# ppt

```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
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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
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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
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•
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Express as rationals:
A) 13.201…
B) .27…
C) .23…
D)4.163333…
• Show that .99…. Represents 1.0
A
•
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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
•
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N = .27…
100N = 27.27
100N – N = 27
99N = 27
N = 27/99
N = 3/11
C
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N = .23
100N = 23.23
100N – N = 23
99N = 23
23/99 = N
D
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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
•
•
•
•
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N = .999…
10N = 9.999
10N – N = 9
9N = 9
N = 1.0
```

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