Download 3.2 Equivalent Fractions: Simplifying and Building

3.2 Equivalent Fractions: Simplifying and Building
Two fractions are said to be equivalent if they have the same value. Naturally, one approach we
could use to determine if two fractions are equivalent is to convert each fraction to a decimal. For
3
15
3
15
example, since = 0.6 and
are equivalent, and we could
= 0.6 , the fractions and
5
25
5
25
3 15
write =
. Alternatively, consider the following forms of the number 1:
5 25
1=
2 3 4
100 n
= = = ... =
=
2 3 4
100 n
Clearly 2 parts out of 2 is equal to 1, as is 100 parts out of 100, or n parts out of n. Now consider
the following property (the Fundamental Property of Fractions):
If a, b, and c are nonzero:
a a•c
=
b b•c
This statement is saying if both the numerator and denominator of a fraction have the same factor
(called a common factor), then that factor can be eliminated resulting in an equivalent fraction.
a a c a•c
a
c
It is true because = • =
, so we are multiplying the fraction
by (a form of 1) to
b b c b•c
b
c
a•c
result in the fraction
. Recall that multiplying a number by 1 does not change its value (the
b•c
3
15
Identity Property of Multiplication). Using our fractions
and
, note that:
5
25
15 3 • 5 3
=
=
25 5 • 5 5
Thus
15
3
15 3
and
are equivalent fractions, or
= .
25
5
25 5
145
Example 1
Determine whether the two fractions are equivalent by using the Fundamental
Property of Fractions.
a.
b.
c.
d.
Solution
9 45
,
10 50
5 60
! ,!
7 84
7 49
,
12 96
4 4xy
,
9 9xy
a.
For the two fractions to be equivalent, there must be a form of 1 (or a
common factor) which can be multiplied by one fraction to create the other.
Note that:
9 5 45
• =
10 5 50
5
Since
is a form of 1, the two fractions are equivalent.
5
b.
We must find a form of 1 (or common factor) which can be multiplied by
one fraction to create the other. Note that:
5 12
60
! •
=!
7 12
84
12
Since
is a form of 1, the two fractions are equivalent.
12
We must find a form of 1 (or common factor) which can be multiplied by
one fraction to create the other. Note that:
7 7 49
• =
12 8 96
7
Since
is not a form of 1, the two fractions are not equivalent. An
8
alternate way to verify this is to convert each fraction to decimal:
7
= 0.583
12
49
= 0.510416
96
Note that these two decimal forms are not the same.
c.
146
d.
We must find a form of 1 (or common factor) which can be multiplied by
one fraction to create the other. Note that:
4 xy 4xy
•
=
9 xy 9xy
xy
Since
is a form of 1, the two fractions are equivalent.
xy
Given a fraction, could you find other fractions which are equivalent to it. For example, given
3
the fraction , what would be some other fractions equivalent to it? We could multiply by
7
different forms of 1:
3 2 6
• =
7 2 14
3 5 15
• =
7 5 35
3 a 3a
• =
7 a 7a
Note that we could list as many equivalent fractions as we can list forms of 1, which is infinite.
Example 2
For each fraction, list three equivalent fractions. Use variables in at least one of
your fractions.
a.
5
8
b.
!
c.
7
12
2x
5
147
Solution
a.
Using the fractions
2 5
ab
, , and
as forms of 1 (yours will probably be
2 5
ab
different):
5 2 10
• =
8 2 16
5 5 25
• =
8 5 40
5 ab 5ab
•
=
8 ab 8ab
10 25
5ab
,
, and
.
16 40
8ab
x2 y
9 100
Using the fractions ,
, and 2 as forms of 1 (yours will probably be
x y
9 100
different):
7 9
63
! • =!
12 9
108
7 100
700
! •
=!
12 100
1200
2
7 x y
7x 2 y
! • 2 =!
12 x y
12x 2 y
Three equivalent fractions are
b.
7x 2 y
63
700
,!
, and !
.
12x 2 y
108
1200
xy
8 15
Using the fractions ,
, and
as forms of 1 (again yours will
xy
8 15
probably be different):
2x 8 16x
• =
5 8 40
2x 15 30x
•
=
5 15
75
2x xy 2x 2 y
•
=
5 xy 5xy
Three equivalent fractions are !
c.
2x 2 y
16x 30x
,
, and
. Note how we
5xy
40 75
multiplied the numbers, and how the exponent was used to represent x • x in
the third fraction.
Three equivalent fractions are
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In addition and subtraction of fractions, it will be necessary to convert a fraction to a specified
5
denominator. For example, given the fraction , how could this fraction be converted to one
6
5
x
with a denominator of 72? That is, what numerator x would result in =
being equivalent?
6 72
12
Since 6 • 12 = 72 (we can find 12 by dividing 6 into 72), the form of 1 to use is
. Thus:
12
5 12 60
•
=
6 12 72
The missing numerator is x = 60 . This idea is often referred to as building fractions.
Example 3
Find the variable such that the two given fractions are equivalent.
a.
b.
c.
d.
Solution
a.
9
x
=
14 70
3
33
! =!
4
y
a
96
=
15 120
5
200
! =!
b
320
Since 70 ÷ 14 = 5 , the form of 1 to use is
5
. Therefore:
5
9 5 45
• =
14 5 70
The missing numerator is x = 45.
b.
Since 33 ÷ 3 = 11 , the form of 1 to use is
3 11
33
• =!
4 11
44
The missing denominator is y = 44.
!
149
11
. Therefore:
11
c.
Since 120 ÷ 15 = 8 , the form of 1 to use is
8
. Instead of multiplying the
8
8
first fraction by , we can alternatively divide the second fraction:
8
96 8 12
÷ =
120 8 15
The missing numerator is a = 12. Note how we used the idea that division is
the inverse of multiplication to do this problem.
d.
40
. Again, we do this problem
40
“backwards” by dividing the second fraction:
200 40
5
!
÷
=!
320 40
8
The missing denominator is b = 8.
Since 200 ÷ 5 = 40 , the form of 1 to use is
This last example leads to the idea of simplifying (or reducing) fractions. That is, given a
32
fraction such as
, can we apply the Fundamental Property of Fractions to reduce the numbers
40
8
to a “simpler” form? Using the form of 1 as , we can write:
8
32 4 • 8 4
=
=
40 5 • 8 5
32
4
4
reduces to . Note that
does not reduce further, since there is no other form
40
5
5
8
of 1 we can use in the Fundamental Property of Fractions. But where did come from? Recall
8
from Chapter 1 that the greatest common factor (GCF) of 32 and 40 is the largest number that
will divide into both 32 and 40, which is precisely the number 8. In other words, using the GCF
of the numerator and denominator as the common factor will always result in the form of 1 to
use. In the past, you may have learned to reduce fractions by dividing the numerator and
denominator by the same number (this is the same as our form of 1). The big problem, however,
is knowing when to stop.
We say that
150
For example, we can attempt to reduce
32
as:
40
32 16 • 2 16
=
=
40 20 • 2 20
However, the result can be reduced further. Thus the GCF becomes the quickest (and safest, in
terms of errors) approach to simplify fractions.
Example 4
Use the greatest common factor to simplify each fraction.
a.
b.
c.
d.
Solution
a.
56
80
25
!
150
48
!
132
5xy
10x
The GCF of 56 and 80 is 8, so the form of 1 to use is
8
. Therefore:
8
56 7 • 8
7
=
=
80 10 • 8 10
b.
The GCF of 25 and 150 is 25, so the form of 1 to use is
25
. Therefore:
25
25
1 • 25
1
=!
=!
150
6 • 25
6
Note how our “invisible” factor of 1 is used in this fraction.
!
151
c.
The GCF of 48 and 132 is 12, so the form of 1 to use is
!
d.
12
. Therefore:
12
48
4 • 12
4
=!
=!
132
11 • 12
11
The GCF of 5xy and 10x is 5x , so the form of 1 to use is
5x
. Therefore:
5x
5xy y • 5x y
=
=
10x 2 • 5x 2
This illustrates how we can simplify fractions with symbols also.
Thus far, we have found the GCF by guessing at it, but recall our alternate approach using
primes, which works particularly well for larger numbers. For example, to reduce the fraction
168
, it would be difficult to guess at the GCF of 168 and 180. We first factor each number into
180
primes:
168 = 8 • 21 = ( 2 • 4 ) • ( 3 • 7 ) = ( 2 • 2 • 2 ) • ( 3 • 7 ) = 2 • 2 • 2 • 3 • 7
180 = 10 • 18 = ( 2 • 5 ) • ( 3 • 6 ) = ( 2 • 5 ) • ( 3 • 2 • 3) = 2 • 2 • 3 • 3 • 5
Instead of finding the GCF, we will use the primes in our fraction, remembering that common
factors of the numerator and denominator will cancel:
168 2 • 2 • 2 • 3 • 7
=
180 2 • 2 • 3 • 3 • 5
2/ • 2/ • 2 • 3/ • 7
=
2/ • 2/ • 3/ • 3 • 5
2•7
=
3• 5
14
=
15
prime factorizations
cancelling common factors
writing the remaining factors
multiplying
For fractions with larger numbers, this is usually the most efficient, and more importantly the
most accurate, approach.
152
Example 5
Use prime numbers to simplify each fraction.
a.
b.
c.
d.
Solution
a.
21
112
90
!
198
168
!
224
5x 2 y 3
4x 4 y
First find the prime factorizations of 21 and 112:
21 = 3 • 7
112 = 4 • 28 = ( 2 • 2 ) • ( 2 • 2 • 7 ) = 2 • 2 • 2 • 2 • 7
Now rewrite the fraction using prime numbers, and simplify:
21
3• 7
=
prime factorizations
112 2 • 2 • 2 • 2 • 7
3 • 7/
=
cancelling common factors
2 • 2 • 2 • 2 • 7/
3
=
writing the remaining factors
2•2•2•2
3
=
multiplying
16
b.
First find the prime factorizations of 90 and 198:
90 = 9 • 10 = ( 3 • 3) • ( 2 • 5 ) = 2 • 3 • 3 • 5
198 = 2 • 99 = 2 • ( 9 • 11) = 2 • 3 • 3 • 11
Now rewrite the fraction using prime numbers, and simplify:
90
2 • 3• 3• 5
!
=!
prime factorizations
198
2 • 3 • 3 • 11
2/ • 3/ • 3/ • 5
=!
cancelling common factors
2/ • 3/ • 3/ • 11
5
=!
writing the remaining factors
11
153
c.
First find the prime factorizations of 168 and 224:
168 = 4 • 42 = ( 2 • 2 ) • ( 2 • 21) = ( 2 • 2 ) • ( 2 • 3 • 7 ) = 2 • 2 • 2 • 3 • 7
224 = 4 • 56 = ( 2 • 2 ) • ( 4 • 14 ) = ( 2 • 2 ) • ( 2 • 2 • 2 • 7 ) = 2 • 2 • 2 • 2 • 2 • 7
Now rewrite the fraction using prime numbers, and simplify:
168
2 • 2 • 2 • 3• 7
!
=!
prime factorizations
224
2•2•2•2•2•7
2/ • 2/ • 2/ • 3 • 7/
=!
cancelling common factors
2/ • 2/ • 2/ • 2 • 2 • 7/
3
=!
writing the remaining factors
2•2
3
=!
multiplying
4
d.
This may not seem to “fit”, but look carefully at the fraction. Treating the
variables as prime numbers, we simplify:
5x 2 y 3
5•x•x•y•y•y
=
prime factorizations
4
4x y 2 • 2 • x • x • x • x • y
5 • x/ • x/ • y • y • y/
=
cancelling common factors
2 • 2 • x/ • x/ • x • x • y/
5•y•y
=
writing the remaining factors
2•2•x•x
5y 2
= 2
multiplying
4x
This example is a typical algebra problem, and it is included here to
illustrate that the ideas we are developing extend directly to algebra.
In the last section, we found that some fractions result in a terminating decimal, while others
result in a repeating decimal. Is there a way to tell in advance which will occur? Consider the
two fractions and their decimal forms:
17
= 0.425
40
19
= 0.63
30
154
Now look at the prime factorizations of the denominators:
17
17
=
= 0.425
40 2 • 2 • 2 • 5
19
19
=
= 0.63
30 2 • 3 • 5
Recall that the decimal system has denominators which are powers of 10 = 2 • 5. Suppose we
17
want to build the first fraction
up to a power of 10 in the denominator. Since each 10 = 2 • 5,
40
we will need to multiply by two additional factors of 5:
17
17
5•5
17 • 25
425
=
•
=
=
= 0.425
40 2 • 2 • 2 • 5 5 • 5 10 • 10 • 10 1000
19
, the prime factor of 3 will always be in the prime
30
factorization of the denominator. Thus we can never build its denominator to a power of 10, and
thus it can’t be represented as a terminating decimal. In summary, only fractions whose
denominators have prime factors of 2 and 5 can be converted to terminating decimals. If a
denominator of a fraction has prime factors other that 2 or 5, it will result in a repeating decimal
(assuming the fraction is simplified). Thus the vast majority of fractions have repeating, rather
than terminating, decimal forms.
However, with the second fraction
Example 6
Determine whether the decimal form of each rational number will be terminating
or repeating. Do not actually convert the fraction to decimal!
a.
b.
c.
d.
29
85
23
!
400
49
320
97
!
440
155
Solution
a.
Finding the prime factorization of 85:
85 = 5 • 17
Since the denominator has a prime factor other than 2 or 5 (17), the decimal
form is a repeating decimal.
b.
Finding the prime factorization of 400:
400 = 4 • 100
= ( 2 • 2 ) • (10 • 10 )
= (2 • 2) • (2 • 5 • 2 • 5)
= 2•2•2•2•5•5
Since the denominator has only 2 and 5 prime factors, the decimal form is a
terminating decimal.
c.
Finding the prime factorization of 320:
320 = 10 • 32
= (2 • 5) • ( 4 • 8)
= (2 • 5) • (2 • 2 • 2 • 2 • 2)
= 2•2•2•2•2•2•5
Since the denominator has only 2 and 5 prime factors, the decimal form is a
terminating decimal.
d.
Finding the prime factorization of 440:
440 = 10 • 44
= ( 2 • 5 ) • ( 4 • 11)
= ( 2 • 5 ) • ( 2 • 2 • 11)
= 2 • 2 • 2 • 5 • 11
Since the denominator has a prime factor other than 2 or 5 (11), the decimal
form is a repeating decimal.
Terminology
equivalent fractions
common factor
simplifying (or reducing) fractions
Fundamental Property of Fractions
building fractions
greatest common factor (GCF)
156
Exercise Set 3.2
Determine whether the two fractions are equivalent by using the Fundamental Property of
Fractions.
1.
3.
5.
7.
9.
5 40
,
7 56
13 52
! ,!
15 75
23 368
! ,!
25 400
5x 25ax
,
4y 20ay
3a
6ab
! ,!
7b 14ab
6 48
,
7 56
5 20
4. ! , !
16 80
15 525
6. ! , !
28 980
7y 35y 2
,
8.
9a 45ay
9w 27awz
10. !
,!
10z
30az
2.
For each fraction, list three equivalent fractions. Use variables in at least one of your fractions.
6
13
5
13. !
12
3x
15. !
5
6a
17.
7b
7
11
8
14. !
15
4a
16. !
7
5s
18.
9t
11.
12.
Find the variable such that the two given fractions are equivalent.
5
x
=
7 56
14
x
21. !
=!
25
500
13 195
=
23.
17
y
7
112
25. ! = !
15
y
6
x
=
13 91
9
x
22. ! = !
15
210
12 216
=
24.
19
y
9
162
26. ! = !
16
y
19.
20.
157
a 60
=
7 84
a
187
29. ! = !
15
255
9 117
31.
=
b 143
12
300
33. ! = !
b
425
a 121
=
12 132
a
198
30. ! = !
13
286
6 108
32.
=
b 198
16
432
34. ! = !
b
567
27.
28.
Use the greatest common factor to simplify each fraction.
35.
37.
39.
41.
43.
45.
47.
49.
12
20
70
!
75
35
72
60
!
100
35
77
36
!
60
12xy
15ax
16abx
!
30axy
36.
38.
40.
42.
44.
46.
48.
50.
158
25
40
32
!
84
49
93
45
!
150
26
143
56
!
78
25ab
40ax
45axy
!
75aby
Use prime numbers to simplify each fraction.
51.
53.
55.
57.
59.
61.
63.
65.
67.
69.
71.
48
72
36
!
40
216
234
270
!
320
81
192
155
!
248
384
432
261
!
290
45x 2 y 5
15x 3 y 3
8a 7b 6
6a 5b 2
16x 8 y12
!
20x 9 y 7
52.
54.
56.
58.
60.
62.
64.
66.
68.
70.
72.
159
70
170
45
!
81
220
242
275
!
286
245
246
153
!
272
364
468
265
!
424
30x 3 y 7
15x 6 y 4
12a 6b 5
18a 4 b 3
36x 6 y11
!
24x 9 y 8
Determine whether the decimal form of each rational number will be terminating or repeating.
Do not actually convert the fraction to decimal!
73.
75.
77.
79.
81.
14
75
19
!
64
67
!
448
21
(Hint: Simplify first)
120
9
!
480
74.
76.
78.
80.
82.
19
90
38
!
125
53
!
480
49
(Hint: Simplify first)
175
21
!
448
Answer each question as true or false. If it is false, give a specific example to show that it is
false. If it is true, explain why.
83. Every fraction has either a terminating or repeating decimal form.
84. Every repeating decimal can be written as a fraction.
a
85. If the numbers a and b are relatively prime, then the fraction is already simplified.
b
a
86. If the numbers a and b are not relatively prime, then the fraction is not simplified.
b
87. There is only one decimal form for every fraction.
(Hint: Refer back to Exercises 95 and 96 of the previous section.)
88. There is only one fraction for every terminating decimal.
160