy

2. PQPQ 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 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.

13. 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!!!

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