4.3 Real numbers: multiplication and division

Contemporary math applications Page 1 / 1

This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. The basic operations with real numbers are presented in this chapter. The concept of absolute value is discussed both geometrically and symbolically. The geometric presentation offers a visual understanding of the meaning of |x|. The symbolic presentation includes a literal explanation of how to use the definition. Negative exponents are developed, using reciprocals and the rules of exponents the student has already learned. Scientific notation is also included, using unique and real-life examples.Objectives of this module: be able to multiply and divide signed numbers.

Overview

Multiplication of Signed Numbers
Division of Signed Numbers

Multiplication of signed numbers

Let us consider first the product of two positive numbers.

Multiply: $3 \cdot 5$ .
$3 \cdot 5$ means $5 + 5 + 5 = 15$ .

This suggests that

$(positive number) \cdot (positive number) = positive number$ .

More briefly, $(+) (+) = +$ .

Now consider the product of a positive number and a negative number.

Multiply: $(3) (- 5)$ .
$(3) (- 5)$ means $(- 5) + (- 5) + (- 5) = - 15$ .

This suggests that

$(positive number) \cdot (negative number) = negative number$

More briefly, $(+) (-) = -$ .

By the commutative property of multiplication, we get

$(negative number) \cdot (positive number) = negative number$

More briefly, $(-) (+) = -$ .

The sign of the product of two negative numbers can be determined using the following illustration: Multiply $- 2$ by, respectively, $4, 3, 2, 1, 0, - 1, - 2, - 3, - 4$ . Notice that when the multiplier decreases by 1, the product increases by 2.

$\begin{array}{l} \begin{array}{l} 4 (- 2) = - 8 \\ 3 (- 2) = - 6 \\ 2 (- 2) = - 4 \\ 1 (- 2) = - 2 \end{array}} \to & As we know, (+) (-) = - . \\ 0 (- 2) = 0 \to & As we know, 0 \cdot (any number) = 0. \end{array}$

$\begin{array}{l} \begin{array}{l} - 1 (- 2) = 2 \\ - 2 (- 2) = 4 \\ - 3 (- 2) = 6 \\ - 4 (- 2) = 8 \end{array}} \to & This pattern suggests (-) (-) = + . \end{array}$

We have the following rules for multiplying signed numbers.

Rules for multiplying signed numbers

To multiply two real numbers that have

the same sign , multiply their absolute values. The product is positive.
$\begin{matrix} (+) (+) = + \\ (-) (-) = + \end{matrix}$
opposite signs , multiply their absolute values. The product is negative.
$\begin{matrix} (+) (-) = - \\ (-) (+) = - \end{matrix}$

Sample set a

Find the following products.

$8 \cdot 6$

$\begin{array}{l} Multiply these absolute values . \\ \begin{array}{l} | 8 | = 8 \\ | 6 | = 6 \end{array}} & 8 \cdot 6 = 48 \\ Since the numbers have the same sign, the product is positive . \\ 8 \cdot 6 = + 48 & or 8 \cdot 6 = 48 \end{array}$

$(- 8) (- 6)$

$\begin{array}{l} Multiply these absolute values . \\ \begin{array}{l} | - 8 | = 8 \\ | - 6 | = 6 \end{array}} & 8 \cdot 6 = 48 \\ Since the numbers have the same sign, the product is positive . \\ (- 8) (- 6) = + 48 & or (- 8) (- 6) = 48 \end{array}$

$(- 4) (7)$

$\begin{array}{l} Multiply these absolute values . \\ \begin{matrix} | - 4 | & = 4 \\ | 7 | & = 7 \end{matrix}} & 4 \cdot 7 = 28 \\ Since the numbers have opposite signs, the product is negative . \\ (- 4) (7) = - 28 \end{array}$

$6 (- 3)$

$\begin{array}{l} Multiply these absolute values . \\ \begin{array}{r} | 6 | & = 6 \\ | - 3 | & = 3 \end{array}} & 6 \cdot 3 = 18 \\ Since the numbers have opposite signs, the product is negative . \\ 6 (- 3) = - 18 \end{array}$

Practice set a

Find the following products.

$3 (- 8)$

$- 24$

$4 (16)$

$(- 6) (- 5)$

$(- 7) (- 2)$

$(- 1) (4)$

$- 4$

$(- 7) 7$

$- 49$

Division of signed numbers

We can determine the sign pattern for division by relating division to multiplication. Division is defined in terms of multiplication in the following way.

If $b \cdot c = a$ , then $\frac{a}{b} = c, b \neq 0$ .

For example, since $3 \cdot 4 = 12$ , it follows that $\frac{12}{3} = 4$ .

Notice the pattern:

Since $\underset{b \cdot c = a}{\underset{︸}{3 \cdot 4}} = 12$ , it follows that $\underset{\frac{a}{b} = c}{\underset{︸}{\frac{12}{3}}} = 4$

The sign pattern for division follows from the sign pattern for multiplication.

Since $\underset{b \cdot c = a}{\underset{︸}{(+) (+)}} = +$ , it follows that $\underset{\frac{a}{b} = c}{\underset{︸}{\frac{(+)}{(+)}}} = +$ , that is,
$\frac{(positive number)}{(positive number)} = positive number$
Since $\underset{b \cdot c = a}{\underset{︸}{(-) (-)}} = +$ , it follows that $\underset{\frac{a}{b} = c}{\underset{︸}{\frac{(+)}{(-)}}} = -$ , that is,
$\frac{(positive number)}{(negative number)} = negative number$
Since $\underset{b \cdot c = a}{\underset{︸}{(+) (-)}} = -$ , it follows that $\underset{\frac{a}{b} = c}{\underset{︸}{\frac{(-)}{(+)}}} = -$ , that is,
$\frac{(negative number)}{(positive number)} = negative number$
Since $\underset{b \cdot c = a}{\underset{︸}{(-) (+)}} = -$ , it follows that $\underset{\frac{a}{b} = c}{\underset{︸}{\frac{(-)}{(-)}}} = +$ , that is
$\frac{(negative number)}{(negative number)} = positive number$

We have the following rules for dividing signed numbers.

Rules for dividing signed numbers

To divide two real numbers that have

the same sign , divide their absolute values. The quotient is positive.
$\begin{matrix} \frac{(+)}{(+)} = + & \frac{(-)}{(-)} = + \end{matrix}$
opposite signs , divide their absolute values. The quotient is negative.
$\begin{matrix} \frac{(-)}{(+)} = - & \frac{(+)}{(-)} = - \end{matrix}$

Sample set b

Find the following quotients.

$\frac{- 10}{2}$

$\begin{array}{l} \begin{array}{l} | - 10 | = & 10 \\ | 2 | = & 2 \end{array}} & Divide these absolute values . \\ \frac{10}{2} = 5 \\ \frac{- 10}{2} = - 5 & Since the numbers have opposite signs, the quotient is negative . \end{array}$

$\frac{- 35}{- 7}$

$\begin{array}{l} \begin{array}{l} | - 35 | = & 35 \\ | - 7 | = & 7 \end{array}} & Divide these absolute values . \\ \frac{35}{7} = 5 \\ \frac{- 35}{- 7} = 5 & Since the numbers have same signs, the quotient is positive . \end{array}$

$\frac{18}{- 9}$

$\begin{array}{l} \begin{array}{l} | 18 | = & 18 \\ | - 9 | = & 9 \end{array}} & Divide these absolute values . \\ \frac{18}{9} = 2 \\ \frac{18}{- 9} = - 2 & Since the numbers have opposite signs, the quotient is negative . \end{array}$

Practice set b

Find the following quotients.

$\frac{- 24}{- 6}$

$\frac{30}{- 5}$

$- 6$

$\frac{- 54}{27}$

$- 2$

$\frac{51}{17}$

Sample set c

Find the value of $\frac{- 6 (4 - 7) - 2 (8 - 9)}{- (4 + 1) + 1}$ .

Using the order of operations and what we know about signed numbers, we get

$\begin{array}{l} \frac{- 6 (4 - 7) - 2 (8 - 9)}{- (4 + 1) + 1} & = & \frac{- 6 (- 3) - 2 (- 1)}{- (5) + 1} \\ = & \frac{18 + 2}{- 5 + 1} \\ = & \frac{20}{- 4} \\ = & - 5 \end{array}$

Find the value of $z = \frac{x - u}{s}$ if $x = 57, u = 51, and s = 2$ .

Substituting these values we get

$z = \frac{57 - 51}{2} = \frac{6}{2} = 3$

Practice set c

Find the value of $\frac{- 7 (4 - 8) + 2 (1 - 11)}{- 5 (1 - 6) - 17}$ .

Find the value of $P = \frac{n (n - 3)}{2 n}$ , if $n = 5$ .

Exercises

Find the value of each of the following expressions.

$(- 2) (- 8)$

$(- 3) (- 9)$

$(- 4) (- 8)$

$(- 5) (- 2)$

$(- 6) (- 9)$

$(- 3) (- 11)$

$(- 8) (- 4)$

$(- 1) (- 6)$

$(3) (- 12)$

$- 36$

$(4) (- 18)$

$8 (- 4)$

$- 32$

$5 (- 6)$

$9 (- 2)$

$- 18$

$7 (- 8)$

$(- 6) 4$

$- 24$

$(- 7) 6$

$(- 10) 9$

$- 90$

$(- 4) 12$

$(10) (- 6)$

$- 60$

$(- 6) (4)$

$(- 2) (6)$

$- 12$

$(- 8) (7)$

$\frac{21}{7}$

$\frac{42}{6}$

$\frac{- 39}{3}$

$- 13$

$\frac{- 20}{10}$

$\frac{- 45}{- 5}$

$\frac{- 16}{- 8}$

$\frac{25}{- 5}$

$- 5$

$\frac{36}{- 4}$

$8 - (- 3)$

$14 - (- 20)$

$20 - (- 8)$

$- 4 - (- 1)$

$0 - 4$

$- 4$

$0 - (- 1)$

$- 6 + 1 - 7$

$- 12$

$15 - 12 - 20$

$1 - 6 - 7 + 8$

$- 4$

$2 + 7 - 10 + 2$

$3 (4 - 6)$

$- 6$

$8 (5 - 12)$

$- 3 (1 - 6)$

$- 8 (4 - 12) + 2$

$- 4 (1 - 8) + 3 (10 - 3)$

$- 9 (0 - 2) + 4 (8 - 9) + 0 (- 3)$

$6 (- 2 - 9) - 6 (2 + 9) + 4 (- 1 - 1)$

$- 140$

$\frac{3 (4 + 1) - 2 (5)}{- 2}$

$\frac{4 (8 + 1) - 3 (- 2)}{- 4 - 2}$

$- 7$

$\frac{- 1 (3 + 2) + 5}{- 1}$

$\frac{- 3 (4 - 2) + (- 3) (- 6)}{- 4}$

$- 3$

$- 1 (4 + 2)$

$- 1 (6 - 1)$

$- 5$

$- (8 + 21)$

$- (8 - 21)$

$- (10 - 6)$

$- (5 - 2)$

$- 3$

$- (7 - 11)$

$- (8 - 12)$

$- 3 [(- 1 + 6) - (2 - 7)]$

$- 2 [(4 - 8) - (5 - 11)]$

$- 4$

$- 5 [(- 1 + 5) + (6 - 8)]$

$- [(4 - 9) + (- 2 - 8)]$

$- 3 [- 2 (1 - 5) - 3 (- 2 + 6)]$

$- 2 [- 5 (- 10 + 11) - 2 (5 - 7)]$

$\begin{matrix} P = R - C . & Find P if R = 2000 and C = 2500 . \end{matrix}$

$\begin{matrix} z = \frac{x - u}{s} . & Find z if x = 23, u = 25, and s = 1. \end{matrix}$

$- 2$

$\begin{matrix} z = \frac{x - u}{s} . & Find z if x = 410, u = 430, and s = 2.5. \end{matrix}$

$\begin{matrix} m = \frac{2 s + 1}{T} . & Find m if s = - 8 and T = 5. \end{matrix}$

$- 3$

$\begin{matrix} m = \frac{2 s + 1}{T} . & Find m if s = - 10 and T = - 5. \end{matrix}$

$\begin{matrix} F = (p_{1} - p_{2}) r^{4} \cdot 9. & Find F if p_{1} = 10, p_{2} = 8, r = 3. \end{matrix}$

1458

$\begin{matrix} F = (p_{1} - p_{2}) r^{4} \cdot 9. & Find F if p_{1} = 12, p_{2} = 7, r = 2. \end{matrix}$

$\begin{matrix} P = n (n - 1) (n - 2) . & Find P if n = - 4. \end{matrix}$

$- 120$

$\begin{matrix} P = n (n - 1) (n - 2) (n - 3) . & Find P if n = - 5. \end{matrix}$

$\begin{matrix} P = \frac{n (n - 2) (n - 4)}{2 n} . & Find P if n = - 6. \end{matrix}$

Exercises for review

( [link] ) What natural numbers can replace $x$ so that the statement $- 4 < x \leq 3$ is true?

( [link] ) Simplify $\frac{{(x + 2 y)}^{5} {(3 x - 1)}^{7}}{{(x + 2 y)}^{3} {(3 x - 1)}^{6}}$ .

${(x + 2 y)}^{2} (3 x - 1)$

( [link] ) Simplify ${(x^{n} y^{3 t})}^{5}$ .

( [link] ) Find the sum. $- 6 + (- 5)$ .

$- 11$

( [link] ) Find the difference. $- 2 - (- 8)$ .

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Source: OpenStax, Contemporary math applications. OpenStax CNX. Dec 15, 2014 Download for free at http://legacy.cnx.org/content/col11559/1.6

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