Real Numbers and their Operations

REAL NUMBERS AND THE NUMBER LINE

LEARNING OBJECTIVES

Construct a number line and graph points on it.
Use a number line to determine the order of real numbers.
Determine the opposite of a real number.
Determine the absolute value of a real number.

DEFINITIONS

A set is a collection of objects, typically grouped within braces $\{ \}$ , where each object is called an element. For example, $\{\text{red}, \text{green}, \text{blue}\}$ is a set of colors. A subset is a set consisting of elements that belong to a given set. For example, $\{\text{green}, \text{blue}\}$ is a subset of the color set above. A set with no elements is called the empty set and has its own special notation, $\{ \}$ or $\varnothing$ .

When studying mathematics, we focus on special sets of numbers. The set of natural (or counting) numbers, denoted $\mathbf{N}$ , is

$\{1, 2, 3, 4, 5, \dots\} \quad \text{\color{blue}Natural Numbers}$

The three periods ($\dots$) is called an ellipsis and indicates that the numbers continue without bound. The set of whole numbers, denoted $\mathbf{W}$ , is the set of natural numbers combined with zero.

$\{0, 1, 2, 3, 4, 5, \dots\} \quad \text{\color{blue}Whole Numbers}$

The set of integers, denoted $\mathbb{Z}$ , consists of both positive and negative whole numbers, as well as zero.

$\{\dots, -3, -2, -1, 0, 1, 2, 3, \dots\} \quad \text{\color{blue}Integers}$

Notice that the sets of natural and whole numbers are both subsets of the set of integers.

Rational numbers, denoted $\mathbb{Q}$ , are defined as any number of the form $\frac{a}{b}$ , where $a$ and $b$ are integers and $b$ is nonzero. Decimals that repeat or terminate are rational. For example,

$0.7 = \frac{7}{10} \quad \text{and} \quad 0.\overline{3} = 0.3333\dots = \frac{1}{3}$

The set of integers is a subset of the set of rational numbers because every integer can be expressed as a ratio of the integer and $1$ . In other words, any integer can be written over $1$ and can be considered a rational number. For example,

$5 = \frac{5}{1}$

Irrational numbers are defined as any number that cannot be written as a ratio of two integers. Nonterminating decimals that do not repeat are irrational. For example,

$\pi = 3.14159\dots \quad \text{and} \quad \sqrt{2} = 1.41421\dots$

The set of real numbers, denoted $\mathbb{R}$ , is defined as the set of all rational numbers combined with the set of all irrational numbers. Therefore, all the numbers defined so far are subsets of the set of real numbers. In summary,

Real Numbers

Rational: $\frac{5}{3}$ $\frac{5}{3}$ , $0.63$ $0.63$ , $0.0\overline{12}$ $0.0 \overline{12}$
- Integers: $\{\dots, -2, -1, 0, 1, 2, \dots\}$ ${\dots, - 2, - 1, 0, 1, 2, \dots}$
  - Whole: $\{0, 1, 2, 3, \dots\}$ ${0, 1, 2, 3, \dots}$
    - Natural: $\{1, 2, 3, \dots\}$
Irrational: $\sqrt{3}$ , $\pi$ , $0.10010001\dots$

Figure 1.1.1: Real Numbers

NUMBER LINE

A real number line, or simply number line, allows us to visually display real numbers by associating them with unique points on a line. The real number associated with a point is called a coordinate. A point on the real number line that is associated with a coordinate is called its graph.

To construct a number line, draw a horizontal line with arrows on both ends to indicate that it continues without bound. Next, choose any point to represent the number zero; this point is called the origin.

Figure 1.1.2

Mark off consistent lengths on both sides of the origin and label each tick mark to define the scale. Positive real numbers lie to the right of the origin and negative real numbers lie to the left. The number zero ($0$) is neither positive nor negative. Typically, each tick represents one unit.

Figure 1.1.3

As illustrated below, the scale need not always be one unit. In the first number line, each tick mark represents two units. In the second, each tick mark represents $\frac{1}{7}$ of a unit.

Figure 1.1.4

The graph of each real number is shown as a dot at the appropriate point on the number line. A partial graph of the set of integers $\mathbb{Z}$ follows:

Figure 1.1.5

EXAMPLE 1.1.1

Graph the following set of real numbers: $\{-\frac{7}{2}, -3, -\frac{5}{2}, -2, -\frac{3}{2}, -1, -\frac{1}{2}, 0, \frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}, 3, \frac{7}{2}\}$

Solution
Graph the numbers on a number line with a scale where each tick mark represents $\frac{1}{2}$ of a unit.

Figure 1.1.6

ORDERING REAL NUMBERS

When comparing real numbers on a number line, the larger number will always lie to the right of the smaller one. It is clear that $15$ is greater than $5$ , but it may not be so clear to see that $-1$ is greater than $-5$ until we graph each number on a number line.

Figure 1.1.7

We use symbols to help us efficiently communicate relationships between numbers on the number line. The symbols used to describe an equality relationship between numbers follow:

$=$ is equal to
$\neq$ is not equal to
$\approx$ is approximately equal to

These symbols are used and interpreted in the following manner:

$5 = 5$ \quad 5 is equal to 5
$0 \neq 5$ \quad 0 is not equal to 5
$\pi \approx 3.14$ \quad pi is approximately equal to 3.14

We next define symbols that denote an order relationship between real numbers.

$<$ *Less than*
$>$ Greater than
$\leq$ Less than or equal to
$\geq$ Greater than or equal to

These symbols allow us to compare two numbers. For example,

Since the graph of $-120$ is to the left of the graph of $-10$ on the number line, that number is less than $-10$ . We could write an equivalent statement as follows:

$-120 < -10$

Similarly, since the graph of zero is to the right of the graph of any negative number on the number line, zero is greater than any negative number.

$0 > -120$

The symbols $<$ and $>$ are used to denote strict inequalities, and the symbols $\leq$ and $\geq$ are used to denote inclusive inequalities. In some situations, more than one symbol can be correctly applied. For example, the following two statements are both true:

$-10 < 0 \quad \text{and} \quad -10 \leq 0$

In addition, the “or equal to” component of an inclusive inequality allows us to correctly write the following:

$-10 \leq -10$

The logical use of the word “or” requires that only one of the conditions need be true: the “less than” or the “equal to.”

EXAMPLE 1.1.2

Fill in the blank with $<, =, \text{or} >$ : $-2$ ____ $-12$ .

Solution
Use $>$ because the graph of $-2$ is to the right of the graph of $-12$ on a number line. Therefore, $-2 > -12$ , which reads “negative two is greater than negative twelve.”

Figure 1.1.8

Answer:
$-2 > -12$

In this text, we will often point out the equivalent notation used to express mathematical quantities electronically using the standard symbols available on a keyboard. We begin with the equivalent textual notation for inequalities:

$\geq$ ">="
$\leq$ "<="
$\neq$ "!="

Many calculators, computer algebra systems, and programming languages use this notation.

OPPOSITES

The opposite of any real number $a$ is $-a$ . Opposite real numbers are the same distance from the origin on a number line, but their graphs lie on opposite sides of the origin and the numbers have opposite signs.

Figure 1.1.9

For example, we say that the opposite of $10$ is $-10$ .

Next, consider the opposite of a negative number. Given the integer $-7$ , the integer the same distance from the origin and with the opposite sign is $+7$ , or just $7$ .

Figure 1.1.10

Therefore, we say that the opposite of $-7$ is $-(-7) = 7$ . This idea leads to what is often referred to as the double-negative property. For any real number $a$ ,

$-(-a) = a$

EXAMPLE 1.1.3

What is the opposite of $-\frac{3}{4}$ ?

Solution
Here we apply the double-negative property.
$-(-\frac{3}{4}) = \frac{3}{4}$

EXAMPLE 1.1.4

Simplify $-(-(-4))$

Solution
Start with the innermost parentheses by finding the opposite of $+4$ .
$-(-(-4)) = -(-(-4))$
$\quad \quad = -(-4)$
$\quad \quad = 4$

Answer
$4$

EXAMPLE 1.1.5

Simplify: $-(-(-(-2)))$ .

Solution
Apply the double-negative property starting with the innermost parentheses.
$-(-(-(-2))) = -(-(-(-2)))$
$\quad \quad \quad = -(-(2))$
$\quad \quad \quad = -2$

Answer
$-2$

TIP
If there is an even number of consecutive negative signs, then the result is positive. If there is an odd number of consecutive negative signs, then the result is negative.

Try this!

EXERCISE 1.1.1

Simplify: $-(-(-(-(-5))))$ .

Answer
$-5$

Procedure:
$-(-(-(-(-5)))) = -(-(-(-(-5))))$
$\quad \quad \quad \quad = -(-(-(-5)))$
$\quad \quad \quad \quad = -(-(5))$
$\quad \quad \quad \quad = -5$

ABSOLUTE VALUE

The absolute value of a real number $a$ , denoted $|a|$ , is defined as the distance between zero (the origin) and the graph of that real number on the number line. Since it is a distance, it is always positive. For example,

$|-4| = 4 \quad \text{and} \quad |4| = 4$

Both $4$ and $-4$ are four units from the origin, as illustrated below:

Figure 1.1.11

EXAMPLE 1.1.6

Simplify:
a. $|-12|$
b. $|12|$

Solution
Both $-12$ and $12$ are twelve units from the origin on a number line. Therefore,
$|-12| = 12 \quad \text{and} \quad |12| = 12$

Answer
a. $12$ b. $12$

Also, it is worth noting that
$|0| = 0$

The absolute value can be expressed textually using the notation $\text{abs}(a)$ . We often encounter negative absolute values, such as $-|3|$ or $-\text{abs}(3)$ . Notice that the negative sign is in front of the absolute value symbol. In this case, work the absolute value first and then find the opposite of the result.

Try not to confuse this with the double-negative property, which states that $-(-7) = +7$ .

EXAMPLE 1.1.7

Simplify: $-|-(-7)|$ .

Solution
First, find the opposite of $-7$ inside the absolute value. Then find the opposite of the result.
$-|-(-7)| = -|7|$
$\quad \quad \quad = -7$

Answer
$-7$

At this point, we can determine what real numbers have a particular absolute value. For example,

$|?| = 5$

Think of a real number whose distance to the origin is $5$ units. There are two solutions: the distance to the right of the origin and the distance to the left of the origin, namely, $\{\pm 5\}$ . The symbol ($\pm$) is read “plus or minus” and indicates that there are two answers, one positive and one negative.

Now consider the following:

$|?| = -5$

Here we wish to find a value for which the distance to the origin is negative. Since negative distance is not defined, this equation has no solution. If an equation has no solution, we say the solution is the empty set: $\varnothing$ .

KEY TAKEAWAYS

Any real number can be associated with a point on a line.
Create a number line by first identifying the origin and marking off a scale appropriate for the given problem.
Negative numbers lie to the left of the origin and positive numbers lie to the right.
Smaller numbers always lie to the left of larger numbers on the number line.
The opposite of a positive number is negative and the opposite of a negative number is positive.
The absolute value of any real number is always positive because it is defined to be the distance from zero (the origin) on a number line.
The absolute value of zero is zero.

EXERCISE 1.1.2

Use set notation to list the described elements.

The hours on a clock.

The days of the week.

The first ten whole numbers.

The first ten natural numbers.

The first five positive even integers.

The first five positive odd integers.

Answer

$\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$

$\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$

$\{2, 4, 6, 8, 10\}$

EXERCISE 1.1.3

Determine whether the following real numbers are integers, rational, or irrational.

$12$

$-3$

$4.5$

$-5$

$0.3\overline{6}$

$0.\overline{3}$

$1.001000100001\dots$

$1.00\overline{1}$

$e = 2.71828\dots$

$\sqrt{7} = 2.645751\dots$

$-7$

$3.14$

$227$

$1.33$

$0$

$8,675,309$

Answer
1: Integer, Rational
3: Rational
5: Rational
7: Irrational
9: Irrational
11: Integer, Rational
13: Rational
15: Integer, Rational

EXERCISE 1.1.4

True or false.

All integers are rational numbers.

All integers are whole numbers.

All rational numbers are whole numbers.

Some irrational numbers are rational.

All terminating decimal numbers are rational.

All irrational numbers are real.

Answer
1: True
3: False
5: True

EXERCISE 1.1.5

Choose an appropriate scale and graph the following sets of real numbers on a number line.

$\{-3, 0, 3\}$

$\{-2, 2, 4, 6, 8, 10\}$

$\{-2, -1/3, 2/3, 5/3\}$

$\{-5/2, -1/2, 0, 1/2, 2\}$

$\{-5/7, 0, 2/7, 1\}$

$\{-5, -2, -1, 0\}$

$\{-3, -2, 0, 2, 5\}$

$\{-2.5, -1.5, 0, 1, 2.5\}$

$\{0, 0.3, 0.6, 0.9, 1.2\}$

$\{-10, 30, 50\}$

$\{-6, 0, 3, 9, 12\}$

$\{-15, -9, 0, 9, 15\}$

Answer
38. $\{-3, 0, 3\}$
Figure 1.1.12
39. $\{-2, -1/3, 2/3, 5/3\}$
Figure 1.1.13
40. $\{-5/7, 0, 2/7, 1\}$
Figure 1.1.14
41. $\{-3, -2, 0, 2, 5\}$
Figure 1.1.15
42. $\{0, 0.3, 0.6, 0.9, 1.2\}$
Figure 1.1.16
43. $\{-6, 0, 3, 9, 12\}$
Figure 1.1.17

EXERCISE 1.1.6

Fill in the blank with $<, =, \text{or} >$ .

$-7$ ____ $0$

$30$ ____ $2$

$10$ ____ $-10$

$-150$ ____ $-75$

$-0.5$ ____ $-1.5$

$0$ ____ $0$

$-500$ ____ $200$

$-1$ ____ $-200$

$-10$ ____ $-10$

$-40$ ____ $-41$

Answer
54. $<$
55. $>$
56. $>$
57. $<$
58. $=$

EXERCISE 1.1.7

True or false.

$5 \neq 7$

$4 = 5$

$1 \neq 1$

$-5 > -10$

$4 \leq 4$

$-12 \geq 0$

$-10 = -10$

$3 > 3$

$-1000 < -20$

$0 = 0$

Answer
69. True
70. False
71. True
72. True
73. True

EXERCISE 1.1.8

List the numbers.

List three integers less than $-5$ .

List three integers greater than $-10$ .

List three rational numbers less than zero.

List three rational numbers greater than zero.

List three integers between $-20$ and $-5$ .

List three rational numbers between $0$ and $1$ .

Answer
80. $-10, -7, -6$ (answers may vary)
81. $-1, -2/3, -1/3$ (answers may vary)
82. $-15, -10, -7$ (answers may vary)

EXERCISE 1.1.9

Translate each statement into an English sentence.

$10 < 20$

$-50 \leq -10$

$-4 \neq 0$

$30 \geq -1$

$0 = 0$

$e \approx 2.718$

Answer
89. Ten is less than twenty.
90. Negative four is not equal to zero.
91. Zero is equal to zero.

EXERCISE 1.1.10

Translate the following into a mathematical statement.

Negative seven is less than zero.

Twenty-four is not equal to ten.

Zero is greater than or equal to negative one.

Four is greater than or equal to negative twenty-one.

Negative two is equal to negative two.

Negative two thousand is less than negative one thousand.

Answer
98. $-7 < 0$
99. $0 \geq -1$
100. $-2 = -2$

EXERCISE 1.1.11

Simplify.

$-(-9)$

$-(-35)$

$-(10)$

$-(3)$

$-(5)$

$-(34)$

$-(-1)$

$-(-(-1))$

$-(-(1))$

$-(-(-3))$

$-(-(-(-11)))$

Answer
112. $9$
113. $-10$
114. $-5$
115. $1$
116. $1$
117. $11$

EXERCISE 1.1.12

Answer the following questions.

What is the opposite of $-12$

What is the opposite of $\pi$ ?

What is the opposite $-0.01$ ?

Is the opposite of $-12$ smaller or larger than $-11$ ?

Is the opposite of $7$ smaller or larger than $-6$ ?

Answer
123. $-\pi$
124. Larger

EXERCISE 1.1.13

Fill in the blank with $<, =, \text{or} >$ .

$-7$ ____ $-(-8)$

$6$ ____ $-(6)$

$13$ ____ $-(-12)$

$-(-5)$ ____ $-(-2)$

$-100$ ____ $-(-(-50))$

$44$ ____ $-(-44)$

Answer
131. $<$
132. $>$
133. $<$

EXERCISE 1.1.14

Simplify.

$|20|$

$|-20|$

$|-33|$

$|-0.75|$

$|-\frac{3}{5}|$

$|38|$

$|0|$

$|1|$

$-|12|$

$-|-20|$

$-|20|$

$-|-8|$

$-|7|$

$-|-316|$

$-(|-\frac{8}{9}|)$

$|-(-2)|$

$-|-(-3)|$

$-(|5|)$

$-(-|-45|)$

$-|-(-21)|$

$\text{abs}(6)$

$\text{abs}(-7)$

$-\text{abs}(5)$

$-\text{abs}(-19)$

$-(-\text{abs}(9))$

$-\text{abs}(-(-12))$

Answer
160. $20$
161. $33$
162. $\frac{3}{5}$
163. $0$
164. $-12$
165. $-20$
166. $-7$
167. $-\frac{8}{9}$
168. $-3$
169. $45$
170. $6$
171. $-5$
172. $9$

EXERCISE 1.1.15

Determine the unknown.

$|?| = 9$

$|?| = 15$

$|?| = 0$

$|?| = 1$

$|?| = -8$

$|?| = -20$

$|?| - 10 = -2$

$|?| + 5 = 14$

Answer
181. $\pm 9$
182. $0$
183. $\varnothing$ , No solution
184. $\pm 8$

EXERCISE 1.1.16

Fill in the blank with $<, =, \text{or} >$ .

$|-2|$ ____ $0$

$|-7|$ ____ $|-10|$

$-10$ ____ $-|-2|$

$|-6|$ ____ $|-(-6)|$

$-|3|$ ____ $|-(-5)|$

$0$ ____ $-|-(-4)|$

Answer
191. $>$
192. $<$
193. $<$

EXERCISE 1.1.17

Discussion Board Topics.

Research and discuss the history of the number zero.

Research and discuss the various numbering systems throughout history.

Research and discuss the definition and history of $\pi$ .

Research the history of irrational numbers. Who is credited with proving that the square root of $2$ is irrational and what happened to him?

Research and discuss the history of absolute value.

Discuss the “just make it positive” definition of absolute value

ADDING AND SUBTRACTING INTEGERS

LEARNING OBJECTIVES

Add and subtract signed integers.
Translate English sentences involving addition and subtraction into mathematical statements.
Calculate the distance between two numbers on a number line.

ADDITION AND SUBTRACTION $(+, -)$

Visualize adding $3 + 2$ on the number line by moving from zero three units to the right then another two units to the right, as illustrated below:

Figure 1.2.1

The illustration shows that $3 + 2 = 5$ . Similarly, visualize adding two negative numbers $(-3) + (-2)$ by first moving from the origin three units to the left and then moving another two units to the left.

Figure 1.2.2

In this example, the illustration shows $(-3) + (-2) = -5$ , which leads to the following two properties of real numbers.

$\text{\color{cyan}positive number} + \text{\color{cyan}positive number} = \text{\color{cyan}positive number}$
$\text{\color{cyan}negative number} + \text{\color{cyan}negative number} = \text{\color{cyan}negative number}$

Next, we will explore addition of numbers with unlike signs. To add $3 + (-7)$ , first move from the origin three units to the right, then move seven units to the left as shown:

Figure 1.2.3

In this case, we can see that adding a negative number is equivalent to subtraction:

$3 + (-7) = 3 - 7 = -4$

It is tempting to say that a positive number plus a negative number is negative, but that is not always true: $7 + (-3) = 7 - 3 = 4$ . The result of adding numbers with unlike signs may be positive or negative. The sign of the result is the same as the sign of the number with the greatest distance from the origin. For example, the following results depend on the sign of the number $12$ because it is farther from zero than $5$ :

$12 + (-5) = 7$
$-12 + 5 = -7$

EXAMPLE 1.2.1

Simplify: $14 + (-25)$ .

Solution
Here $-25$ is the greater distance from the origin. Therefore, the result is negative.
$14 + (-25) = 14 - 25$
$\quad \quad \quad \quad \quad = -11$

Answer
$-11$

PROPERTIES OF ADDITION

Given any real numbers $a$ , $b$ , and $c$ , we have the following properties of addition:

Additive Identity Property:
$a + 0 = 0 + a = a \tag{1.2.1}$

Additive Inverse Property:
$a + (-a) = (-a) + a = 0 \tag{1.2.2}$

Associative Property:
$(a + b) + c = a + (b + c) \tag{1.2.3}$

Commutative Property:
$a + b = b + a \tag{1.2.4}$

Below are some examples of these properties in action.

EXAMPLE 1.2.2

Simplify:
a. $5 + 0$
b. $10 + (-10)$

Solution
a. Adding zero to any real number results in the same real number.
$5 + 0 = 5$
b. Adding opposites results in zero.
$10 + (-10) = 0$

Answer
a. $5$; b. $0$

EXAMPLE 1.2.3

Simplify:
a. $(3 + 7) + 4$
b. $3 + (7 + 4)$

Solution
Parentheses group the operations that are to be performed first.
a.
$({\color{cyan}3 + 7}) + 4 = {\color{cyan}10} + 4$
$\quad \quad \quad \quad \quad= 14$
b.
$3 + ({\color{cyan}7 + 4}) = 3 + {\color{cyan}11}$
$\quad \quad \quad \quad \quad= 14$

These two examples both result in $14$ : changing the grouping of the numbers does not change the result.
$(3 + 7) + 4 = 3 + (7 + 4) = 14$

Answer
a. $14$; b. $14$

At this point, we highlight the fact that addition is commutative: the order in which we add does not matter and yields the same result.

$2 + 9 = 9 + 2$
$11 = 11$

On the other hand, subtraction is not commutative.

$2 - 9 \neq 9 - 2$
$-7 \neq 7$

We will use these properties, along with the double-negative property for real numbers, to perform more involved sequential operations. To simplify things, we will make it a general rule to first replace all sequential operations with either addition or subtraction and then perform each operation in order from left to right.

EXAMPLE 1.2.4

Simplify: $4 - (-10) + (-5)$ .

Solution
Replace the sequential operations and then perform them from left to right.
$4 - (-10) + (-5) = 4 + 10 - 5 \quad \text{\color{cyan}Replace } -(-) \text{ \color{cyan}with addition } (+).$
$\quad \quad \quad \quad \quad \quad \quad = 14 - 5 \quad \text{\color{cyan}Replace } +(-) \text{ \color{cyan}with addition } (-).$
$\quad \quad \quad \quad \quad \quad \quad = 9$

Answer
$9$

EXAMPLE 1.2.5

Simplify: $-3 + (-8) - (-7)$ .

Solution
$-3 + (-8) - (-7) = -3 - 8 + 7 \quad \text{\color{cyan}Replace } +(-) \text{ \color{cyan}with } (-).$
$\quad \quad \quad \quad \quad \quad \quad = -11 + 7 \quad \text{\color{cyan}Replace } -(-) \text{ \color{cyan}with } (+).$
$\quad \quad \quad \quad \quad \quad \quad = -4$

Answer
$-4$

EXAMPLE 1.2.6

Try this!

Simplify: $12 - (-9) + (-6)$ .

Solution
(click to see video)

Often we find the need to translate English sentences involving addition and subtraction to mathematical statements. Listed below are some key words that translate to the given operation.

Table 1.2.2

Key Words	Operation
Sum, increased by, more than, plus, added to, total	$\mathbf{+}$
Difference, decreased by, subtracted from, less, minus	$\mathbf{-}$

EXAMPLE 1.2.7

What is the difference of $7$ and $-3$ ?

Solution
The key word “difference” implies that we should subtract the numbers.
$7 - (-3) = 7 + 3$
$\quad \quad \quad = 10$

Answer
The difference of $7$ and $-3$ is $10$ .

EXAMPLE 1.2.8

What is the sum of the first five positive integers?

Solution
The initial key word to focus on is “sum”; this means that we will be adding the five numbers. The first five positive integers are $\{1, 2, 3, 4, 5\}$ . Recall that $0$ is neither positive nor negative.
$1 + 2 + 3 + 4 + 5 = 15$

Answer
The sum of the first five positive integers is $15$ .

EXAMPLE 1.2.9

What is $10$ subtracted from the sum of $8$ and $6$ ?

Solution
We know that subtraction is not commutative; therefore, we must take care to subtract in the correct order. First, add $8$ and $6$ and then subtract $10$ as follows:

$\overbrace{(8 + 6)}^{\text{\color{cyan}the sum of 8 and 6}} \quad \overbrace{- \quad 10}^{\text{\color{cyan}subtract 10 from the sum}}$

It is important to notice that the phrase “$10$ subtracted from” does not translate to a mathematical statement in the order it appears. In other words, $10 - (8 + 6)$ would be an incorrect translation and leads to an incorrect answer. After translating the sentence, perform the operations.
$(8 + 6) - 10 = 14 - 10$
$\quad \quad \quad \quad \quad \quad = 4$

Answer
Ten subtracted from the sum of $8$ and $6$ is $4$ .

DISTANCE ON A NUMBER LINE

One application of the absolute value is to find the distance between any two points on a number line. For real numbers $a$ and $b$ , the distance formula for a number line is given as,

$d = |b - a|$

Figure 1.2.4

EXAMPLE 1.2.10

Determine the distance between $2$ and $7$ on a number line.

Solution
On the graph we see that the distance between the two given integers is $5$ units.

Figure 1.2.5

Using the distance formula we obtain the same result.
$d = |7 - 2|$
$\quad = |5|$
$\quad = 5$

Answer
$5$ units

EXAMPLE 1.2.11

Determine the distance between $-4$ and $7$ on a number line.

Solution
Use the distance formula for a number line $d = |b - a|$ , where $a = -4$ and $b = 7$ .
$d = |7 - (-4)|$
$\quad = |7 + 4|$
$\quad = |11|$
$\quad = 11$

Answer
$11$ units

It turns out that it does not matter which points are used for $a$ and $b$; the absolute value always ensures a positive result.

Table 1.2.3

Using $a = -4$ and $b = 7$	Using $a = 7$ and $b = -4$
$d =	7 - (-4)

EXERCISE 1.2.1

Determine the distance between $-12$ and $-9$ on the number line.

Answer
$3$

KEY TAKEAWAYS

A positive number added to a positive number is positive. A negative number added to a negative number is negative.
The sign of a positive number added to a negative number is the same as the sign of the number with the greatest distance from the origin.
Addition is commutative and subtraction is not.
When simplifying, it is a best practice to first replace sequential operations and then work the operations of addition and subtraction from left to right.
The distance between any two numbers on a number line is the absolute value of their difference. In other words, given any real numbers $a$ and $b$ , use the formula $d = |b - a|$ to calculate the distance $d$ between them.

EXERCISE 1.2.2

Add and subtract.

$24 + (-18)$

$9 + (-11)$

$-31 + 5$

$-12 + 15$

$-30 + (-8)$

$-50 + (-25)$

$-7 + (-7)$

$-13 - (-13)$

$8 - 12 + 5$

$-3 - 7 + 4$

$-1 - 2 - 3 - 4$

$6 - (-5) + (-10) - 14$

$-5 + (-3) - (-7)$

$2 - 7 + (-9)$

$-30 + 20 - 8 - (-18)$

$10 - (-12) + (-8) - 20$

$5 - (-2) + (-6)$

$-3 + (-17) - (-13)$

$-10 + (-12) - (-20)$

$-13 + (-5) - (-25)$

$20 - (-4) - (-5)$

$17 + (-12) - (-2)$

Answer
1: $6$
3: $-26$
5: $-38$
7: $-14$
9: $1$
11: $-10$
13: $-1$
15: $0$
17: $1$
19: $-2$
21: $29$

EXERCISE 1.2.3

Translate each sentence to a mathematical statement and then simplify.

Find the sum of $3$ , $7$ , and $-8$ .

Find the sum of $-12$ , $-5$ , and $7$ .

Determine the sum of the first ten positive integers.

Determine the sum of the integers in the set $\{-2, -1, 0, 1, 2\}$ .

Find the difference of $10$ and $6$ .

Find the difference of $10$ and $-6$ .

Find the difference of $-16$ and $-5$ .

Find the difference of $-19$ and $7$ .

Subtract $12$ from $10$ .

Subtract $-10$ from $-20$ .

Subtract $5$ from $-31$ .

Subtract $-3$ from $27$ .

Two less than $8$ .

Five less than $-10$ .

Subtract $8$ from the sum of $4$ and $7$ .

Subtract $-5$ from the sum of $10$ and $-3$ .

Subtract $2$ from the difference of $8$ and $5$ .

Subtract $6$ from the difference of $-1$ and $7$ .

Mandy made a $\$200$ deposit into her checking account on Tuesday. She then wrote $4$ checks for $\$50.00$ , $\$125.00$ , $\$60.00$ , and $\$45.00$ . How much more than her deposit did she spend?

The quarterback ran the ball three times in last Sunday’s football game. He gained $7$ yards on one run but lost $3$ yards and $8$ yards on the other two. What was his total yardage running for the game?

The revenue for a local photographer for the month is $\$1,200$ . His costs include a studio rental of $\$600$ , props costing $\$105$ , materials fees of $\$135$ , and a make-up artist who charges $\$120$ . What is his total profit for the month?

An airplane flying at $30,000$ feet lost $2,500$ feet in altitude and then rose $1,200$ feet. What is the new altitude of the plane?

The temperature was $22^\circ$ at $6:00$ p.m. and dropped $26^\circ$ by midnight. What was the temperature at midnight?

A nurse has $30$ milliliters of saline solution but needs $75$ milliliters of the solution. How much more does she need?

The width of a rectangle is $2$ inches less than its length. If the length measures $16$ inches, determine the width.

The base of a triangle is $3$ feet shorter than its height. If the height measures $5$ feet, find the length of the base.

Answer
1: $2$
3: $55$
5: $4$
7: $-11$
9: $-2$
11: $-36$
13: $6$
15: $3$
17: $1$
19: $\$80$
21: $\$240$
23: $-4^\circ$
25: $14$ inches

EXERCISE 1.2.4

Find the distance between the given numbers on a number line.

$-3$ and $12$

$8$ and $-13$

$-25$ and $-10$

$-100$ and $-130$

$-7$ and $-20$

$0$ and $-33$

$-10$ and $10$

$-36$ and $36$

The coldest temperature on earth, $-129^\circ\text{F}$ , was recorded in 1983 at Vostok Station, Antarctica. The hottest temperature on earth, $136^\circ\text{F}$ , was recorded in 1922 at Al 'Aziziyah, Libya. Calculate earth’s temperature range.

The daily high temperature was recorded as $91^\circ\text{F}$ and the low was recorded as $63^\circ\text{F}$ . What was the temperature range for the day?

A student earned $67$ points on his lowest test and $87$ points on his best. Calculate his test score range.

On a busy day, a certain website may have $12,500$ hits. On a slow day, it may have as few as $750$ hits. Calculate the range of the number of hits.

Answer
1: $15$ units
3: $15$ units
5: $13$ units
7: $20$ units
9: $265^\circ\text{F}$
11: $20$ points

EXERCISE 1.2.5

Discussion Board Topics

Share an example of adding signed numbers in a real-world application.

Demonstrate the associative property of addition with any three real numbers.

Show that subtraction is not commutative

MULTIPLYING AND DIVIDING INTEGERS

LEARNING OBJECTIVES

Multiply and divide signed integers.
Translate English sentences involving multiplication and division into mathematical statements.
Determine the prime factorization of composite numbers.
Interpret the results of quotients involving zero.

MULTIPLICATION AND DIVISION

We begin with a review of what it means to multiply and divide signed numbers. The result of multiplying real numbers is called the product and the result of dividing is called the quotient. Recall that multiplication is equivalent to adding:

$3 \cdot 4 = 4 + 4 + 4 = 12$

Clearly, the product of two positive numbers is positive. Similarly, the product of a positive number and negative number can be written as shown:

$3(-4) = (-4) + (-4) + (-4) = -12$

We see that the product of a positive number and a negative number is negative. Next, explore the results of multiplying two negative numbers. Consider the products in the following illustration and try to identify the pattern:

$$
\begin{array}{rl}
3(-3) = -9 & \
2(-3) = -6 & \text{\color{cyan}The products increase by 3.} \
1(-3) = -3 & \
& \
0(-3) = 0 & \text{\color{cyan}Zero times any real number is zero.} \
& \
(-1)(-3) = 3 & \
(-2)(-3) = 6 & \text{\color{cyan}The pattern continues by increasing the products by 3.} \
(-3)(-3) = 9 &
\end{array}
$$

This shows that the product of two negative numbers is positive. To summarize,

$\text{\color{cyan}positive} \times \text{\color{cyan}positive} = \text{\color{cyan}positive}$
$\text{\color{cyan}positive} \times \text{\color{cyan}negative} = \text{\color{cyan}negative}$
$\text{\color{cyan}negative} \times \text{\color{cyan}negative} = \text{\color{cyan}positive}$

The rules for division are the same because division can always be rewritten as multiplication:

$10 \div (-2) = 10 \cdot \left(-\frac{1}{2}\right) = -5$

The rules for multiplication and division should not be confused with the fact that the sum of two negative numbers is negative.

EXAMPLE 1.3.1

Simplify:
a. $(-3) + (-5)$
b. $(-3)(-5)$

Solution
Here we add and multiply the same two negative numbers.
a. The result of adding two negative numbers is negative.
$(-3) + (-5) = -3 - 5$
$\quad \quad \quad \quad \quad = -8$
b. The result of multiplying two negative numbers is positive.
$(-3)(-5) = 15$

Answer
a. $-8$ b. $15$

Given any real numbers $a$ , $b$ , and $c$ , we have the following properties of multiplication:


Zero factor property:	$a \cdot 0 = 0 \cdot a = 0$
Multiplicative identity property:	$a \cdot 1 = 1 \cdot a = a$
Associative property:	$(a \cdot b) \cdot c = a \cdot (b \cdot c)$
Commutative property:	$a \cdot b = b \cdot a$

EXAMPLE 1.3.2

Simplify:
a. $5 \cdot 0$
b. $10 \cdot 1$

Solution
a. Multiplying by zero results in zero.
$5 \cdot 0 = 0$
b. Multiplying any real number by one results in the same real number.
$10 \cdot 1 = 10$

Answers:
a. $0$ b. $10$

EXAMPLE 1.3.3

Simplify:
a. $(3 \cdot 7) \cdot 2$
b. $3 \cdot (7 \cdot 2)$

Solution
a.
$({\color{cyan}3 \cdot 7}) \cdot 2 = {\color{cyan}21} \cdot 2$
$\quad \quad \quad \quad = 42$
b.
$3 \cdot ({\color{cyan}7 \cdot 2}) = 3 \cdot {\color{cyan}14}$
$\quad \quad \quad \quad = 42$

The value of each expression is 42. Changing the grouping of the numbers does not change the result.
$({\color{cyan}3 \cdot 7}) \cdot 2 = 3 \cdot ({\color{cyan}7 \cdot 2}) = 42$

Answer
a. $42$ b. $42$

At this point, we highlight that multiplication is commutative: the order in which we multiply does not matter and yields the same result.

$2 \cdot 9 = 9 \cdot 2$
$18 = 18$

On the other hand, division is not commutative.

$10 \div 5 \neq 5 \div 10$
$2 \neq \frac{1}{2}$

Use these properties to perform sequential operations involving multiplication and division. When doing so, it is important to perform these operations in order from left to right.

EXAMPLE 1.3.4

Simplify: $5(-3)(-2)(-4)$

Solution
Multiply two numbers at a time as follows:
$\underbrace{5(-3)}(-2)(-4) = \underbrace{-15(-2)}(-4)$
$\quad \quad \quad \quad \quad \quad \quad = \underbrace{30(-4)}$
$\quad \quad \quad \quad \quad \quad \quad = -120$

Figure 1.3.1

Answer
$-120$

Because multiplication is commutative, the order in which we multiply does not affect the final answer. When sequential operations involve multiplication and division, order does matter; hence we must work the operations from left to right to obtain a correct result.

EXAMPLE 1.3.5

Simplify: $10 \div (-2)(-5)$

Solution
Perform the division first; otherwise the result will be incorrect.

Correct!
$10 \div (-2)(-5) = \underbrace{10 \div (-2)}_{\text{\color{green}division first}}(-5)$
$\quad \quad \quad \quad \quad \quad = -5(-5)$
$\quad \quad \quad \quad \quad \quad = 25 \quad \text{\color{green}\checkmark}$

Incorrect!
$10 \div (-2)(-5) = 10 \div \underbrace{(-2)(-5)}_{\text{\color{red}multiplication first}}$
$\quad \quad \quad \quad \quad \quad = 10 \div 10$
$\quad \quad \quad \quad \quad \quad = 1 \quad \text{\color{red}\times}$

Figure 1.3.2

Notice that the order in which we multiply and divide does affect the result. Therefore, it is important to perform the operations of multiplication and division as they appear from left to right.

Answer
$25$

EXAMPLE 1.3.6

Simplify: $-6(3) \div (-2)(-3)$

Solution
Work the operations one at a time from left to right.
$-6(3) \div (-2)(-3)$
$= -18 \div (-2)(-3)$
$= 9(-3)$
$= -27$

EXAMPLE 1.3.7

Try this! Simplify: $-5 \div 5 \cdot 2(-3)$

Video Solution:
(click to see video)

Within text-based applications, the symbol used for multiplication is the asterisk (*) and the symbol used for division is the forward slash (/).

$5 * 3 \quad \text{and} \quad 14 / 2 = 7$

The set of even integers is the set of all integers that are evenly divisible by $2$ . We can also obtain the set of even integers by multiplying each integer by $2$ .

$\{\dots, -6, -4, -2, 0, 2, 4, 6, \dots\} \quad \text{\color{cyan}Even integers}$

The set of odd integers is the set of all integers that are not evenly divisible by $2$ .

$\{\dots, -5, -3, -1, 1, 3, 5, \dots\} \quad \text{\color{cyan}Odd integers}$

A prime number is an integer greater than $1$ that is divisible only by $1$ and itself. The smallest prime number is $2$ and the rest are necessarily odd.

$\{2, 3, 5, 7, 11, 13, 17, 19, 23, \dots\} \quad \text{\color{cyan}Prime numbers}$

Any integer greater than $1$ that is not prime is called a composite number and can be written as a product of primes. When a composite number, such as $30$ , is written as a product, $30 = 2 \cdot 15$ , we say that $2 \cdot 15$ is a factorization of $30$ and that $2$ and $15$ are factors. Note that factors divide the number evenly. We can continue to write composite factors as products until only a product of primes remains.

\begin{array}{c} 30 \\ / \quad \backslash \\ = 2 \cdot 15 \\ \quad \quad / \quad \backslash \\ = 2 \cdot 3 \cdot 5 \end{array}

Figure 1.3.3

The prime factorization of $30$ is $2 \cdot 3 \cdot 5$ .

EXAMPLE 1.3.8

Determine the prime factorization of $70$ .

Solution
Begin by writing $70$ as a product with $2$ as a factor. Then express any composite factor as a product of prime numbers.
$70 = 2 \cdot 35$
$\quad = 2 \cdot 5 \cdot 7$

Since the prime factorization is unique, it does not matter how we choose to initially factor the number because the end result is the same.
$70 = 7 \cdot 10$
$\quad = 7 \cdot 2 \cdot 5$
$\quad = 2 \cdot 5 \cdot 7$

Answer
The prime factorization of $70$ is $2 \cdot 5 \cdot 7$ .

Some tests (called divisibility tests) useful for finding prime factors of composite numbers follow:

If the integer is even, then $2$ is a factor.
If the sum of the digits is evenly divisible by $3$ , then $3$ is a factor.
If the last digit is a $5$ or $0$ , then $5$ is a factor.

Often we find the need to translate English sentences that include multiplication and division terms to mathematical statements. Listed below are some key words that translate to the given operation.

Key Words	Operation
Product, multiplied by, of, times	$*$ or $\cdot$
Quotient, divided by, ratio, per	$/$ or $\div$

EXAMPLE 1.3.9

Calculate the quotient of $20$ and $-10$ .

Solution
The key word “quotient” implies that we should divide.
$20 \div (-10) = -2$

Answer:
The quotient of $20$ and $-10$ is $-2$ .

EXAMPLE 1.3.10

What is the product of the first three positive even integers?

Solution
The first three positive even integers are $\{2, 4, 6\}$ and the key word “product” implies that we should multiply.
$2 \cdot 4 \cdot 6 = 8 \cdot 6$
$\quad \quad \quad = 48$

Answer
The product of the first three positive even integers is $48$ .

EXAMPLE 1.3.11

Joe is able to drive $342$ miles on $18$ gallons of gasoline. How many miles per gallon of gas is this?

Solution
The key word “per” indicates that we must divide the number of miles driven by the number of gallons used:
$\frac{342 \text{ miles}}{18 \text{ gallons}} = 19 \text{ miles per gallon (mpg)}$

Answer
Joe gets $19$ miles per gallon from his vehicle.

In everyday life, we often wish to use a single value that typifies a set of values. One way to do this is to use what is called the arithmetic mean or average. To calculate an average, divide the sum of the values in the set by the number of values in that set.

EXAMPLE 1.3.12

A student earns $75$ , $86$ , and $94$ on his first three exams. What is the student’s test average?

Solution
Add the scores and divide the sum by $3$ .
$\frac{75 + 86 + 94}{3} = \frac{255}{3}$
$\quad \quad \quad \quad \quad = 85$

Answer
The student’s test average is $85$ .

ZERO AND DIVISION

Recall the relationship between multiplication and division:

$\underbrace{12}_{\text{\color{cyan}dividend}} \div \underbrace{6}_{\text{\color{cyan}divisor}} = \underbrace{2}_{\text{\color{cyan}quotient}} \quad \text{because} \quad 6 \cdot 2 = 12$

Figure 1.3.4

In this case, the dividend $12$ is evenly divided by the divisor $6$ to obtain the quotient, $2$ . It is true in general that if we multiply the divisor by the quotient we obtain the dividend. Now consider the case where the dividend is zero and the divisor is nonzero:

$0 \div 6 = 0 \quad \text{Since } 6 \cdot 0 = 0$

This demonstrates that zero divided by any nonzero real number must be zero. Now consider a nonzero number divided by zero:

$12 \div 0 = ? \quad \text{or } 0 \cdot ? = 12$

There is no real number that, when multiplied by zero, results in $12$ . Therefore, the quotient $12 \div 0$ is undefined. Now consider zero divided by zero:

$0 \div 0 = ? \quad \text{or } 0 \cdot ? = 0$

Here any real number seems to work. For example, $0 \cdot 5 = 0$ and $0 \cdot 3 = 0$ . Therefore, the quotient is uncertain or indeterminate.

$0 \div 0 = \frac{0}{0} \quad \text{\color{cyan}Is indeterminate.}$

In this course, we state that $0 \div 0$ is undefined.

KEY TAKEAWAYS

A positive number multiplied by a negative number is negative. A negative number multiplied by a negative number is positive.
Multiplication is commutative and division is not.
When simplifying, work the operations of multiplication and division in order from left to right.
Even integers are numbers that are evenly divisible by $2$ or multiples of $2$ , and all other integers are odd.
A prime number is an integer greater than $1$ that is divisible only by $1$ and itself.
Composite numbers are integers greater than $1$ that are not prime. Composite numbers can be written uniquely as a product of primes.
The prime factorization of a composite number is found by continuing to divide it into factors until only a product of primes remains.
To calculate an average of a set of numbers, divide the sum of the values in the set by the number of values in the set.
Zero divided by any nonzero number is zero. Any number divided by zero is undefined.

EXERCISE 1.3.1

Multiply and divide.

$5(-7)$

$-3(-8)$

$2(-4)(-9)$

$-3 \cdot 2 \cdot 5$

$-12(3)(0)$

$0(-12)(-5)$

$(-1)(-1)(-1)(-1)$

$(-1)(-1)(-1)$

$-100 \div 25$

$25 \div 5(-5)$

$-15(-2) \div 10(-3)$

$-5 \cdot 10 \div 2(-5)$

$(-3)(25) \div (-5)$

$6 * (-3) / (-9)$

$20 / (-5) * 2$

$-50 / 2 * 5$

$-44 / 11 * 2$

$-5 * 3 / (-15)$

$4 * 3 * 2 / 6$

$-6 * 7 / (-2)$

Determine the quotient of $-36$ and $3$ .

Determine the quotient of $26$ and $-13$ .

Calculate the product of $3$ and $-8$ divided by $-2$ .

Calculate the product of $-1$ and $-3$ divided by $3$ .

Determine the product of the first three positive even integers.

Determine the product of the first three positive odd integers.

Answer
1: $-35$
3: $72$
5: $0$
7: $1$
9: $-4$
11: $-9$
13: $15$
15: $-8$
17: $-33$
19: $-60$
21: $-12$
23: $12$
25: $48$

EXERCISE 1.3.2

Determine the prime factorization of the following integers.

$105$

$78$

$138$

$154$

$165$

$330$

Answer
1: $3 \cdot 5 \cdot 7$
3: $2 \cdot 3 \cdot 23$
5: $3 \cdot 5 \cdot 11$

EXERCISE 1.3.3

Calculate the average of the numbers in each of the following sets.

$\{50, 60, 70\}$

$\{9, 12, 30\}$

$\{3, 9, 12, 30, 36\}$

$\{72, 84, 69, 71\}$

The first four positive even integers.

The first four positive odd integers.

Answer
1: $60$
2: $18$
5: $5$

EXERCISE 1.3.4

The distance traveled $D$ is equal to the average rate $r$ times the time traveled $t$ at that rate: $D = rt$ . Determine the distance traveled given the rate and the time.

$60$ miles per hour for $3$ hours

$55$ miles per hour for $3$ hours

$15$ miles per hour for $5$ hours

$75$ feet per second for $5$ seconds

$60$ kilometers per hour for $10$ hours

$60$ meters per second for $30$ seconds

A student club ran a fund-raiser in the quad selling hot dogs. The students sold $122$ hot dog meals for $\$3.00$ each. Their costs included $\$50.00$ for the hot dogs and buns, $\$25.00$ for individually wrapped packages of chips, and $\$35.00$ for the sodas. What was their profit?

A $230$-pound man loses $4$ pounds each week for $8$ weeks. How much does he weigh at the end of $8$ weeks?

Mary found that she was able to drive $264$ miles on $12$ gallons of gas. How many miles per gallon does her car get?

After filling his car with gasoline, Bill noted that his odometer reading was $45,346$ miles. After using his car for a week, he filled up his tank with $14$ gallons of gas and noted that his odometer read $45,724$ miles. In that week, how many miles per gallon did Bill’s car get?

Answer
1: $180$ miles
3: $75$ miles
5: $600$ kilometers
7: $\$256.00$
9: $22$ miles per gallon
11: $0$

EXERCISE 1.3.5

Perform the operations.

$0 \div 9$

$15 \div 0$

$4(-7) \div 0$

$7(0) \div (-15)$

$-5(0) \div 9(0)$

$5 \cdot 2(-3)(-5)$

$-8 - 5 + (-13)$

$-4(-8) \div 16(-2)$

$50 \div (-5) \div (-10)$

$49 \div 7 \div (-1)$

$3 \cdot 4 \div 12$

$0 - (-8) - 12$

$-8 \cdot 4(-3) \div 2$

$0 / (-3 * 8 * 5)$

$(-4 * 3) / (2 * (-3))$

$-16 / (-2 * 2) * 3$

$-44 / 11 * 2$

$-5 * 3 / (-15)$

$4 * 3 * 2 / 6$

$-6 * 7 / (-2)$

During $5$ consecutive winter days, the daily lows were $-7^\circ$ , $-3^\circ$ , $0^\circ$ , $-5^\circ$ , and $-10^\circ$ . Calculate the average low temperature.

On a very cold day the temperature was recorded every 4 hours with the following results: $-16^\circ$ , $-10^\circ$ , $2^\circ$ , $6^\circ$ , $-5^\circ$ , and $-13^\circ$ . Determine the average temperature.

A student earns $9$ , $8$ , $10$ , $7$ , and $6$ points on the first $5$ chemistry quizzes. What is her quiz average?

A website tracked hits on its homepage over the Thanksgiving holiday. The number of hits for each day from Thursday to Sunday was $12,250$; $4,400$; $7,750$; and $10,200$ , respectively. What was the average number of hits per day over the holiday period?

Answer
1: $0$
3: Undefined
5: $0$
7: $-26$
9: $1$
11: $1$
13: $48$
15: $2$
17: $-8$
19: $4$
21: $-5^\circ$
23: $8$ points

Discussion Board Topics.

Demonstrate the associative property of multiplication with any three real numbers.

Show that division is not commutative.

Discuss the importance of working multiplication and division operations from left to right. Make up an example where order does matter and share the solution.

Discuss division involving $0$ . With examples, explain why the result is sometimes $0$ and why it is sometimes undefined.

Research and discuss the fundamental theorem of arithmetic.

Research and discuss other divisibility tests. Provide an example for each test.

The arithmetic mean is one way to typify a set of values. Research other methods used to typify a set of values

FRACTIONS

LEARNING OBJECTIVES

Reduce a fraction to lowest terms.
Multiply and divide fractions.
Add and subtract fractions.

REDUCING

A fraction is a real number written as a quotient, or ratio, of two integers $a$ and $b$ , where $b \neq 0$ .

\begin{array}{r@{\quad\longrightarrow\quad}l} \text{Numerator} & a \\ \text{Denominator} & b \end{array}

Figure 1.4.1

The integer above the fraction bar is called the numerator and the integer below is called the denominator. The numerator is often called the “part” and the denominator is often called the “whole.” Equivalent fractions are two equal ratios expressed using different numerators and denominators. For example,

$\frac{50}{100} = \frac{1}{2}$

Fifty parts out of $100$ is the same ratio as $1$ part out of $2$ and represents the same real number. Consider the following factorizations of $50$ and $100$ :

$50 = 2 \cdot 25$
$100 = 4 \cdot 25$

The numbers $50$ and $100$ share the factor $25$ . A shared factor is called a common factor. We can rewrite the ratio $\frac{50}{100}$ as follows:

$\frac{50}{100} = \frac{2 \cdot 25}{4 \cdot 25}$

Making use of the multiplicative identity property and the fact that $\frac{25}{25} = 1$ , we have

$\frac{50}{100} = \frac{2 \cdot \cancel{25}}{4 \cdot \cancel{25}} = \frac{2}{4} \cdot {\color{cyan}1} = \frac{2}{4}$

Figure 1.4.2

Dividing $\frac{25}{25}$ and replacing this factor with a $1$ is called canceling. Together, these basic steps for finding equivalent fractions define the process of reducing. Since factors divide their product evenly, we achieve the same result by dividing both the numerator and denominator by $25$ as follows:

$\frac{50 \div {\color{cyan}25}}{100 \div {\color{cyan}25}} = \frac{2}{4}$

Figure 1.4.3

Finding equivalent fractions where the numerator and denominator have no common factor other than $1$ is called reducing to lowest terms. When learning how to reduce to lowest terms, it is helpful to first rewrite the numerator and denominator as a product of primes and then cancel. For example,

$\frac{50}{100} = \frac{\cancel{2}^1 \cdot \cancel{5}^1 \cdot \cancel{5}^1}{2 \cdot \cancel{2}_1 \cdot \cancel{5}_1 \cdot \cancel{5}_1} = \frac{1}{2}$

Figure 1.4.4

We achieve the same result by dividing the numerator and denominator by the greatest common factor (GCF). The GCF is the largest number that divides both the numerator and denominator evenly. One way to find the GCF of $50$ and $100$ is to list all the factors of each and identify the largest number that appears in both lists. Remember, each number is also a factor of itself.

$\{1, 2, 5, 10, 25, \mathbf{50}\} \quad \text{\color{cyan}Factors of 50}$
$\{1, 2, 4, 5, 10, 20, 25, \mathbf{50}, 100\} \quad \text{\color{cyan}Factors of 100}$

Common factors are listed in bold, and we see that the greatest common factor is $\mathbf{50}$ . We use the following notation to indicate the GCF of two numbers: $\text{GCF}(50, 100) = \mathbf{50}$ . After determining the GCF, reduce by dividing both the numerator and the denominator as follows:

$\frac{50 \div {\color{cyan}50}}{100 \div {\color{cyan}50}} = \frac{1}{2}$

Figure 1.4.5

EXAMPLE 1.4.1

Reduce to lowest terms: $\frac{105}{300}$ .

Solution
Rewrite the numerator and denominator as a product of primes and then cancel.

$\frac{105}{300} = \frac{\cancel{3}^1 \cdot \cancel{5}^1 \cdot 7}{2 \cdot 2 \cdot \cancel{3}_1 \cdot \cancel{5}_1 \cdot 5}$
$\quad = \frac{7}{2 \cdot 2 \cdot 5}$
$\quad = \frac{7}{20}$

Figure 1.4.6

Alternatively, we achieve the same result if we divide both the numerator and denominator by the $\text{GCF}(105, 300)$ . A quick way to find the GCF of the two numbers requires us to first write each as a product of primes. The GCF is the product of all the common prime factors.

$$
\left.
\begin{array}{l}
105 = {\color{cyan}3} \cdot {\color{cyan}5} \cdot 7 \
300 = 2 \cdot 2 \cdot {\color{cyan}3} \cdot {\color{cyan}5} \cdot 5
\end{array}
\right} \quad \text{\color{cyan}GCF}(105, 300) = {\color{cyan}3 \cdot 5} = 15
$$

Figure 1.4.7

In this case, the common prime factors are $\mathbf{3}$ and $\mathbf{5}$ and the greatest common factor of $105$ and $300$ is $\mathbf{15}$ .

$\frac{105}{300} = \frac{105 \div {\color{cyan}15}}{300 \div {\color{cyan}15}} = \frac{7}{20}$

Figure 1.4.8

Answer:
$\frac{7}{20}$

EXAMPLE 1.4.2

Try this! Reduce to lowest terms: $\frac{32}{96}$ .

An improper fraction is one where the numerator is larger than the denominator. A mixed number is a number that represents the sum of a whole number and a fraction. For example, $5 \frac{1}{2}$ is a mixed number that represents the sum $5 + \frac{1}{2}$ . Use long division to convert an improper fraction to a mixed number; the remainder is the numerator of the fractional part.

EXAMPLE 1.4.3

Write $\frac{23}{5}$ as a mixed number.

Solution
Notice that $5$ divides into $23$ four times with a remainder of $3$ .

$$
\begin{array}{r}
4 \
5 \overline{) 23} \
\underline{-20} \
3
\end{array}
$$

Figure 1.4.9

We then can write

$\frac{23}{5} = 4 + \frac{3}{5}$
$\quad = 4 \frac{3}{5}$

Note that the denominator of the fractional part of the mixed number remains the same as the denominator of the original fraction.

Answer
$4 \frac{3}{5}$

To convert mixed numbers to improper fractions, multiply the whole number by the denominator and then add the numerator; write this result over the original denominator.

EXAMPLE 1.4.4

Write $3 \frac{5}{7}$ as an improper fraction.

Solution
Obtain the numerator by multiplying $7$ times $3$ and then add $5$ .

$3 \frac{5}{7} = \frac{7 \cdot 3 + 5}{7}$
$\quad = \frac{21 + 5}{7}$
$\quad = \frac{26}{7}$

Answer
$\frac{26}{7}$

It is important to note that converting to a mixed number is not part of the reducing process. We consider improper fractions, such as $\frac{26}{7}$ , to be reduced to lowest terms. In algebra it is often preferable to work with improper fractions, although in some applications, mixed numbers are more appropriate.

EXAMPLE 1.4.5

Try this! Convert $10 \frac{1}{2}$ to an improper fraction.

Solution
(click to see video)

MULTIPLYING AND DIVIDING FRACTIONS

In this section, assume that $a, b, c,$ and $d$ are all nonzero integers. The product of two fractions is the fraction formed by the product of the numerators and the product of the denominators. In other words, to multiply fractions, multiply the numerators and multiply the denominators:

$\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}$

EXAMPLE 1.4.6

Multiply: $\frac{2}{3} \cdot \frac{5}{7}$

Solution
Multiply the numerators and multiply the denominators.

$\frac{2}{3} \cdot \frac{5}{7} = \frac{2 \cdot 5}{3 \cdot 7}$
$\quad \quad = \frac{10}{21}$

Answer:
$\frac{10}{21}$

EXAMPLE 1.4.7

Multiply: $\frac{5}{9} (-\frac{1}{4})$

Solution
Recall that the product of a positive number and a negative number is negative.

$\frac{5}{9} \left( -\frac{1}{4} \right) = -\frac{5 \cdot 1}{9 \cdot 4}$
$\quad \quad \quad = -\frac{5}{36}$

Answer:
$-\frac{5}{36}$

EXAMPLE 1.4.8

Multiply: $\frac{2}{3} \cdot 5 \frac{3}{4}$

Solution
Begin by converting $5 \frac{3}{4}$ to an improper fraction.

$\frac{2}{3} \cdot 5 \frac{3}{4} = \frac{2}{3} \cdot \frac{23}{4}$
$= \frac{\cancel{2}^1}{3} \cdot \frac{23}{\cancel{4}_2}$
$= \frac{23}{6}$
$= 3 \frac{5}{6}$

Figure 1.4.10

In this example, we noticed that we could reduce before we multiplied the numerators and the denominators. Reducing in this way is called cross canceling, and can save time when multiplying fractions.

Answer
$3 \frac{5}{6}$

Two real numbers whose product is $1$ are called reciprocals. Therefore, $\frac{a}{b}$ and $\frac{b}{a}$ are reciprocals because $\frac{a}{b} \cdot \frac{b}{a} = \frac{ab}{ab} = 1$ . For example,

$\frac{2}{3} \cdot \frac{3}{2} = \frac{6}{6} = 1$

Because their product is $1$ , $\frac{2}{3}$ and $\frac{3}{2}$ are reciprocals. Some other reciprocals are listed below:

$\frac{5}{8} \text{ and } \frac{8}{5} \quad\quad 7 \text{ and } \frac{1}{7} \quad\quad -\frac{4}{5} \text{ and } -\frac{5}{4}$

This definition is important because dividing fractions requires that you multiply the dividend by the reciprocal of the divisor.

$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c} = \frac{ad}{bc}$

EXAMPLE 1.4.9

Divide: $\frac{2}{3} \div \frac{5}{7}$

Solution
Multiply $\frac{2}{3}$ by the reciprocal of $\frac{5}{7}$ .

$\frac{2}{3} \div \frac{5}{7} = \frac{2}{3} \cdot \frac{7}{5}$
$\quad \quad \quad = \frac{2 \cdot 7}{3 \cdot 5}$
$\quad \quad \quad = \frac{14}{15}$

Answer:
$\frac{14}{15}$

You also need to be aware of other forms of notation that indicate division: / and —. For example,

$5 / (1/2) = 5 * (2/1) = (5/1) * (2/1) = 10/1 = 10$

Or

$\frac{\frac{7}{8}}{\frac{2}{3}} = \frac{7}{8} \div \frac{2}{3} = \frac{7}{8} \cdot \frac{3}{2} = \frac{21}{16}$

The latter is an example of a complex fraction, which is a fraction whose numerator, denominator, or both are fractions.

NOTE
Students often ask why dividing is equivalent to multiplying by the reciprocal of the divisor. A mathematical explanation comes from the fact that the product of reciprocals is $1$ . If we apply the multiplicative identity property and multiply numerator and denominator by the reciprocal of the denominator, then we obtain the following:

$\frac{\frac{7}{8}}{\frac{2}{3}} = \frac{\frac{7}{8}}{\frac{2}{3}} \cdot {\color{green}1} = \frac{\frac{7}{8}}{\frac{2}{3}} \cdot \frac{\color{green}\frac{3}{2}}{\color{green}\frac{3}{2}} = \frac{\frac{7}{8} \cdot \frac{3}{2}}{1} = \frac{7}{8} \cdot \frac{3}{2}$

Figure 1.4.11

Before multiplying, look for common factors to cancel; this eliminates the need to reduce the end result.

EXAMPLE 1.4.10

Divide: $\frac{\frac{5}{2}}{\frac{7}{4}}$ .

Solution

$\frac{\frac{5}{2}}{\frac{7}{4}} = \frac{5}{2} \div \frac{7}{4}$
$\quad \quad = \frac{5}{2} \cdot \frac{4}{7}$
$\quad \quad = \frac{5}{\cancel{2}^1} \cdot \frac{\cancel{4}^2}{7}$
$\quad \quad = \frac{5 \cdot 2}{1 \cdot 7}$
$\quad \quad = \frac{10}{7}$

Figure 1.4.12

Answer
$\frac{10}{7}$

When dividing by an integer, it is helpful to rewrite it as a fraction over $1$ .

EXAMPLE 1.4.11

Divide: $\frac{2}{3} \div 6$

Solution
Rewrite 6 as $\frac{6}{1}$ and multiply by its reciprocal.

$\frac{2}{3} \div 6 = \frac{2}{3} \div \frac{6}{1}$
$\quad \quad = \frac{2}{3} \cdot \frac{1}{6}$
$\quad \quad = \frac{\cancel{2}^1}{3} \cdot \frac{1}{\cancel{6}_3}$
$\quad \quad = \frac{1 \cdot 1}{3 \cdot 3}$
$\quad \quad = \frac{1}{9}$

Figure 1.4.13

Answer:
$\frac{1}{9}$

Also, note that we only cancel when working with multiplication. Rewrite any division problem as a product before canceling.

EXAMPLE 1.4.12

Try this! Divide: $5 \div 2 \frac{2}{3}$

Video Solution:
(click to see video)

ADDING AND SUBTRACTING FRACTIONS

Negative fractions are indicated with the negative sign in front of the fraction bar, in the numerator, or in the denominator. All such forms are equivalent and interchangeable.

$-\frac{3}{4} = \frac{-3}{4} = \frac{3}{-4}$

Adding or subtracting fractions requires a common denominator. In this section, assume the common denominator $c$ is a nonzero integer.

$\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c} \quad\text{and}\quad \frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}$

It is good practice to use positive common denominators by expressing negative fractions with negative numerators. In short, avoid negative denominators.

EXAMPLE 1.4.13

Subtract: $\frac{12}{15} - \frac{3}{15}$

Solution
The two fractions have a common denominator $15$ . Therefore, subtract the numerators and write the result over the common denominator:

$\frac{12}{15} - \frac{3}{15} = \frac{12 - 3}{15} \quad \text{\color{cyan}Subtract the numerators.}$
$\quad \quad \quad \quad = \frac{9}{15}$
$\quad \quad \quad \quad = \frac{9 \div {\color{cyan}3}}{15 \div {\color{cyan}3}} \quad \text{\color{cyan}Reduce.}$
$\quad \quad \quad \quad = \frac{3}{5}$

Answer
$\frac{3}{5}$

Most problems that you are likely to encounter will have unlike denominators. In this case, first find equivalent fractions with a common denominator before adding or subtracting the numerators. One way to obtain equivalent fractions is to divide the numerator and the denominator by the same number. We now review a technique for finding equivalent fractions by multiplying the numerator and the denominator by the same number. It should be clear that $5/5$ is equal to $1$ and that $1$ multiplied times any number is that number:

$\frac{1}{2} = \frac{1}{2} \cdot {\color{cyan}1} = \frac{1}{2} \cdot {\color{cyan}\frac{5}{5}} = \frac{5}{10}$

We have equivalent fractions $\frac{1}{2} = \frac{5}{10}$ . Use this idea to find equivalent fractions with a common denominator to add or subtract fractions. The steps are outlined in the following example.

EXAMPLE 1.4.14

Subtract: $\frac{7}{15} - \frac{3}{10}$

Solution
Step 1: Determine a common denominator. To do this, use the least common multiple (LCM) of the given denominators. The LCM of $15$ and $10$ is indicated by $\text{LCM}(15, 10)$ . Try to think of the smallest number that both denominators divide into evenly. List the multiples of each number:

$\{10, 20, \mathbf{30}, 40, 50, \mathbf{60}, \dots\} \quad \text{\color{cyan}Multiples of 10}$
$\{15, \mathbf{30}, 45, \mathbf{60}, 75, 90, \dots\} \quad \text{\color{cyan}Multiples of 15}$

Common multiples are listed in bold, and the least common multiple is $\mathbf{30}$ .

$\text{LCM}(10, 15) = 30$

Step 2: Multiply the numerator and the denominator of each fraction by values that result in equivalent fractions with the determined common denominator.

$\frac{7}{15} - \frac{3}{10} = \frac{7 \cdot {\color{cyan}2}}{15 \cdot {\color{cyan}2}} - \frac{3 \cdot {\color{cyan}3}}{10 \cdot {\color{cyan}3}}$
$\quad \quad \quad = \frac{14}{30} - \frac{9}{30}$

Step 3: Add or subtract the numerators, write the result over the common denominator and then reduce if possible.

$\frac{14}{30} - \frac{9}{30} = \frac{14 - 9}{30}$
$\quad \quad \quad = \frac{5}{30}$
$\quad \quad \quad = \frac{5 \div {\color{cyan}5}}{30 \div {\color{cyan}5}}$
$\quad \quad \quad = \frac{1}{6}$

Answer:
$\frac{1}{6}$

The least common multiple of the denominators is called the least common denominator (LCD). Finding the LCD is often the difficult step. It is worth finding because if any common multiple other than the least is used, then there will be more steps involved when reducing.

EXAMPLE 1.4.15

Add: $\frac{5}{10} + \frac{1}{18}$

Solution
First, determine that the $\text{LCM}(10, 18)$ is $90$ and then find equivalent fractions with $90$ as the denominator.

$\frac{5}{10} + \frac{1}{18} = \frac{5 \cdot {\color{cyan}9}}{10 \cdot {\color{cyan}9}} + \frac{1 \cdot {\color{cyan}5}}{18 \cdot {\color{cyan}5}}$
$\quad \quad \quad = \frac{45}{90} + \frac{5}{90}$
$\quad \quad \quad = \frac{45 + 5}{90}$
$\quad \quad \quad = \frac{50}{90}$
$\quad \quad \quad = \frac{50 \div {\color{cyan}10}}{90 \div {\color{cyan}10}}$
$\quad \quad \quad = \frac{5}{9}$

Answer
$\frac{5}{9}$

EXAMPLE 1.4.16

Try this! Add: $\frac{2}{30} + \frac{5}{21}$

Video Solution:
(click to see video)

EXAMPLE 1.4.17

Simplify: $2 \frac{1}{3} + \frac{3}{5} - \frac{1}{2}$

Solution
Begin by converting $2 \frac{1}{3}$ to an improper fraction.

$2 \frac{1}{3} + \frac{3}{5} - \frac{1}{2} = \frac{7}{3} + \frac{3}{5} - \frac{1}{2} \quad \text{\color{cyan}Convert to improper fractions.}$
$\quad \quad \quad \quad \quad = \frac{7 \cdot {\color{cyan}10}}{3 \cdot {\color{cyan}10}} + \frac{3 \cdot {\color{cyan}6}}{5 \cdot {\color{cyan}6}} - \frac{1 \cdot {\color{cyan}15}}{2 \cdot {\color{cyan}15}} \quad \text{\color{cyan}LCM}(3, 5, 2) = 30$
$\quad \quad \quad \quad \quad = \frac{70}{30} + \frac{18}{30} - \frac{15}{30} \quad \begin{array}{l}\text{\color{cyan}Equivalent fractions} \\ \text{\color{cyan}with a common denominator}\end{array}$
$\quad \quad \quad \quad \quad = \frac{70 + 18 - 15}{30}$
$\quad \quad \quad \quad \quad = \frac{73}{30}$
$\quad \quad \quad \quad \quad = 2 \frac{13}{30}$

Answer:
$2 \frac{13}{30}$

In general, it is preferable to work with improper fractions. However, when the original problem involves mixed numbers, if appropriate, present your answers as mixed numbers. Also, mixed numbers are often preferred when working with numbers on a number line and with real-world applications.

EXAMPLE 1.4.18

Subtract: $\frac{6}{7} - 2 \frac{1}{7}$

EXAMPLE 1.4.19

How many $\frac{1}{2}$ inch thick paperback books can be stacked to fit on a shelf that is $1 \frac{1}{2}$ feet in height?

Solution
First, determine the height of the shelf in inches. To do this, use the fact that there are $12$ inches in $1$ foot and multiply as follows:

$1 \frac{1}{2} \text{ ft} = 1 \frac{1}{2} \text{ ft} \cdot \left( \frac{12 \text{ in}}{1 \text{ ft}} \right)$
$\quad \quad = \frac{3}{\cancel{2}_1} \cdot \frac{\cancel{12}^6}{1} \text{ in}$
$\quad \quad = 18 \text{ in}$

Next, determine how many notebooks will fit by dividing the height of the shelf by the thickness of each book.

$18 \text{ in} \div \frac{1}{2} \text{ in} = 18 \cdot \frac{2}{1}$
$\quad \quad \quad \quad \quad = 18 \cdot 2$
$\quad \quad \quad \quad \quad = 36$

Answer
$36$ books can be stacked on the shelf.

KEY TAKEAWAYS:

Fractions are not unique; there are many ways to express the same ratio. Find equivalent fractions by multiplying or dividing the numerator and the denominator by the same real number.
Equivalent fractions in lowest terms are generally preferred. It is a good practice to always reduce.
In algebra, improper fractions are generally preferred. However, in real-life applications, mixed number equivalents are often preferred. We may present answers as improper fractions unless the original question contains mixed numbers, or it is an answer to a real-world or geometric application.
Multiplying fractions does not require a common denominator; multiply the numerators and multiply the denominators to obtain the product. It is a best practice to cancel any common factors in the numerator and the denominator before multiplying.
Reciprocals are rational numbers whose product is equal to $1$ . Given a fraction $\frac{a}{b}$ , its reciprocal is $\frac{b}{a}$ .
Divide fractions by multiplying the dividend by the reciprocal of the divisor. In other words, multiply the numerator by the reciprocal of the denominator.
Rewrite any division problem as a product before canceling.
Adding or subtracting fractions requires a common denominator. When the denominators of any number of fractions are the same, simply add or subtract the numerators and write the result over the common denominator.
Before adding or subtracting fractions, ensure that the denominators are the same by finding equivalent fractions with a common denominator. Multiply the numerator and the denominator of each fraction by the appropriate value to find the equivalent fractions.
Typically, it is best to convert all mixed numbers to improper fractions before beginning the process of adding, subtracting, multiplying, or dividing.

EXERCISE 1.4.1

Reduce each fraction to lowest terms.

$\frac{5}{30}$

$\frac{6}{24}$

$\frac{30}{70}$

$\frac{18}{27}$

$\frac{44}{84}$

$\frac{54}{90}$

$\frac{135}{30}$

$\frac{105}{300}$

$\frac{18}{6}$

$\frac{256}{16}$

$\frac{126}{45}$

$\frac{52}{234}$

$\frac{54}{162}$

$\frac{2000}{3000}$

$\frac{270}{360}$

Answer
1: $1/6$
3: $3/7$
5: $11/21$
7: $9/2$
9: $3$
11: $14/5$
13: $1/3$
15: $3/4$

EXERCISE 1.4.2

Rewrite as an improper fraction.

$4 \frac{3}{4}$

$2 \frac{1}{2}$

$5 \frac{7}{15}$

$1 \frac{1}{2}$

$3 \frac{5}{8}$

$1 \frac{3}{4}$

$-2 \frac{1}{2}$

$-1 \frac{3}{4}$

Answer
1: $19/4$
3: $82/15$
5: $29/8$
7: $-5/2$

EXERCISE 1.4.3

Rewrite as a mixed number.

$\frac{15}{2}$

$\frac{9}{2}$

$\frac{40}{13}$

$\frac{103}{25}$

$\frac{73}{10}$

$-\frac{52}{7}$

$-\frac{59}{6}$

Answer
2: $4 \frac{1}{2}$
4: $4 \frac{3}{25}$
6: $-7 \frac{3}{7}$

EXERCISE 1.4.4

Multiply and reduce to lowest terms.

$\frac{2}{3} \cdot \frac{5}{7}$

$\frac{1}{5} \cdot \frac{4}{8}$

$\frac{1}{2} \cdot \frac{1}{3}$

$\frac{3}{4} \cdot \frac{20}{9}$

$\frac{5}{7} \cdot \frac{49}{10}$

$\frac{2}{3} \cdot \frac{9}{12}$

$\frac{6}{14} \cdot \frac{21}{12}$

$\frac{44}{15} \cdot \frac{15}{11}$

$3 \frac{3}{4} \cdot 2 \frac{1}{3}$

$2 \frac{7}{10} \cdot 5 \frac{5}{9}$

$\frac{3}{11} \left(-\frac{5}{2}\right)$

$-\frac{4}{5} \left(\frac{9}{2}\right)$

$\left(-\frac{9}{8}\right) \left(-\frac{3}{10}\right)$

$\frac{6}{7} \left(-\frac{14}{3}\right)$

$\left(-\frac{9}{12}\right) \left(-\frac{4}{8}\right)$

$-\frac{3}{8} \left(-\frac{4}{15}\right)$

$\frac{1}{7} \cdot \frac{1}{2} \cdot \frac{1}{3}$

$\frac{3}{8} \cdot \frac{15}{21} \cdot \frac{7}{27}$

$\frac{2}{5} \cdot 3 \frac{1}{8} \cdot \frac{4}{5}$

$2 \frac{4}{9} \cdot \frac{2}{5} \cdot 2 \frac{5}{11}$

Answer
1: $10/21$
3: $1/6$
5: $7/2$
7: $3/4$
9: $35/4$
11: $-15/22$
13: $27/50$
15: $3/8$
17: $1/42$
19: $1$

EXERCISE 1.4.5

Determine the reciprocal of the following numbers.

$\frac{1}{2}$

$\frac{8}{5}$

$-\frac{2}{3}$

$-\frac{4}{3}$

$10$

$-4$

$2 \frac{1}{3}$

$1 \frac{5}{8}$

Answer
1: $2$
3: $-3/2$
5: $1/10$
7: $3/7$
9: $3/4$

EXERCISE 1.4.6

Divide and reduce to lowest terms.

$\frac{1}{2} \div \frac{2}{3}$

$\frac{5}{9} \div \frac{1}{3}$

$\frac{5}{8} \div \left(-\frac{4}{5}\right)$

$\left(-\frac{2}{7}\right) \div \frac{15}{3}$

$\frac{-\frac{6}{11}}{-\frac{7}{22}}$

$\frac{\frac{1}{2}}{\frac{1}{10}}$

$\frac{-\frac{3}{4}}{-\frac{5}{20}}$

$\frac{\frac{2}{3}}{\frac{9}{2}}$

$\frac{\frac{30}{50}}{\frac{5}{3}}$

$\frac{\frac{1}{2}}{2}$

$\frac{5}{\frac{2}{5}}$

$\frac{-6}{\frac{5}{8}}$

$2 \frac{1}{2} \div \frac{5}{3}$

$4 \frac{2}{3} \div 3 \frac{1}{2}$

$5 \div 2 \frac{3}{5}$

$4 \frac{3}{8} \div 23$

Answer
1: $3/4$
3: $-25/32$
5: $12/7$
7: $3$
9: $9/25$
11: $25/2$
13: $3/2$
15: $25/13$

EXERCISE 1.4.7

Add or subtract and reduce to lowest terms.

$\frac{17}{20} - \frac{5}{20}$

$\frac{4}{9} - \frac{13}{9}$

$\frac{3}{5} + \frac{1}{5}$

$\frac{11}{15} + \frac{9}{15}$

$\frac{5}{7} - \frac{2}{7}$

$\frac{1}{2} + \frac{1}{3}$

$\frac{1}{5} - \frac{1}{4}$

$\frac{3}{4} - \frac{5}{2}$

$\frac{3}{8} + \frac{7}{16}$

$\frac{7}{15} - \frac{3}{10}$

$\frac{3}{10} + \frac{2}{14}$

$\frac{2}{30} + \frac{5}{21}$

$\frac{3}{18} - \frac{1}{24}$

$5 \frac{1}{2} + 2 \frac{1}{3}$

$1 \frac{3}{4} + 2 \frac{1}{10}$

$\frac{1}{2} + \frac{1}{3} + \frac{1}{6}$

$\frac{2}{3} + \frac{3}{5} - \frac{2}{9}$

$\frac{7}{3} - \frac{3}{2} + \frac{2}{15}$

$\frac{9}{4} - \frac{3}{2} + \frac{3}{8}$

$\frac{2}{3} - 4 \frac{1}{2} + 3 \frac{1}{6}$

$1 - \frac{6}{16} + \frac{3}{18}$

$3 - \frac{1}{21} - \frac{1}{15}$

Answer
1: $3/5$
3: $4/5$
5: $3/7$
7: $5/6$
9: $-7/4$
11: $1/6$
13: $32/105$
15: $47/6$
17: $1$
19: $29/30$
21:
23: $19/24$

EXERCISE 1.4.8

Perform the operations. Reduce answers to lowest terms.

$\frac{3}{14} \cdot \frac{7}{3} + \frac{1}{8}$

$\frac{1}{2} \cdot \left(-\frac{4}{5}\right) + \frac{14}{15}$

$\frac{1}{2} \div \frac{3}{4} \cdot \frac{1}{5}$

$-\frac{5}{9} \div \frac{5}{3} \cdot \frac{5}{2}$

$\frac{4}{5} \div 4 \cdot \frac{1}{2}$

$\frac{5}{3} \div 15 \cdot \frac{2}{3}$

What is the product of $\frac{3}{16}$ and $\frac{4}{9}$ ?

What is the product of $-\frac{24}{5}$ and $\frac{25}{8}$ ?

What is the quotient of $\frac{5}{9}$ and $\frac{25}{3}$ ?

What is the quotient of $-\frac{16}{5}$ and $32$ ?

Subtract $\frac{1}{6}$ from the sum of $\frac{9}{2}$ and $\frac{2}{3}$ .

Subtract $\frac{1}{4}$ from the sum of $\frac{3}{4}$ and $\frac{6}{5}$ .

What is the total width when $3$ boards, each with a width of $2 \frac{5}{8}$ inches, are glued together?

The precipitation in inches for a particular 3-day weekend was published as $\frac{3}{10}$ inches on Friday, $1 \frac{1}{2}$ inches on Saturday, and $\frac{3}{4}$ inches on Sunday. Calculate the total precipitation over this period.

A board that is $5 \frac{1}{4}$ feet long is to be cut into $7$ pieces of equal length. What is length of each piece?

How many $\frac{3}{4}$ inch thick notebooks can be stacked into a box that is $2$ feet high?

In a mathematics class of $44$ students, one-quarter of the students signed up for a special Saturday study session. How many students signed up?

Determine the length of fencing needed to enclose a rectangular pen with dimensions $35 \frac{1}{2}$ feet by $20 \frac{2}{3}$ feet.

Each lap around the track measures $\frac{1}{4}$ mile. How many laps are required to complete a $2 \frac{1}{2}$ mile run?

A retiree earned a pension that consists of three-fourths of his regular monthly salary. If his regular monthly salary was $\$5,200$ , then what monthly payment can the retiree expect from the pension plan?

Answer
1: $5/8$
3: $2/15$
5:
7: $1/10$
9: $1/12$
11: $1/15$
13: $5$
15: $63/8$ inches
17: $3/4$ feet
19: $11$ students
21: $10$ laps

Discussion Board Topics

Does $0$ have a reciprocal? Explain.

Explain the difference between the LCM and the GCF. Give an example.

Explain the difference between the LCM and LCD.

Why is it necessary to find an LCD in order to add or subtract fractions?

Explain how to determine which fraction is larger, $\frac{7}{16}$ or $\frac{1}{2}$ .

EXPONENTS AND SQUARE ROOTS

LEARNING OBJECTIVES

Interpret exponential notation with positive integer exponents.
Calculate the $n\text{th}$ power of a real number.
Calculate the exact and approximate value of the square root of a real number.

EXPONENTIAL NOTATION AND POSITIVE INTEGER EXPONENTS

If a number is repeated as a factor numerous times, then we can write the product in a more compact form using exponential notation. For example,

$5 \cdot 5 \cdot 5 \cdot 5 = 5^4$

The base is the factor, and the positive integer exponent indicates the number of times the base is repeated as a factor. In the above example, the base is $5$ and the exponent is $4$ . In general, if $a$ is the base that is repeated as a factor $n$ times, then

$a^n = \underbrace{a \cdot a \cdot a \cdot \dots \cdot a}_{n \text{ factors of } a}$

Figure 1.6.1

When the exponent is $2$ , we call the result a square. For example,

$3^2 = 3 \cdot 3 = 9$

The number $3$ is the base and the integer $2$ is the exponent. The notation $3^2$ can be read two ways: “three squared” or “$3$ raised to the second power.” The base can be any real number.

It is important to study the difference between the ways the last two examples are calculated. In the example $(-7)^2$ , the base is $-7$ as indicated by the parentheses. In the example $-5^2$ , the base is $5$ , not $-5$ , so only the $5$ is squared and the result remains negative. To illustrate this, write

$-5^2 = -1 \cdot 5^2 = -1 \cdot 5 \cdot 5 = -25$

This subtle distinction is very important because it determines the sign of the result.

The textual notation for exponents is usually denoted using the caret (^) symbol as follows:

$8^2 = 8 \text{ \^{} } 2 = 8 * 8 = 64$
$-5.1^2 = -5.1 \text{ \^{} } 2 = -5.1 * 5.1 = -26.01$

The square of an integer is called a perfect square. The ability to recognize perfect squares is useful in our study of algebra. The squares of the integers from $1$ to $15$ should be memorized. A partial list of perfect squares follows:

$\{0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, \dots\}$

EXERCISE 1.6.1

Simplify
$(-12)^2.$

Answer
$144$

When the exponent is $3$ we call the result a cube. For example,

$3^3 = 3 \cdot 3 \cdot 3 = 27$

The notation $3^3$ can be read two ways: “three cubed” or “$3$ raised to the third power.” As before, the base can be any real number.

Note that the result of cubing a negative number is negative. The cube of an integer is called a perfect cube. The ability to recognize perfect cubes is useful in our study of algebra. The cubes of the integers from $1$ to $10$ should be memorized. A partial list of perfect cubes follows:

$\{0, 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, \dots\}$

EXERCISE 1.6.2

Simplify $(-2)^3$ .

Answer
$-8$

If the exponent is greater than $3$ , then the notation $a^n$ is read “$a$ raised to the $n\text{th}$ power.”

$10^6 = 10 \cdot 10 \cdot 10 \cdot 10 \cdot 10 \cdot 10 = 1,000,000$
$(-1)^4 = (-1)(-1)(-1)(-1) = 1$
$\left(\frac{1}{3}\right)^5 = \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{243}$

Notice that the result of a negative base with an even exponent is positive. The result of a negative base with an odd exponent is negative. These facts are often confused when negative numbers are involved. Study the following four examples carefully:

Table 1.6.1

The base is $(-2)$	The base is $2$
$(-2)^4 = (-2) \cdot (-2) \cdot (-2) \cdot (-2) = +16$	$-2^4 = -2 \cdot 2 \cdot 2 \cdot 2 = -16$
$(-2)^3 = (-2) \cdot (-2) \cdot (-2) = -8$	$-2^3 = -2 \cdot 2 \cdot 2 = -8$

The parentheses indicate that the negative number is to be used as the base.

EXAMPLE 1.6.1

Calculate:
a. $\left(-\frac{1}{3}\right)^3$
b. $\left(-\frac{1}{3}\right)^4$

Solution:
The base is $-\frac{1}{3}$ for both problems.
a. Use the base as a factor three times.
$\left(-\frac{1}{3}\right)^3 = \left(-\frac{1}{3}\right)\left(-\frac{1}{3}\right)\left(-\frac{1}{3}\right)$
$\quad \quad = -\frac{1}{27}$
b. Use the base as a factor four times.
$\left(-\frac{1}{3}\right)^4 = \left(-\frac{1}{3}\right)\left(-\frac{1}{3}\right)\left(-\frac{1}{3}\right)\left(-\frac{1}{3}\right)$
$\quad \quad = +\frac{1}{81}$

Answer:
a. $-\frac{1}{27}$; b. $\frac{1}{81}$

EXERCISE 1.6.3

Simplify:
$-10^4 \text{ and } (-10)^4.$

Answer
$-10,000$ and $10,000$

SQUARE ROOT OF A REAL NUMBER

Think of finding the square root of a number as the inverse of squaring a number. In other words, to determine the square root of $25$ the question is, “What number squared equals $25$ ?” Actually, there are two answers to this question, $5$ and $-5$ .

$5^2 = 25 \quad \text{and} \quad (-5)^2 = 25$

When asked for the square root of a number, we implicitly mean the principal (nonnegative) square root. Therefore we have,

$\sqrt{a^2} = a, \text{ if } a \geq 0 \text{ or more generally } \sqrt{a^2} = |a|$

As an example, $\sqrt{25} = 5$ , which is read “square root of $25$ equals $5$ .” The symbol $\sqrt{\quad}$ is called the radical sign and $25$ is called the radicand.

The alternative textual notation for square roots follows:

$\sqrt{16} = \text{sqrt}(16) = 4$

It is also worthwhile to note that

$\sqrt{1} = 1 \quad \text{and} \quad \sqrt{0} = 0$

This is the case because $1^2 = 1$ and $0^2 = 0$ .

EXAMPLE 1.6.2

Simplify:
$\sqrt{10,000}.$

Solution:
$10,000$ is a perfect square because $100 \cdot 100 = 10,000$ .
$\sqrt{10,000} = \sqrt{(100)^2}$
$\quad \quad \quad = 100$

Answer:
$100$

EXAMPLE 1.6.3

Simplify:
$\sqrt{\frac{1}{9}}.$

Solution:
Here we notice that $\frac{1}{9}$ is a square because $\frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9}$ .
$\sqrt{\frac{1}{9}} = \sqrt{\left(\frac{1}{3}\right)^2}$
$\quad = \frac{1}{3}$

Answer:
$\frac{1}{3}$

Given $a$ and $b$ as positive real numbers, use the following property to simplify square roots whose radicands are not squares:

$\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$

The idea is to identify the largest square factor of the radicand and then apply the property shown above. As an example, to simplify $\sqrt{8}$ notice that $8$ is not a perfect square. However, $8 = 4 \cdot 2$ and thus has a perfect square factor other than $1$ . Apply the property as follows:

$\sqrt{8} = \sqrt{4 \cdot 2}$
$\quad = \sqrt{4} \cdot \sqrt{2}$
$\quad = 2 \cdot \sqrt{2}$
$\quad = 2\sqrt{2}$

Here $2\sqrt{2}$ is a simplified irrational number. You are often asked to find an approximate answer rounded off to a certain decimal place. In that case, use a calculator to find the decimal approximation using either the original problem or the simplified equivalent.

$\sqrt{8} = 2\sqrt{2} \approx 2.83$

On a calculator, try $2.83 \text{ \^{} } 2$ . What do you expect? Why is the answer not what you would expect?

It is important to mention that the radicand must be positive. For example, $\sqrt{-9}$ is undefined since there is no real number that when squared is negative. Try taking the square root of a negative number on your calculator. What does it say?

NOTE
Taking the square root of a negative number is defined later in the course.

EXAMPLE 1.6.4

Simplify and give an approximate answer rounded to the nearest hundredth:
$\sqrt{75}.$

Solution:
The radicand $75$ can be factored as $25 \cdot 3$ where the factor $25$ is a perfect square.
$\sqrt{75} = \sqrt{25 \cdot 3} \quad \text{\color{cyan}The largest perfect square factor of 75 is 25.}$
$\quad = \sqrt{25} \cdot \sqrt{3}$
$\quad = 5 \cdot \sqrt{3} \quad \text{\color{cyan}Exact answer}$
$\quad \approx 8.66 \quad \text{\color{cyan}Approximate answer}$

Answer:
$\sqrt{75} \approx 8.66$

As a check, calculate $\sqrt{75}$ and $5\sqrt{3}$ on a calculator and verify that the both results are approximately $8.66$ .

EXAMPLE 1.6.5

Simplify:
$\sqrt{180}.$

Solution:
$\sqrt{180} = \sqrt{36 \cdot 5}$
$\quad \quad = \sqrt{36} \cdot \sqrt{5}$
$\quad \quad = 6 \cdot \sqrt{5}$
$\quad \quad = 6\sqrt{5}$

Since the question did not ask for an approximate answer, we present the exact answer.

Answer:
$6\sqrt{5}$

EXAMPLE 1.6.6

Simplify:
$-5\sqrt{162}.$

Solution:
$-5\sqrt{162} = -5 \cdot \sqrt{81 \cdot 2}$
$\quad \quad \quad = -5 \cdot \sqrt{81} \cdot \sqrt{2}$
$\quad \quad \quad = -5 \cdot 9 \cdot \sqrt{2}$
$\quad \quad \quad = -45 \cdot \sqrt{2}$
$\quad \quad \quad = -45\sqrt{2}$

Answer:
$-45\sqrt{2}$

EXERCISE 1.6.4

Simplify and give an approximate answer rounded to the nearest hundredth:
$\sqrt{128}.$

Answer
$8\sqrt{2} \approx 11.31$

A right triangle is a triangle where one of the angles measures $90^\circ$ . The side opposite the right angle is the longest side, called the hypotenuse, and the other two sides are called legs. Numerous real-world applications involve this geometric figure. The Pythagorean theorem states that given any right triangle with legs measuring $a$ and $b$ units, the square of the measure of the hypotenuse $c$ is equal to the sum of the squares of the measures of the legs: $a^2 + b^2 = c^2$ . In other words, the hypotenuse of any right triangle is equal to the square root of the sum of the squares of its legs.

$c = \sqrt{a^2 + b^2}$

Figure 1.6.1

EXAMPLE 1.6.7

If the two legs of a right triangle measure $3$ units and $4$ units, then find the length of the hypotenuse.

Solution:
Given the lengths of the legs of a right triangle, use the formula $c = \sqrt{a^2 + b^2}$ to find the length of the hypotenuse.

Figure 1.6.2

$c = \sqrt{a^2 + b^2}$
$c = \sqrt{3^2 + 4^2}$
$\quad = \sqrt{9 + 16}$
$\quad = \sqrt{25}$
$\quad = 5$

Answer:
$c = 5$ units

When finding the hypotenuse of a right triangle using the Pythagorean theorem, the radicand is not always a perfect square.

EXAMPLE 1.6.8

If the two legs of a right triangle measure $2$ units and $6$ units, find the length of the hypotenuse.

Solution:

Figure 1.6.3

$c = \sqrt{a^2 + b^2}$
$\quad = \sqrt{2^2 + 6^2}$
$\quad = \sqrt{4 + 36}$
$\quad = \sqrt{40}$
$\quad = \sqrt{4 \cdot 10}$
$\quad = \sqrt{4} \cdot \sqrt{10}$
$\quad = 2 \cdot \sqrt{10}$

Answer:
$c = 2\sqrt{10}$ units

KEY TAKEAWAYS

When using exponential notation $a^n$ , the base $a$ is used as a factor $n$ times.
When the exponent is $2$ , the result is called a square. When the exponent is $3$ , the result is called a cube.
Memorize the squares of the integers up to $15$ and the cubes of the integers up to $10$ . They will be used often as you progress in your study of algebra.
When negative numbers are involved, take care to associate the exponent with the correct base. Parentheses group a negative number raised to some power.
A negative base raised to an even power is positive.
A negative base raised to an odd power is negative.
The square root of a number is a number that when squared results in the original number. The principal square root is the positive square root.
Simplify a square root by looking for the largest perfect square factor of the radicand. Once a perfect square is found, apply the property $\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$ , where $a$ and $b$ are nonnegative, and simplify.
Check simplified square roots by calculating approximations of the answer using both the original problem and the simplified answer on a calculator to verify that the results are the same.
Find the length of the hypotenuse of any right triangle given the lengths of the legs using the Pythagorean theorem.

EXERCISE 1.6.5 SQUARE OF A NUMBER

Simplify.

$10^2$

$12^2$

$(-9)^2$

$-12^2$

$11^2$

$(-20)^2$

$0^2$

$1^2$

$-(-8)^2$

$-(13)^2$

$\left(\frac{1}{2}\right)^2$

$\left(-\frac{2}{3}\right)^2$

$0.5^2$

$1.25^2$

$(-2.6)^2$

$-(-5.1)^2$

$\left(2\frac{1}{3}\right)^2$

$\left(5\frac{1}{2}\right)^2$

Answer

$100$

$81$

$121$

$0$

$-64$

$\frac{1}{4}$

$.25$

$6.76$

$5\frac{4}{9}$

EXERCISE 1.6.6 SQUARE OF A NUMBER

If $s$ is the length of the side of a square, then the area is given by $A = s^2$ .

Determine the area of a square given that a side measures $5$ inches.

Determine the area of a square given that a side measures $2.3$ feet.

List all the squares of the integers $0$ through $15$ .

List all the squares of the integers from $-15$ to $0$ .

List the squares of all the rational numbers in the set $\{0, \frac{1}{3}, \frac{2}{3}, 1, \frac{4}{3}, \frac{5}{3}, 2\}$ .

List the squares of all the rational numbers in the set $\{0, \frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}\}$ .

Answer
16. $25$ square inches
17. $\{0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225\}$
18. $\{0, \frac{1}{9}, \frac{4}{9}, 1, \frac{16}{9}, \frac{25}{9}, 4\}$

EXERCISE 1.6.7 INTEGER EXPONENTS

Simplify.

$5^3$

$2^6$

$(-1)^4$

$(-3)^3$

$-1^4$

$(-2)^4$

$-7^3$

$(-7)^3$

$-(-3)^3$

$-(-10)^4$

$(-1)^{20}$

$(-1)^{21}$

$(-6)^3$

$-3^4$

$1^{100}$

$0^{100}$

$-\left(\frac{1}{2}\right)^3$

$\left(\frac{1}{2}\right)^6$

$\left(\frac{5}{2}\right)^3$

$-\left(\frac{3}{4}\right)^4$

List all the cubes of the integers $-5$ through $5$ .

List all the cubes of the integers from $-10$ to $0$ .

List all the cubes of the rational numbers in the set $\{-\frac{2}{3}, -\frac{1}{3}, 0, \frac{1}{3}, \frac{2}{3}\}$ .

List all the cubes of the rational numbers in the set $\{-\frac{3}{7}, -\frac{1}{7}, 0, \frac{1}{7}, \frac{3}{7}\}$ .

Answer
43. $125$
44. $1$
45. $-1$
46. $-343$
47. $27$
48. $1$
49. $-216$
50. $1$
51. $-\frac{1}{8}$
52. $\frac{125}{8}$
53. $\{-125, -64, -27, -8, -1, 0, 1, 8, 27, 64, 125\}$
54. $\{-\frac{8}{27}, -\frac{1}{27}, 0, \frac{1}{27}, \frac{8}{27}\}$

EXERCISE 1.6.8 SQUARE ROOT OF A NUMBER

Determine the exact answer in simplified form.

$\sqrt{121}$

$\sqrt{81}$

$\sqrt{100}$

$\sqrt{169}$

$-\sqrt{25}$

$-\sqrt{144}$

$\sqrt{12}$

$\sqrt{27}$

$\sqrt{45}$

$\sqrt{50}$

$\sqrt{98}$

$\sqrt{2000}$

$\sqrt{\frac{1}{4}}$

$\sqrt{\frac{9}{16}}$

$\sqrt{\frac{5}{9}}$

$\sqrt{\frac{8}{36}}$

$\sqrt{0.64}$

$\sqrt{0.81}$

$\sqrt{30^2}$

$\sqrt{15^2}$

$\sqrt{(-2)^2}$

$\sqrt{(-5)^2}$

$\sqrt{-9}$

$\sqrt{-16}$

$3\sqrt{16}$

$5\sqrt{18}$

$-2\sqrt{36}$

$-3\sqrt{32}$

$6\sqrt{200}$

$10\sqrt{27}$

Answer
85. $11$
86. $10$
87. $-5$
88. $2\sqrt{3}$
89. $3\sqrt{5}$
90. $7\sqrt{2}$
91. $\frac{1}{2}$
92. $\frac{\sqrt{5}}{3}$
93. $0.8$
94. $30$
95. $2$
96. Not real
97. $12$
98. $-12$
99. $60\sqrt{2}$

EXERCISE 1.6.9 SQUARE ROOT OF A NUMBER

Approximate the following to the nearest hundredth.

$\sqrt{2}$

$\sqrt{3}$

$\sqrt{10}$

$\sqrt{15}$

$2\sqrt{3}$

$5\sqrt{2}$

$-6\sqrt{5}$

$-4\sqrt{6}$

$\sqrt{79}$

$\sqrt{54}$

$-\sqrt{162}$

$-\sqrt{86}$

If the two legs of a right triangle measure $6$ units and $8$ units, then find the length of the hypotenuse.

If the two legs of a right triangle measure $5$ units and $12$ units, then find the length of the hypotenuse.

If the two legs of a right triangle measure $9$ units and $12$ units, then find the length of the hypotenuse.

If the two legs of a right triangle measure $\frac{3}{2}$ units and $2$ units, then find the length of the hypotenuse.

If the two legs of a right triangle both measure $1$ unit, then find the length of the hypotenuse.

If the two legs of a right triangle measure $1$ unit and $5$ units, then find the length of the hypotenuse.

If the two legs of a right triangle measure $2$ units and $4$ units, then find the length of the hypotenuse.

If the two legs of a right triangle measure $3$ units and $9$ units, then find the length of the hypotenuse.

Answer
120. $1.41$
121. $3.16$
122. $3.46$
123. $-13.42$
124. $8.89$
125. $-12.73$
126. $10$ units
127. $15$ units
128. $\sqrt{2}$ units
129. $2\sqrt{5}$ units

EXERCISE 1.6.10 DISCUSSION BOARD TOPICS

Why is the result of an exponent of $2$ called a square? Why is the result of an exponent of $3$ called a cube?

Research and discuss the history of the Pythagorean theorem.

Research and discuss the history of the square root.

Discuss the importance of the principal square root.

Answer
134. Answers may vary
135. Answers may vary

Real Numbers and their Operations

REAL NUMBERS AND THE NUMBER LINE

LEARNING OBJECTIVES

DEFINITIONS

NUMBER LINE

ORDERING REAL NUMBERS

OPPOSITES

ABSOLUTE VALUE

KEY TAKEAWAYS

ADDING AND SUBTRACTING INTEGERS

LEARNING OBJECTIVES

ADDITION AND SUBTRACTION (+,−)(+, -)(+,−)

DISTANCE ON A NUMBER LINE

KEY TAKEAWAYS

MULTIPLYING AND DIVIDING INTEGERS

LEARNING OBJECTIVES

MULTIPLICATION AND DIVISION

ZERO AND DIVISION

KEY TAKEAWAYS

FRACTIONS

LEARNING OBJECTIVES

REDUCING

MULTIPLYING AND DIVIDING FRACTIONS

ADDING AND SUBTRACTING FRACTIONS

KEY TAKEAWAYS:

EXPONENTS AND SQUARE ROOTS

LEARNING OBJECTIVES

EXPONENTIAL NOTATION AND POSITIVE INTEGER EXPONENTS

SQUARE ROOT OF A REAL NUMBER

KEY TAKEAWAYS

ADDITION AND SUBTRACTION $(+, -)$