OpenStax-CNX module: m34927
1
Introduction to Fractions and
Multiplication and Division of
Fractions: Equivalent Fractions,
Reducing Fractions to Lowest
Terms, and Raising Fractions to
∗
Higher Terms
Wade Ellis
Denny Burzynski
This work is produced by OpenStax-CNX and licensed under the
Creative Commons Attribution License 3.0
†
Abstract
This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr.
This
module discusses equivalent fractions, reducing fractions to lowest terms, and raising fractions to higher
terms.
By the end of the module students should be able to recognize equivalent fractions, reduce a
fraction to lowest terms and be able to raise a fraction to higher terms.
1 Section Overview
• Equivalent Fractions
• Reducing Fractions to Lowest Terms
• Raising Fractions to Higher Terms
2 Equivalent Fractions
Let's examine the following two diagrams.
∗ Version
1.3: Aug 20, 2010 11:24 pm +0000
† http://creativecommons.org/licenses/by/3.0/
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OpenStax-CNX module: m34927
Notice that both
2
3
and
Equivalent Fractions
2
4
6
represent the same part of the whole, that is, they represent the same number.
Fractions that have the same value are called equivalent fractions. Equivalent fractions may look dierent,
but they are still the same point on the number line.
There is an interesting property that equivalent fractions satisfy.
A Test for Equivalent Fractions Using the Cross Product
These pairs of products are called cross products.
If the cross products are equal, the fractions are equivalent. If the cross products are not equal, the
fractions are not equivalent.
Thus, 32 and 46 are equivalent, that is, 32 = 46 .
2.1 Sample Set A
Determine if the following pairs of fractions are equivalent.
Example 1
3
4
and 86 . Test for equality of the cross products.
The fractions
3
4
The cross products are equals.
and 68 are equivalent, so 34 = 68 .
Example 2
3
8
9
and 16
. Test for equality of the cross products.
The fractions
3
8
The cross products are not equal.
9
and 16
are not equivalent.
2.2 Practice Set A
Determine if the pairs of fractions are equivalent.
Exercise 1
1
2
,
3
6
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(Solution on p. 13.)
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Exercise 2
4
5
,
,
,
15
,
12
(Solution on p. 13.)
5
40
Exercise 5
3
(Solution on p. 13.)
8
Exercise 4
1
8
(Solution on p. 13.)
12
15
Exercise 3
2
3
3
(Solution on p. 13.)
1
4
3 Reducing Fractions to Lowest Terms
It is often very useful to conver t one fraction to an equivalent fraction that has reduced values in the
numerator and denominator. We can suggest a method for doing so by considering the equivalent fractions
9
9
9
and 35 . First, divide both the numerator and denominator of 15
by 3. The fractions 15
and 35 are
15
equivalent.
9
9
(Can you prove this?) So, 15
= 35 . We wish to convert 15
to 53 . Now divide the numerator and
9
denominator of 15 by 3, and see what happens.
9÷3
15÷3
=
3
5
9
The fraction 15
is converted to 35 .
A natural question is "Why did we choose to divide by 3?" Notice that
9
15
=
3·3
5·3
We can see that the factor 3 is common to both the numerator and denominator.
Reducing a Fraction
From these observations we can suggest the following method for converting one fraction to an equivalent
fraction that has reduced values in the numerator and denominator. The method is called reducing a
fraction.
A fraction can be reduced by dividing both the numerator and denominator by the same nonzero whole
number.
Consider the collection of equivalent fractions
5
4
3
, 16
, 12
, 28 , 41
20
Reduced to Lowest Terms
Notice that each of the rst four fractions can be reduced to the last fraction, 41 , by dividing both the
numerator and denominator by, respectively, 5, 4, 3, and 2. When a fraction is converted to the fraction
that has the smallest numerator and denominator in its collection of equivalent fractions, it is said to be
7
reduced to lowest terms. The fractions 41 , 38 , 25 , and 10
are all reduced to lowest terms.
Observe a very important property of a fraction that has been reduced to lowest terms. The only whole
number that divides both the numerator and denominator without a remainder is the number 1. When 1 is
the only whole number that divides two whole numbers, the two whole numbers are said to be relatively
prime.
Relatively Prime
A fraction is reduced to lowest terms if its numerator and denominator are relatively
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prime.
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3.1 Methods of Reducing Fractions to Lowest Terms
Method 1: Dividing Out Common Primes
1. Write the numerator and denominator as a product of primes.
2. Divide the numerator and denominator by each of the common prime factors. We often indicate
this division by drawing a slanted line through each divided out factor. This process is also called
cancelling common factors.
3. The product of the remaining factors in the numerator and the product of remaining factors of the
denominator are relatively prime, and this fraction is reduced to lowest terms.
3.1.1 Sample Set B
Reduce each fraction to lowest terms.
Example 3
1
6
=
18
1
)2·)3
)2·)3·3
1
=
1
3
1 and 3 are relatively prime.
1
Example 4
1
1
16
20
=
)2·)2·2·2
)2·)2·5
1
=
4
5
4 and 5 are relatively prime.
1
Example 5
1
56
104
=
1
1
)2·)2·)2·7
)2·)2·)2·13
1
1
=
7
13
7 and 13 are relatively prime (and also truly prime)
1
Example 6
1
315
336
=
1
)3·3·5·)7
2·2·2·2·)3·)7·
1
=
15
16
15 and 16 are relatively prime.
1
Example 7
8
15
No common prime factors, so 8 and 15 are relatively prime.
8
The fraction 15
is reduced to lowest terms.
=
2·2·2
3·5
3.1.2 Practice Set B
Reduce each fraction to lowest terms.
Exercise 6
(Solution on p. 13.)
4
8
Exercise 7
(Solution on p. 13.)
6
15
Exercise 8
(Solution on p. 13.)
6
48
Exercise 9
(Solution on p. 13.)
Exercise 10
(Solution on p. 13.)
Exercise 11
(Solution on p. 13.)
21
48
72
42
135
243
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Method 2: Dividing Out Common Factors
1. Mentally divide the numerator and the denominator by a factor that is common to each. Write the
quotient above the original number.
2. Continue this process until the numerator and denominator are relatively prime.
3.1.3 Sample Set C
Reduce each fraction to lowest terms.
Example 8
25
30
. 5 divides into both 25 and 30.
5
)25
)30
=
5
6
5 and 6 are relatively prime.
6
Example 9
18
24
. Both numbers are even so we can divide by 2.
9
)18
)24
12
Now, both 9 and 12 are divisible by 3.
3
)9
)18
)24
)12
=
3
4
3 and 4 are relatively prime.
4
Example 10
7
)21
)210
)150
= 75 . 7 and 5 are relatively prime.
)15
5
Example 11
36
96
=
18
48
=
9
24
= 38 . 3 and 8 are relatively prime.
3.1.4 Practice Set C
Reduce each fraction to lowest terms.
Exercise 12
(Solution on p. 13.)
Exercise 13
(Solution on p. 13.)
12
16
9
24
Exercise 14
(Solution on p. 13.)
Exercise 15
(Solution on p. 13.)
Exercise 16
(Solution on p. 13.)
Exercise 17
(Solution on p. 13.)
21
84
48
64
63
81
150
240
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4 Raising Fractions to Higher Terms
Equally as important as reducing fractions is raising fractions to higher terms. Raising a fraction to higher
terms is the process of constructing an equivalent fraction that has higher values in the numerator and
denominator than the original fraction.
9
9
are equivalent, that is, 35 = 15
. Notice also,
The fractions 35 and 15
3·3
5·3
=
9
15
Notice that 33 = 1 and that 53 · 1 = 53 . We are not changing the value of 35 .
From these observations we can suggest the following method for converting one fraction to an equivalent
fraction that has higher values in the numerator and denominator. This method is called raising a fraction
to higher terms.
Raising a Fraction to Higher Terms
A fraction can be raised to an equivalent fraction that has higher terms in the numerator and denominator
by multiplying both the numerator and denominator by the same nonzero whole number.
24
by multiplying both the numerator and denominator by 8.
The fraction 34 can be raised to 32
Most often, we will want to convert a given fraction to an equivalent fraction with a higher specied
denominator. For example, we may wish to convert 58 to an equivalent fraction that has denominator 32,
that is,
5
8
=
?
5
8
=
5·4
8·4
5
8
=
20
32
32
This is possible to do because we know the process. We must multiply both the numerator and denominator of 85 by the same nonzero whole number in order to 8 obtain an equivalent fraction.
We have some information. The denominator 8 was raised to 32 by multiplying it by some nonzero whole
number. Division will give us the proper factor. Divide the original denominator into the new denominator.
32 ÷ 8 = 4
Now, multiply the numerator 5 by 4.
5 · 4 = 20
Thus,
So,
=
20
32
4.1 Sample Set D
Determine the missing numerator or denominator.
Example 12
3
7
. Divide the original denominator into the new denominator.
35 ÷ 7 = 5 The quotient is 5. Multiply the original numerator by 5.
3
3·5
15
7 = 7·5 = 35 The missing numerator is 15.
=
?
35
Example 13
5
6
? . Divide the original numerator into the new numerator.
45 ÷ 5 = 9 The quotient is 9. Multiply the original denominator by 9.
5
5·9
45
6 = 6·9 = 54 The missing denominator is 45.
=
45
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4.2 Practice Set D
Determine the missing numerator or denominator.
Exercise 18
4
5
=
Exercise 19
3
7
=
=
28
10
24
=
15
=
(Solution on p. 13.)
45
?
Exercise 22
8
(Solution on p. 13.)
?
Exercise 21
3
(Solution on p. 13.)
?
Exercise 20
1
6
(Solution on p. 13.)
?
40
(Solution on p. 13.)
?
165
5 Exercises
For the following problems, determine if the pairs of fractions are equivalent.
Exercise 23
(Solution on p. 13.)
1 5
2 , 10
Exercise 24
2 8
3 , 12
Exercise 25
(Solution on p. 13.)
, 10
12 24
5
Exercise 26
1 3
2, 6
Exercise 27
3
5,
(Solution on p. 14.)
12
15
Exercise 28
1 7
6 , 42
Exercise 29
16
25
(Solution on p. 14.)
49
, 75
Exercise 30
5
28
20
, 112
Exercise 31
(Solution on p. 14.)
, 36
10 110
3
Exercise 32
6
10
18
, 32
Exercise 33
5
8,
(Solution on p. 14.)
15
24
Exercise 34
10
16
15
, 24
Exercise 35
4 3
5, 4
Exercise 36
5
7,
15
21
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(Solution on p. 14.)
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Exercise 37
8
(Solution on p. 14.)
, 11
11
9
9
For the following problems, determine the missing numerator or denominator.
Exercise 38
1
3
=
?
12
Exercise 39
1
5
=
(Solution on p. 14.)
?
30
Exercise 40
2
3
=
?
9
Exercise 41
1
5
=
(Solution on p. 14.)
?
30
Exercise 42
2
3
=
?
9
Exercise 43
3
4
=
(Solution on p. 14.)
?
16
Exercise 44
5
6
=
?
18
Exercise 45
4
5
=
(Solution on p. 14.)
?
25
Exercise 46
1
2
=
4
?
Exercise 47
9
25
=
(Solution on p. 14.)
27
?
Exercise 48
3
2
=
18
?
Exercise 49
5
3
=
(Solution on p. 14.)
80
?
Exercise 50
1
8
=
3
?
Exercise 51
4
5
=
(Solution on p. 14.)
?
100
Exercise 52
1
2
=
25
?
Exercise 53
3
16
=
(Solution on p. 14.)
?
96
Exercise 54
15
16
=
225
?
Exercise 55
11
12
=
(Solution on p. 14.)
?
168
Exercise 56
9
13
=
?
286
Exercise 57
32
33
=
?
1518
Exercise 58
19
20
=
1045
?
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(Solution on p. 14.)
OpenStax-CNX module: m34927
Exercise 59
37
50
=
9
(Solution on p. 14.)
1369
?
For the following problems, reduce, if possible, each of the fractions to lowest terms.
Exercise 60
6
8
Exercise 61
(Solution on p. 14.)
8
10
Exercise 62
5
10
Exercise 63
(Solution on p. 14.)
6
14
Exercise 64
3
12
Exercise 65
(Solution on p. 14.)
4
14
Exercise 66
1
6
Exercise 67
(Solution on p. 14.)
4
6
Exercise 68
18
14
Exercise 69
(Solution on p. 14.)
20
8
Exercise 70
4
6
Exercise 71
(Solution on p. 14.)
10
6
Exercise 72
6
14
Exercise 73
(Solution on p. 14.)
14
6
Exercise 74
10
12
Exercise 75
(Solution on p. 14.)
16
70
Exercise 76
40
60
Exercise 77
(Solution on p. 15.)
20
12
Exercise 78
32
28
Exercise 79
36
10
Exercise 80
36
60
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(Solution on p. 15.)
OpenStax-CNX module: m34927
Exercise 81
10
(Solution on p. 15.)
12
18
Exercise 82
18
27
Exercise 83
(Solution on p. 15.)
18
24
Exercise 84
32
40
Exercise 85
(Solution on p. 15.)
11
22
Exercise 86
27
81
Exercise 87
(Solution on p. 15.)
17
51
Exercise 88
16
42
Exercise 89
(Solution on p. 15.)
39
13
Exercise 90
44
11
Exercise 91
(Solution on p. 15.)
66
33
Exercise 92
15
1
Exercise 93
(Solution on p. 15.)
15
16
Exercise 94
15
40
Exercise 95
(Solution on p. 15.)
36
100
Exercise 96
45
32
Exercise 97
(Solution on p. 15.)
30
75
Exercise 98
121
132
Exercise 99
(Solution on p. 15.)
72
64
Exercise 100
30
105
Exercise 101
46
60
Exercise 102
75
45
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(Solution on p. 15.)
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11
Exercise 103
(Solution on p. 15.)
40
18
Exercise 104
108
76
Exercise 105
(Solution on p. 15.)
7
21
Exercise 106
6
51
Exercise 107
(Solution on p. 15.)
51
12
Exercise 108
8
100
Exercise 109
(Solution on p. 15.)
51
54
Exercise 110
A ream of paper contains 500 sheets. What fraction of a ream of paper is 200 sheets? Be sure to
reduce.
Exercise 111
(Solution on p. 15.)
There are 24 hours in a day. What fraction of a day is 14 hours?
Exercise 112
A full box contains 80 calculators. How many calculators are in
Exercise 113
There are 48 plants per at. How many plants are there in
1
3
1
4
of a box?
of a at?
(Solution on p. 15.)
Exercise 114
A person making $18,000 per year must pay $3,960 in income tax. What fraction of this person's
yearly salary goes to the IRS?
For the following problems, nd the mistake.
Exercise 115
3
24
=
)3
)3·8
(Solution on p. 15.)
0
8
=
=0
Exercise 116
8
10
=
)2+6
)2+8
=
6
8
=
3
4
Exercise 117
7
15
=
)7
)7+8
=
(Solution on p. 15.)
1
8
Exercise 118
6
7
=
)5+1
)5+2
=
1
2
Exercise 119
)9
)9
=
0
0
=0
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(Solution on p. 15.)
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12
5.1 Exercises for Review
Exercise 120
( here1 ) Round 816 to the nearest thousand.
Exercise 121
( here2 ) Perform the division: 0 ÷ 6.
(Solution on p. 15.)
Exercise 122
( here3 ) Find all the factors of 24.
Exercise 123
( here4 ) Find the greatest common factor of 12 and 18.
Exercise 124
( here5 ) Convert
15
8
(Solution on p. 15.)
to a mixed number.
1 "Addition and Subtraction of Whole Numbers: Rounding Whole Numbers"
<http://legacy.cnx.org/content/m34780/latest/>
2 "Multiplication and Division of Whole Numbers: Concepts of Division of Whole Numbers"
<http://legacy.cnx.org/content/m34864/latest/>
3 "Exponents, Roots, Factorization of Whole Numbers: Prime Factorization of Natural Numbers"
<http://legacy.cnx.org/content/m34873/latest/>
4 "Exponents, Roots, and Factorizations of Whole Numbers: The Greatest Common Factor"
<http://legacy.cnx.org/content/m34874/latest/>
5 "Introduction to Fractions and Multiplication and Division of Fractions: Proper Fractions, Improper Fractions, and Mixed
Numbers" <http://legacy.cnx.org/content/m34912/latest/>
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OpenStax-CNX module: m34927
Solutions to Exercises in this Module
Solution to Exercise (p. 2)
, yes
Solution to Exercise (p. 3)
, yes
Solution to Exercise (p. 3)
30 6= 24, no
Solution to Exercise (p. 3)
, yes
Solution to Exercise (p. 3)
, yes
Solution to Exercise (p. 4)
1
2
Solution to Exercise (p. 4)
2
5
Solution to Exercise (p. 4)
1
8
Solution to Exercise (p. 4)
7
16
Solution to Exercise (p. 4)
12
7
Solution to Exercise (p. 4)
5
9
Solution to Exercise (p. 5)
3
4
Solution to Exercise (p. 5)
3
8
Solution to Exercise (p. 5)
1
4
Solution to Exercise (p. 5)
3
4
Solution to Exercise (p. 5)
7
9
Solution to Exercise (p. 5)
5
8
Solution to Exercise (p. 7)
32
Solution to Exercise (p. 7)
12
Solution to Exercise (p. 7)
4
Solution to Exercise (p. 7)
150
Solution to Exercise (p. 7)
88
Solution to Exercise (p. 7)
equivalent
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13
OpenStax-CNX module: m34927
Solution to Exercise (p. 7)
equivalent
Solution to Exercise (p. 7)
not equivalent
Solution to Exercise (p. 7)
not equivalent
Solution to Exercise (p. 7)
not equivalent
Solution to Exercise (p. 7)
equivalent
Solution to Exercise (p. 7)
not equivalent
Solution to Exercise (p. 8)
not equivalent
Solution to Exercise (p. 8)
6
Solution to Exercise (p. 8)
12
Solution to Exercise (p. 8)
12
Solution to Exercise (p. 8)
20
Solution to Exercise (p. 8)
75
Solution to Exercise (p. 8)
48
Solution to Exercise (p. 8)
80
Solution to Exercise (p. 8)
18
Solution to Exercise (p. 8)
154
Solution to Exercise (p. 8)
1,472
Solution to Exercise (p. 9)
1,850
Solution to Exercise (p. 9)
4
5
Solution to Exercise (p. 9)
3
7
Solution to Exercise (p. 9)
2
7
Solution to Exercise (p. 9)
2
3
Solution to Exercise (p. 9)
5
2
Solution to Exercise (p. 9)
5
3
Solution to Exercise (p. 9)
7
3
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15
Solution to Exercise (p. 9)
8
35
Solution to Exercise (p. 9)
5
3
Solution to Exercise (p. 9)
18
5
Solution to Exercise (p. 10)
2
3
Solution to Exercise (p. 10)
3
4
Solution to Exercise (p. 10)
1
2
Solution to Exercise (p. 10)
1
3
Solution to Exercise (p. 10)
3
Solution to Exercise (p. 10)
2
Solution to Exercise (p. 10)
already reduced
Solution to Exercise (p. 10)
9
25
Solution to Exercise (p. 10)
2
5
Solution to Exercise (p. 10)
9
8
Solution to Exercise (p. 10)
23
30
Solution to Exercise (p. 11)
20
9
Solution to Exercise (p. 11)
1
3
Solution to Exercise (p. 11)
17
4
Solution to Exercise (p. 11)
17
18
Solution to Exercise (p. 11)
7
12
Solution to Exercise (p. 11)
16
Solution to Exercise (p. 11)
Should be 18 ; the cancellation is division, so the numerator should be 1.
Solution to Exercise (p. 11)
Cancel factors only, not addends;
Solution to Exercise (p. 11)
Same as Exercise ; answer is
1
1
Solution to Exercise (p. 12)
7
15
or 1.
0
Solution to Exercise (p. 12)
6
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is already reduced.