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Let us consider first, the product of two positive numbers. Multiply: $\text{3}\cdot \text{5}$ .
$\text{3}\cdot \text{5}$ means $5+5+5=\text{15}$
This suggests In later mathematics courses, the word "suggests" turns into the word "proof." One example does not prove a claim. Mathematical proofs are constructed to validate a claim for all possible cases. that
$(\text{positive number})\cdot (\text{positive number})=(\text{positive number})$
More briefly,
$(+)(+)=(+)$
Now consider the product of a positive number and a negative number. Multiply: $(3)(-5)$ .
$(3)(-5)$ means $(-5)+(-5)+(-5)=-\text{15}$
This suggests that
$(\text{positive number})\cdot (\text{negative number})=(\text{negative number})$
More briefly,
$(+)(-)=(-)$
By the commutative property of multiplication, we get
$(\text{negative number})\cdot (\text{positive number})=(\text{negative number})$
More briefly,
$(-)(+)=(-)$
The sign of the product of two negative numbers can be suggested after observing the following illustration.
Multiply -2 by, respectively, 4, 3, 2, 1, 0, -1, -2, -3, -4.
We have the following rules for multiplying signed numbers.
Find the following products.
$\text{8}\cdot \text{6}$
$\left(\begin{array}{ccc}\left|8\right|& =& 8\\ \left|6\right|& =& 6\end{array}\right\}$ Multiply these absolute values.
$8\cdot 6=\text{48}$
Since the numbers have the same sign, the product is positive.
Thus, $8\cdot 6\text{=+}\text{48}$ , or $8\cdot 6=\text{48}$ .
$(-8)(-6)$
$\left(\begin{array}{ccc}|-8|& =& 8\\ |-6|& =& 6\end{array}\right\}$ Multiply these absolute values.
$8\cdot 6=\text{48}$
Since the numbers have the same sign, the product is positive.
Thus, $(-8)(-6)\text{=+}\text{48}$ , or $(-8)(-6)=\text{48}$ .
$(-4)(7)$
$\left(\begin{array}{ccc}|-4|& =& 4\\ \left|7\right|& =& 7\end{array}\right\}$ Multiply these absolute values.
$4\cdot 7=\text{28}$
Since the numbers have opposite signs, the product is negative.
Thus, $(-4)(7)=-\text{28}$ .
$6(-3)$
$\left(\begin{array}{ccc}\left|6\right|& =& 6\\ |-3|& =& 3\end{array}\right\}$ Multiply these absolute values.
$6\cdot 3=\text{18}$
Since the numbers have opposite signs, the product is negative.
Thus, $6(-3)=-\text{18}$ .
Find the following products.
To determine the signs in a division problem, recall that
$\frac{\text{12}}{3}=4$ since $\text{12}=3\cdot 4$
This suggests that
What is $\frac{\text{12}}{-3}$ ?
$-\text{12}=(-3)(-4)$ suggests that $\frac{\text{12}}{-3}=-4$ . That is,
What is $\frac{-\text{12}}{3}$ ?
$-\text{12}=(3)(-4)$ suggests that $\frac{-\text{12}}{3}=-4$ . That is,
What is $\frac{-\text{12}}{-3}$ ?
$-\text{12}=(-3)(4)$ suggests that $\frac{-\text{12}}{-3}=4$ . That is,
We have the following rules for dividing signed numbers.
Find the following quotients.
$\frac{-\text{10}}{2}$
$\left(\begin{array}{ccc}|-10|& =& 10\\ \left|2\right|& =& 2\end{array}\right\}$ Divide these absolute values.
$\frac{\text{10}}{2}=5$
Since the numbers have opposite signs, the quotient is negative.
Thus $\frac{-\text{10}}{2}=-5$ .
$\frac{-\text{35}}{-7}$
$\left(\begin{array}{ccc}|-35|& =& 35\\ |-7|& =& 7\end{array}\right\}$ Divide these absolute values.
$\frac{\text{35}}{7}=5$
Since the numbers have the same signs, the quotient is positive.
Thus, $\frac{-\text{35}}{-7}=5$ .
$\frac{\text{18}}{-9}$
$\left(\begin{array}{ccc}\left|18\right|& =& 18\\ |-9|& =& 9\end{array}\right\}$ Divide these absolute values.
$\frac{\text{18}}{9}=2$
Since the numbers have opposite signs, the quotient is negative.
Thus, $\frac{\text{18}}{-9}=2$ .
Find the following quotients.
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,
$\begin{array}{ccc}\hfill \frac{-6(4-7)-2(8-9)}{-(4+1)+1}& =& \frac{-6(-3)-2(-1)}{-\left(5\right)+1}\hfill \\ & =& \frac{18+2}{-5+1}\hfill \\ & =& \frac{20}{-}\hfill \\ & =& -5\hfill \end{array}$
Find the value of $\frac{-5(2-6)-4(-8-1)}{2(3-\text{10})-9(-2)}$ .
14
Calculators with the key can be used for multiplying and dividing signed numbers.
Use a calculator to find each quotient or product.
$(-\text{186})\cdot (-\text{43})$
Since this product involves a $(\text{negative})\cdot (\text{negative})$ , we know the result should be a positive number. We'll illustrate this on the calculator.
Display Reads | ||
Type | 186 | 186 |
Press | -186 | |
Press | × | -186 |
Type | 43 | 43 |
Press | -43 | |
Press | = | 7998 |
Thus, $(-\text{186})\cdot (-\text{43})=\mathrm{7,}\text{998}$ .
$\frac{\text{158}\text{.}\text{64}}{-\text{54}\text{.}3}$ . Round to one decimal place.
Display Reads | ||
Type | 158.64 | 158.64 |
Press | ÷ | 158.64 |
Type | 54.3 | 54.3 |
Press | -54.3 | |
Press | = | -2.921546961 |
Rounding to one decimal place we get -2.9.
Use a calculator to find each value.
$(0\text{.}\text{006})\cdot (-0\text{.}\text{241})$ . Round to three decimal places.
-0.001
Find the value of each of the following. Use a calculator to check each result.
$\left(-3\right)\left(-9\right)$
$\left(-5\right)\left(-2\right)$
$\left(4\right)\left(-\text{18}\right)$
$\left(-6\right)\left(4\right)$
$\left(-8\right)\left(7\right)$
$\frac{\text{42}}{6}$
$\frac{-\text{20}}{\text{10}}$
$\frac{-\text{16}}{-8}$
$\frac{\text{36}}{-4}$
$\text{14}-\left(-\text{20}\right)$
$-4-\left(-1\right)$
$0-\left(-1\right)$
$\text{15}-\text{12}-\text{20}$
$2+7-\text{10}+2$
$8\left(5-\text{12}\right)$
$-8\left(4-\text{12}\right)+2$
$-9\left(0-2\right)+4\left(8-9\right)+0\left(-3\right)$
$6\left(-2-9\right)-6\left(2+9\right)+4\left(-1-1\right)$
-140
$\frac{3\left(4+1\right)-2\left(5\right)}{-2}$
$\frac{-1\left(3+2\right)+5}{-1}$
$\frac{-3\left(4-2\right)+\left(-3\right)\left(-6\right)}{-4}$
-3
$-1\left(4+2\right)$
$-\left(8+\text{21}\right)$
( [link] ) Use the order of operations to simplify $\left({5}^{2}+{3}^{2}+2\right)\xf7{2}^{2}$ .
( [link] ) Find $\frac{3}{8}\text{of}\frac{\text{32}}{9}$ .
$\frac{4}{3}=1\frac{1}{3}$
( [link] ) Write this number in decimal form using digits: “fifty-two three-thousandths”
( [link] ) The ratio of chlorine to water in a solution is 2 to 7. How many mL of water are in a solution that contains 15 mL of chlorine?
$\text{52}\frac{1}{2}$
( [link] ) Perform the subtraction $-8-\left(-\text{20}\right)$
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