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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. - PowerPoint PPT Presentation
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Decimal Expansion of FractionsBrent Murphy
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).
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 = 4.169N*10^3 = 4,169N/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 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.
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!!!
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