Another major rule that applies to addition involves how the addends are grouped when you have three or more numbers. You can lump the addends together any way you want, and you’ll always end up with the same result. The simplest case of this rule, called the associative law for addition, involves sums of three integers.
It works when you add
Here’s how a mathematician would state the associative law for three addends. For any three integersa, b, and c
(a+b)+c=a+ (b+c) For instance:
(3 + 5) + 7 = 8 + 7 = 15 and
3+ (5 + 7) = 3 + 12 = 15
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60 Addition and Subtraction
This works whether the numbers are positive, negative, or 0. It also works if there are four or more numbers in a sum.
It fails when you subtract
In subtraction, the grouping does matter. It’s easy to see why you can’t apply the associative law to subtraction and get away with it. Consider this:
(3 − 5) − 7 = −2− 7 = −9 but
3 − (5 − 7) = 3 − (−2)
= 3 + 2 = 5
In formal terms, we would say that for any three integers a, b, and c,it is not always true that (a−b)−c=a− (b−c)
The associative law hardly ever works with subtraction.
Mixing the signs
How about mixed addition and subtraction? Let’s try it with some integers.
(3 + 5) − 7 = 8 − 7 = 1 and
3+ (5 − 7) = 3 + (−2)= 1
It works in this case. What happens if we switch the positions of the signs?
(3 − 5) + 7 = −2+ 7 = 5 but
3− (5 + 7) = 3 − 12 = −9
This time, it fails! Now we know that with mixed signs, the associative law sometimes works, but not always. If something is to be called a law in mathematics, “sometimes” does not suf-fice. “Usually” won’t do the job either. Even “almost always” is not good enough. In order to be a law, something has to work all the time.
It’s easy to prove that a law does not hold in every possible case. You only have to find one case where it fails, called a counterexample, to show that something can’t be called a law. Prov-ing that a law always works is more difficult. You can’t do it usProv-ing specific integers, because you’d have to try an infinite number of cases one at a time. You have to use airtight logic.
That’s what proofs are all about.
Add, then subtract
The associative law can work indirectly when subtraction is involved, if you change every sub-traction into addition. It’s the same trick as with the commutative law. For any three integers a, b, andc
(a+b)−c= (a+b)+ (−c)
=a+ [b+ (−c)]
=a+ (b−c)
We use square parentheses, called brackets, to indicate a grouping with another group-ing inside. This looks messy, but a little junk is easy to tolerate when you realize that we’ve just proved something significant. We’ve shown that you can always use the associative law with mixed signs when the first operation is addition and the second one is subtraction.
Subtract, then add
Now let’s look the general situation where we have seen that direct application of the asso-ciative law does not always work. How can we modify it to make it work? We can rewrite an expression where the first operation is subtraction and the second one is addition, so it becomes all addition, like this:
(a−b)+c= [a+ (−b)]+c We can use the associative law for addition to get
a+ [(−b)+c]
Now we can use the commutative law for addition inside the brackets to get a+ [c+ (−b)]
We can simplify this to
a+ (c−b)
Now we know that when the first operation is subtraction and the second is addition, (a−b)+c=a+ (c−b)
Subtract, then subtract again
Finally, let’s explore the situation where we subtract twice. We can rearrange an expression like that as follows:
(a−b)−c= [a+ (−b)]+ (−c)
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62 Addition and Subtraction
We can use the associative law for addition to get a+ [(−b)+ (−c)]
We can simplify this to
a+ (−b−c)
Now we know that when both operations are subtraction, (a−b)−c=a+ (−b−c)
Are you confused?
We haven’t started to dissect the anatomy of multiplication yet. That will come in the next chapter. But you’ve had basic multiplication in your arithmetic classes, so let’s “cheat” for a moment and take advantage of that. What are you actually doing when you change c to −c? Here’s an alternative to the “number reflec-tor” idea. When you want to find the negative (also called its additive inverse) of any integer, multiply by
−1. It works like this:
c × (−1)= −c and
−c × (−1)=c
Here’s a challenge!
Based on the commutative law for the sum of two integers and the associative law for the sum of three integers, show that for any three integers a, b, and c
a+b+c=c+b+a
Solution
If you can manipulate the left-hand side of this equation to get the expression on the right-hand side, that’s good enough. Because these statements are not very complicated and the proof is not too hard, you can write it as a table with statements on the left and reasons on the right. Table 4-1 shows how it’s done. This is a simple statements/reasons (S/R) proof.
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. Evaluate and compare these two sums:
a= |−3+ 4 + (−5)+ 6|
and
b= |−3|+ |4| + |−5|+ |6|
What general fact can you deduce from the results?
2. To illustrate the importance of the placement of parentheses in a mixed sum and difference, evaluate the following two expressions. In long strings of sums and
differences, you should first perform the operations inside the parentheses from left to right, and then perform the operations outside the parentheses from left to right. Here are the expressions:
(3+ 5) − (7 + 9) − (11 + 13) − 15 and
3+ (5 − 7) + (9 − 11) + (13 − 15)
3. Using the rules explained in the previous exercise, how should you evaluate the string of sums and differences if there are no parentheses at all? Here it is:
3+ 5 − 7 + 9 − 11 + 13 − 15
4. Suppose someone tells you that there was a significant trend in the mid-winter average temperatures in the town of Hoodopolis during the period 1998 through 2005. You want to find out if this is true. You come across some old heating bills from the utility company that show how much warmer or cooler a given month was, on the average,
Table 4-1. Here is a proof that shows how you can reverse the order in which three integers a,b, and c are added, and get the same sum. As you read down the left-hand column,
each statement is equal to all the statements above it.
Statements Reasons
Q.E.D. Latin Quod erat demonstradum, translated into English as “Which was to be proved”
Practice Exercises 63
64 Addition and Subtraction
compared to the same month in the previous year. You look at the records for January and find out that in Hoodopolis,
• January 2005 averaged 5 degrees cooler than January 2004.
• January 2004 averaged 2 degrees warmer than January 2003.
• January 2003 averaged 1 degree cooler than January 2002.
• January 2002 averaged 7 degrees warmer than January 2001.
• January 2001 averaged the same temperature as January 2000.
• January 2000 averaged 6 degrees cooler than January 1999.
• January 1999 averaged 3 degrees warmer than January 1998.
What was the difference in the average temperature between January 2005 and January 1998 in the town of Hoodopolis?
5. Show at least one situation where can you say that a−b=b−a
Don’t use the trivial case where a and b are both equal to 0.
6. Show at least one situation where you can say that (a−b)+c=a− (b+c)
where a, b, and c are integers. Don’t use the trivial case where a, b, andc are all 0.
7. Show at least one situation where you say that (a−b)−c=a− (b−c)
where a,b, and c are integers. Don’t use the trivial case where a,b, and c are all 0.
8. Based on the commutative law for the sum of two integers and the associative law for the sum of three integers, construct an S/R proof showing that for any four integers a,b,c, and d
a+b+c+d=d+c+b+a
Here’s a hint: use the solution to the last “challenge” problem in this chapter as a shortcut. A previously proved fact, when used to prove something new, is called a lemma.
9. Based on the associative law for the sum of three integers, prove that for any four integersa, b, c, and d
(a+b+c)+d=a+ (b+c+d)
Do this in narrative form. Don’t use the S/R table method. Here’s a hint: “zip up” the sumb+c, and call it by another name.
10. Simplify and compare these expressions:
(a+b−c)+ (a−b+c) and
a+ (b−c)+ (a−b)+c
In this chapter we’ll explore the mechanics of multiplication and division. Multiplication is a shortcut for repeated addition. Division is something like “undoing” multiplication. These two operations, like addition and subtraction, obey certain rules as long as they aren’t “stretched”
too far!