Download Section 5.3 The Rational Numbers Defining the Rational Numbers

Section 5.3
The Rational Numbers
Defining the Rational Numbers
Objectives
1. Define the rational numbers.
2. Reduce rational numbers.
3. Convert between mixed numbers and improper
fractions.
4. Express rational numbers as decimals.
5. Express decimals in the form a / b.
6. Multiply and divide rational numbers.
7. Add and subtract rational numbers.
8. Apply the density property of rational numbers.
9. Solve problems involving rational numbers.
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Section 5.3
• The set of rational numbers is the set of all numbers
which can be expressed in the form a , where a and b
are integers and b is not equal to 0. b
• The integer a is called the numerator.
• The integer b is called the denominator.
Examples: The following are examples of rational numbers:
¼, -½, ¾, 5, 0
1
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Reducing a Rational Number
Section 5.3
Reducing a Rational Number
a
• If b is a rational number and c is any number other than
0,
a⋅c a
=
b⋅c b
Section 5.3
3
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Section 5.3
• A mixed number consists of the sum of an integer and a
rational number, expressed without the use of an
addition sign.
Example:
130 130 ÷ 65 2
=
=
455 455 ÷ 65 7
There are no common divisors of 2 and 7 other than 1.
Thus, the rational number 2 is in its lowest terms.
• An improper fraction is a rational number whose
numerator is greater than its denominator.
Example:
7
19
5
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Section 5.3
4
Mixed Numbers and Improper Fractions
Divide the numerator and the denominator of the given
rational number by 5 · 13 or 65.
or
130
to lowest terms.
455
Thus, 130 = 2 · 5 · 13, and 455 = 5 · 7 · 13. The greatest
common divisor is 5 · 13 or 65.
Reducing a Rational Number
Example Continued
130 2 ⋅ 5 ⋅ 13 2
=
=
455 5 ⋅ 7 ⋅ 13 7
Example: Reduce
Solution: Begin by finding the greatest common divisor of
130 and 455.
• The rational numbers a and a ⋅ c are called equivalent
b
b⋅c
fractions.
• To reduce a rational number to its lowest terms, divide
both the numerator and denominator by their greatest
common divisor.
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2
5
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19 is larger than 5
Section 5.3
6
1
Mixed Numbers and Improper Fractions
Mixed Numbers and Improper Fractions
Converting from Mixed Number to an Improper Fraction
Converting from an Improper Fraction to a Mixed Number
1. Multiply the denominator of the rational number by the
integer and add the numerator to this product.
2. Place the sum in step 1 over the denominator of the
mixed number.
Example: Convert
to an improper fraction.
1. Divide the denominator into the numerator. Record the
quotient and the remainder.
2. Write the mixed number using the following form:
Solution:
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Section 5.3
7
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Mixed Numbers and Improper Fractions
42
to a mixed number.
5
•
Any rational number can be expressed as a decimal by
dividing the denominator into the numerator.
Example: Express each rational number as a decimal.
a. 5
b. 7
Solution: Step 1. Divide the denominator into the
numerator.
8
Section 5.3
9
Rational Numbers and Decimals
Example Continued
a.
0.625
8 5.000
b. 0.6363
11 7.000
48
66
20
16
40
33
40
0
Notice the decimal stops.
This is called a
terminating decimal.
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Section 5.3
10
Expressing Decimals as a Quotient of Two Integers
Notice the digits 63
repeat over and over
indefinitely. This is
called a repeating
decimal.
70
66
40
11
Solution: In each case, divide the denominator into the
numerator.
Step 2. Write the mixed number with the
Thus,
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8
Rational Numbers and Decimals
Converting from an Improper Fraction to a Mixed Number
Example: Convert
Section 5.3
• Terminating decimals can be
expressed with denominators of 10,
100, 1000, 10,000, and so on.
• Using the chart, the digits to the
right of the decimal point are the
numerator of the rational number.
40
Example: Express each terminating decimal as a
quotient of integers:
33
70
a. 0.7
b. 0.49
c. 0.048.
M
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Section 5.3
11
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Section 5.3
12
2
Expressing Decimals as a Quotient of Two Integers
Expressing Decimals as a Quotient of Two Integers
Example Continued
Repeating Decimals
Example: Express 0.6 as a quotient of integers.
Solution:
7 because the 7 is in the tenths position.
a. 0.7 =
10
Solution: Step1. Let n equal the repeating decimal such
that n = 0 .6 , or 0.6666…
Step 2. If there is one repeating digit, multiply both sides of
the equation in step 1 by 10.
49
b. 0.49 =
because the digit on the right, 9, is in the
100
hundredths position.
48
c. 0.048 =
because the digit on the right, 8, is in the
1000
thousandths position. Reducing to lowest terms,
n = 0.66666…
10n = 10(0.66666…)
10n = 6.66666…
48
48 ÷ 8
6
=
=
1000 1000 ÷ 8 125
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Section 5.3
13
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Multiplying by
10 moves the
decimal point
one place to
the right.
Section 5.3
14
Expressing Decimals as a Quotient of Two Integers
Repeating Decimals
Example Continued
Multiplying Rational Numbers
Step 3. Subtract the equation in step 1 from the equation in
step 2.
• The product of two rational numbers is the product of
their numerators divided by the product of their
denominators.
a c a⋅c
• If a and c are rational numbers, then ⋅ =
.
b d b⋅d
b
d
Step 4. Divide both sides of the equation in step 3 by the
number in front of n and solve for n.
We solve 9n = 6 for n:
9n = 6
2
9n 6
Thus, 0.6 =
.
=
Example: Multiply. If possible, reduce the product to its
lowest terms:
3
9
9
6 2
n= =
9 3
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 2  9 
 −  − 
 3  4 
Section 5.3
15
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Section 5.3
Multiplying Rational Numbers
Example Continued
Dividing Rational Numbers
Solution:
• The quotient of two rational numbers is a product of the
first number and the reciprocal of the second number.
If a and c are rational numbers, then
1
 2  9  (− 2 )(− 9 ) 18 3 ⋅ 6 3
=
=
=
or 1
 −  −  =
3⋅ 4
12 2 ⋅ 6 2
2
 3  4 
Multiply across.
b
d
a c a d a⋅d
÷ = ⋅ =
b d b c b⋅c
Simplify to
lowest terms.
Example: Divide. If possible, reduce the quotient to its
lowest terms:
−
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Section 5.3
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17
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3 7
÷
5 11
Section 5.3
18
3
Dividing Rational Numbers
Example Continued
Adding and Subtracting Rational Numbers
Identical Denominators
Solution:
The sum or difference of two rational numbers with
identical denominators is the sum or difference of their
numerators over the common denominator.
a
c
If
and are rational numbers, then
3 7
3 11
3 ⋅ 11
33
=−
− ÷ =− ⋅ =−
5 11
5 7
5⋅7
35
Change to
multiplication
by using the
reciprocal.
b
b
a c a+c
+ =
b b
b
Multiply across.
and
a c a−c
− =
b b
b
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Section 5.3
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Example: Perform the indicated operations:
a. 3 + 2
b. 11 − 5
c. − 5 1 −  − 2 3 
7

4 
12 12
7
7

4
a a c a⋅c
= ⋅ =
b b c b⋅c
7
b. 11 − 5 = 11 − 5 = 6 = 1 ⋅ 6 1
=
12 12
12
12
2⋅6
20
• If the rational numbers to be added or subtracted have
different denominators, we use the least common
multiple of their denominators to rewrite the rational
numbers.
• The least common multiple of their denominators is
called the least common denominator or LCD.
Solution:
a. 3 + 2 = 3 + 2 = 5
7
Section 5.3
Adding and Subtracting Rational Numbers
Unlike Denominators
Adding and Subtracting Rational Numbers
Identical Denominators
7
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2
c. − 5 1 −  − 2 3  = − 21 −  − 11  = − 21 + 11 = − 21 + 11 = − 10 = − 5 or − 2 1
2
4 
4
4

4
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4
4
4
4
2
Section 5.3
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Adding and Subtracting Rational Numbers
Unlike Denominators
Example: Find the sum of
3 1
+ .
4 6
We multiply the first rational
3
1
3 3
1 2
+
=
⋅ +
⋅
number by 3/3 and the second one
4
6
4 3
6 2
by 2/2 to obtain 12 in the
denominator for each number.
9
2
=
+
Notice, we have 12 in the
12
12
denominator for each number.
11
=
Thus, we add across to obtain the answer.
12
Section 5.3
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Density of Rational Numbers
Solution: Find the least common multiple of 4 and 6 so that
the denominators will be identical. LCM of 4 and 6 is 12.
Hence, 12 is the LCD.
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If r and t represent rational numbers, with r < t, then there is
a rational number s such that s is between r and t:
r < s < t.
Example: Find a rational number halfway between ½ and
¾.
Solution: First add ½ and ¾.
1 3 1 2 3 2 3 5
+ = ⋅ + = + =
2 4 2 2 4 4 4 4
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Section 5.3
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4
Density of Rational Numbers
Example Continued
Problem Solving with Rational Numbers
• A common application of rational numbers involves
preparing food for a different number of servings than
what the recipe gives.
• The amount of each ingredient can be found as follows:
Next, divide this sum by 2.
5
5 2 5 1 5 ⋅1 5
=
÷2 = ÷ = ⋅ =
4
4 1 4 2 4⋅2 8
The number
5
is halfway between ½ and ¾. Thus,
8
amount of ingredient needed =
desired serving size
× ingredient amount
recipe serving size
in the recipe
1 5 3
< < .
2 8 4
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Section 5.3
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Problem Solving with Rational Numbers
Changing the Size of a Recipe
26
The amount of sugar, in cups, needed is determined by
multiplying the rational numbers:
Solution:
desired serving size
amount of sugar needed =
× ingredient amount
recipe serving size
in the recipe
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Section 5.3
Problem Solving with Rational Numbers
Changing the Size of a Recipe
A chocolate-chip recipe for five dozen cookies requires ¾
cup of sugar. If you want to make eight dozen cookies,
how much sugar is needed?
=
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8 3 8 ⋅ 3 24 6
1
× =
=
= =1
5 4 5 ⋅ 4 20 5
5
1
Thus, 1 5 cups of sugar is needed.
8 dozen 3
× cup
5 dozen 4
Section 5.3
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5