4
MULTIPLYING AND DIVIDING
SIGNED NUMBERS
We can only do arithmetic in the usual way. To calculate 5(−2), we have to do 5· 2 = 10 -- and then decide on the sign. Is it +10 or −10? To decide, we have the followng Rule of Signs.
1. What is the Rule of Signs for multiplying, dividing, and
fractions?
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Do the problem yourself first!
Like signs produce a positive number;
unlike signs, a negative number.
| Example 1. | −5(−2) | = | 10. | Like signs. |
| 5(−2) | = | −10. | Unlike signs. | |
| −12 −4 |
= | 3. | Like signs. | |
| 12 −4 |
= | −3. | Unlike signs. | |
For an explaination of these rules, see below.
2. Write the formal Rule of Signs as it applies to fractions.
| −a −b |
= | a b |
−a b |
= − | a b |
a −b |
= − | a b |
| Examples. | −2 −6 |
= | 1 3 |
|
| −2 6 |
= | − | 1 3 |
|
| 1 −3 |
= | − | 1 3 |
|
Problem 1. Calculate the following.
| a) | 7(−8) = −56 | b) | (−7)8 = −56 | c) | 8(−7) = −56 | d) | −8(−7) = 56 | |||
| e) | (−3)7 = −21 | f) | 5(−9) = −45 | g) | −6(−9) = 54 | h) | −8(−4) = 32 | |||
Problem 2. Evaluate the following. (Be careful to distinguish the operations.)
| a) | 4 − 6 = −2 | b) | 4(−6) = −24 | |
| c) | (−4) − 6 = −10 | d) | (−4)(−6) = 24 | |
| e) | 5 − 8· 2 = −11 | f) | (5 − 8)· 2 = −6 | |
| g) | 5 − 8 + 2 = −1 | h) | 5 − (8 + 2) = −5 | |
| i) | 2 − 3(−6) = 20 | j) | (2 − 3)(−6) = 6 | |
Example 2. The form a − b(−c). Consider a problem in this form:
3 − 5(−2).
We are to subtract 5 times −2:
| 3 − 5(−2) | = | 3 − (−10) |
| = | 3 + 10 | |
| = | 13. | |
And so even though the problem means to subtract (5 times −2), we may interpret it to mean: −5 times −2 = +10. We may simply write
| 3 − 5(−2) | = | 3 + 10 |
| = | 13. | |
In other words, any problem that looks like this --
a − b(−c)
-- we may evaluate like this:
a + bc.
Problem 3. Evaluate the following.
| a) | 8 − 2(−4) = 16 | b) | 9 − 5(−2) = 19 | |
| c) | −20 − 3(−5) = −5 | d) | −70 − 9(−7) = −7 | |
| e) | 3 + 4(−9) = −33 | f) | −6 + 5(−2) = −16 | |
| g) | −10 − 2(4 −8) = −2 | h) | (−10 − 2)(4 −8) = (−12)(−4) = 48 | |
Problem 4. Two variables. Let the value of y depend on the value
of x as follows:
y = 3x − 6.
Calculate the value of y that corresponds to each value of x:
When x = 0, y = 3· 0 − 6 = 0 − 6 = −6.
When x = 1, y = 3· 1 − 6 = 3 − 6 = −3 .
When x = −1, y = 3· −1 − 6 = −3 − 6 = −9.
When x = 2, y = 3· 2 − 6 = 6 − 6 = 0.
When x = −2, y = 3· −2 − 6 = −6 − 6 = −12.
When x = 3, y = 3· 3 − 6 = 9 − 6 = 3.
When x = −3, y = 3· −3 − 6 = −9 − 6 = −15.
Problem 5. Negative factors.
| a) | (−2)(−2) = 4 | b) | (−2)(−2)(−2) = −8 | |
| c) | (−2)(−2)(−2)(−2) = 16 | d) | (−2)(−2)(−2)(−2)(−2) = −32 | |
Problem 6. According to the previous problem:
An even number of negative factors produces a positive number. While an odd number of negative factors produces a negative number.
Problem 7. Calculate.
| a) | 2(−3)4 = −24 | b) | (−2)3(−4) = 24 | c) | 2(−3 −4) = −14 | ||
| d) | (−3)(−4)(−5) = −60 | e) | (−1)(−2)(−3)(−4) = 24 | ||||
| f) | (−2)13(−5)2 = 260 | g) | 25(−8)(−3)(−4) = −100· 24 = −2400 | ||||
| h) | (−1)(−1)(−1) = −1 | i) | (−1)(−1)(−1)(−1) = 1 | ||||
Problem 8. Evaluate each of the following as a positive or negative fraction in lowest terms, or as an integer.
| a) | −24 6 |
= − | 4 | b) | 24 −6 |
= − | 4 | c) | −24 −6 |
= | 4 | ||
| d) | 3 −12 |
= − | 1 4 |
e) | −8 −20 |
= | 2 5 |
f) | −18 42 |
= − | 3 7 |
||
| g) | −2 3 |
= − | 2 3 |
h) | 2 −3 |
= − | 2 3 |
i) | −2 −3 |
= | 2 3 |
||
| j) | −12 3 |
= − | 4 | k) | −5 −20 |
= | 1 4 |
l) | 3 −4 |
= − | 3 4 |
||
| Example 3. Multiplying fractions. |  |
To multiply fractions, multiply the numerators, and
multiply the denominators.
Problem 9. Multiply.
| a) | −3 5 |
· | 7 8 |
= |
−21 40 |
= | − |
21 40 |
b) | 1 2 |
· | −x 4 |
= |
−x 8 |
= | − |
x 8 |
| c) | −2 3 |
· | x −8 |
= |
−2x −24 |
= |
x 12 |
d) | x −6 |
· | 3 −5 |
= |
3x 30 |
= |
x 10 |
|||
An explanation of the Rule of Signs
The inclusion of a negative factor changes the sign of a product. That is, if ab is positive, then (−a)b cannot also be positive. It must be negative.
For, suppose (−a)b were also positive. Then ab + (−a)b would be positive. But that sum is actually 0
ab + (−a)b = [a + (−a)]b = 0· b = 0
Therefore, (−a)b cannot be positive. It must be negative.
In fact, since
ab + (−a)b = 0,
(−a)b must be the negative of ab.
(−a)b = −ab.
Similarly, the inclusion of two negative factors will change the sign twice -- the sign will return to positive
(−a)(−b) = ab.
Next Lesson: Some rules of algebra
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