When you come across a sum or difference that is multiplied by a single number or variable, you will sometimes want to expand it into a sum or difference of products. To do this, you can use the distributive laws.
Multiplication over addition
Suppose you have three integers a, b, and c arranged so that you must multiply a by the sum of b and c. The left-hand distributive law of multiplication over addition tells you that this is equal to the product ab plus the product bc. This comes out simpler if you write it down:
a(b+c)=ab+ac
Now imagine multiplying the sum of a and b by c. The right-hand distributive law of multiplication over addition says that
(a+b)c=ac+bc
Table 5-2. Here is a proof that shows how you can reverse the order in which three integers a,b, and c are multiplied, and get the same product. As you read down the left-hand column, each statement is
equal to all the statements above it.
Statements Reasons
abc Begin here
a(bc) Group the second two integers
a(cb) Commutative law for the product of b and c (cb)a Commutative law for the product of a and (cb) cba Ungroup the first two integers
Q.E.D. Mission accomplished
78 Multiplication and Division
Multiplication over subtraction
The distributive laws also work with subtraction. For any three integers a, b, and c a(b−c)=ab−ac
and
(a−b)c=ac−bc
It’s not hard to show how these follow from the laws for addition. If you want to do a rigor-ous job, the process is rather long. Table 5-3 breaks the derivation down into an S/R process, showing every logical step, for the left-hand distributive law of multiplication over subtraction.
If you want to do a proof for the right-hand distributive law of multiplication over subtraction, consider it a bonus exercise!
The left-hand law fails with division
The left-hand distributive laws do not work for division over addition or for division over subtraction. To see why, all you have to do is produce examples of failure. That’s easy! Con-sider this:
24/(4+ 2) = 24/6 = 4 but
24/4+ 24/2 = 6 + 12 = 18 And this:
24/(4− 2) = 24/2 = 12
The Distributive Laws 79
Table 5-3. Derivation of the left-hand distributive law for multiplication over subtraction. As you read down, each statement is equal to all the statements above it.
Warning: Don’t mistake the expression (–1) for the subtraction of 1!
The parentheses emphasize that –1 is a factor in a product.
Statements Reasons
a(b−c) Begin here
a[b+ (−c)] Convert the subtraction to the addition of a negative ab+a(−c) Left-hand distributive law of multiplication over addition ab+ac(−1) Principle of the sign-changing element
ab+a(−1)c Commutative law for multiplication ab+ (−1)ac Commutative law for multiplication (again)
ab+ (−ac) Principle of sign-changing element (the other way around) ab−ac Convert the addition of a negative to a subtraction
Q.E.D. Mission accomplished
but
24/4− 24/2 = 6 − 12 = −6
The right-hand law works with division
If you have a sum or difference as the dividend and the single number as the divisor, you can use the distributive law for division over addition or division over subtraction. If a and b are any integers, and if c is any nonzero integer, then
(a+b)/c=a/c+b/c and
(a−b)/c=a/c−b/c
Are you confused?
To help get rid of possible confusion about how the distributive laws operate when integers are negative, try an example where a= −2,b= −3, and c= −4. First, work out the expression where you multiply a times (b+c):
−2× [−3+ (−4)]= −2× (−7)= 14 Now work out the expression where you add ab and ac:
[−2× (−3)]+ [−2× (−4)]= 6 + 8 = 14 Here’s a challenge!
With the aid of the commutative and distributive laws, prove that for any four integers a,b, c, and d (a+b)(c+d)=ac+ad+bc+bd
Solution
Let’s do this as a narrative. Even though S/R proofs look neat, narrative proofs are often preferred by mathematicians. We’ll start with the left-hand side of the above equation:
(a+b)(c+d)
Let’s think of the sum a+b as a single unit, and call it e. Now we can rewrite this as e(c+d)
The left-hand distributive law for multiplication over addition allows us to rewrite this as ec+ed
80 Multiplication and Division
Next, we expand e back into its original form and substitute it into the above expression twice, getting (a+ b)c+(a+ b)d
We can apply the right-hand distributive law twice, and rewrite this as ac+bc+ad+bd
Now we employ the commutative law in a generalized way, obtaining ac+ad+bc+bd
We know this is equal to the expression we began with, because we just got done “morphing” it using known tactics, one step at a time. Therefore
(a+b)(c+d)=ac+ad+bc+bd Q.E.D.
Practice Exercises
This is an open-book quiz. You may (and should) refer to the text as you solve these problems.
Don’t hurry! You’ll find worked-out answers in App. A. The solutions in the appendix may not represent the only way a problem can be figured out. If you think you can solve a particular problem in a quicker or better way than you see there, by all means try it!
1. How does the absolute value change if you multiply an integer by −3?
2. How does the absolute value change if you start with an integer and multiply by −3 over and over without end?
3. Evaluate the following expression:
4+ 32 / 8 × (−2)+ 20 / 5 / 2 − 8 4. Do an S/R proof showing that for any four integers a, b, c, and d
abcd= dcba
Here’s a hint: you solved a problem like this for addition in Chap. 4. This proof proceeds in the same way.
5. Start with the integer −15, multiply by −45, then divide that result by −25, then multiply that result by −9, then divide that result by −81, and finally multiply that result by −5. What do you end up with?
6. Show at least one situation where you can say that a/(b/c)= (a/b)/c
where a, b, and c are integers. Do not use the trivial case a= 1, b= 1, and c= 1. Find something more interesting!
Practice Exercises 81
82 Multiplication and Division
7. Show at least one situation where you can say that (ab)/c=a(b/c)
where a, b, and c are integers. Do not use the trivial case a= 1, b= 1, and c= 1. Find something more interesting!
8. Suppose you have learned the left-hand distributive law for multiplication over addition, but you have never heard about the right-hand version. Show that for any three integers n, m, andp, the right-hand distributive law for multiplication over addition will always work:
(m+n)p=mp+np Use the narrative form, not an S/R table.
9. Construct an S/R table showing that for any two integers d and g
−(d+g)= −d−g
10. Prove that when you want to find the negative of the subtraction of one quantity from another, you can simply switch the order of the subtraction. To do this, put together an S/R table showing that for any two integers h and k
−(h−k)=k−h
You can always divide an integer by another integer, except when the divisor is 0. Then you get a fraction. A fraction might not be an integer, but it’s still a number.