HOW AND WHEN TO USE THE CONTRADICTION METHOD With the contradiction method, you begin by assuming that A is true, just

as you do in the forward-backward method. However, to reach the desired conclusion that B is true, you proceed by asking yourself the simple question,

“Why can’t B be false?” After all, if B is supposed to be true, then there must be some reason why B cannot be false. The objective of the contradiction method is to discover that reason.

In other words, the idea of a proof by contradiction is to assume that A is true and B is false, and see why this cannot happen. So what does it mean to “see why this cannot happen?” Suppose, for example, that, as a result of assuming A is true and B is false (hereafter written as N OT B), you were somehow able to reach the conclusion that 1 = 0! Would that not convince you that it is impossible for A to be true and B to be false simultaneously?

Thus, in a proof by contradiction, you assume that A is true and N OT B is true, using the techniques in Chapter 8 to write the statement N OT B.

You must use this information to reach a contradiction to something that you absolutely know is true.

Another way of viewing the contradiction method is to recall from Table 1.1 on page 4 that the statement “A implies B” is true in all cases except when A is true and B is false. With a proof by contradiction, you rule out this one unfavorable case by assuming that it actually does happen, and then reaching a contradiction. At this point, several natural questions arise:

1. What contradiction should you be looking for?

2. Exactly how do you use the assumption that A is true and B is false to reach the contradiction?

3. Why and when should you use this approach instead of the forward-backward method?

9.2 HOW AND WHEN TO USE THE CONTRADICTION METHOD 103

Fig. 9.1 Comparing the forward-backward and contradiction methods.

The first question is by far the hardest to answer because there are no specific guidelines. Each problem gives rise to its own contradiction. It takes creativity, insight, persistence, and sometimes luck to produce a contradiction.

As to the second question, the most common approach to finding a con-tradiction is to work forward from the assumptions that A and N OT B are true, as is illustrated in a moment.

The foregoing discussion also indicates why you might wish to use contra-diction instead of the forward-backward method. With the forward-backward method you assume only that A is true, while in the contradiction method you assume that both A and N OT B are true. Thus, you have two statements from which to reason forward instead of just one (see Figure 9.1). On the other hand, the disadvantage of the contradiction method is that you do not know what the contradiction will be and therefore cannot work backward.

As a general rule, use contradiction when the statement N OT B gives you useful information. There are at least two recognizable instances when this happens. Recall the statement B associated with Proposition 12, “n is an even integer.” Because an integer is either odd or even, when you assume that B is not true—that is, that n is not an even integer—it must be the case that n is odd, resulting in some useful information. In general, when the statement B is one of two possible alternatives, the contradiction method is likely to be effective because, by assuming N OT B, you will know that the other case must happen, and that should help you to reach a contradiction.

A second instance when contradiction is likely to be successful is when the statement B contains the keyword “no” or “not”. This is because, as you learned in Chapter 8, assuming N OT B eliminates the word “no” or “not,”

which can result in a useful statement to work forward from, as shown now.

Proposition 13 If r is a real number such that r²= 2, then r is irrational.

Analysis of Proof. It is important to note that you can rewrite the conclu-sion of Proposition 13 to read “r is not rational.” In this form, the appearance of the keyword “not” now suggests using the contradiction method, whereby you assume that A and N OT B are both true—in this case,

A:

A1 (NOT B):

r²= 2, and

r is a rational number.

A contradiction must now be reached using this information.

Working forward from A1 by using Definition 7 on page 26 for a rational number, you can state that

A2: There are integers p and q with q6= 0 such that r = p/q.

There is still the unanswered question of where the contradiction arises, and this takes a lot of creativity. A crucial observation here really helps—it is possible to assume that

A3: p and q have no common divisor (that is, there is no integer that divides both p and q).

The reason for this is that, if p and q did have a common divisor, you could divide this integer out of both p and q.

Now you can reach a contradiction to A3 by showing that 2 is a common divisor of p and q. This is done by working forward to show that p and q are even, and hence 2 divides them both.

Working forward by squaring both sides of the equality in A2, it follows that

A4: r²= p²/q².

But from A you also know that r²= 2, so A5: 2 = p²/q².

The rest of the forward process is mostly rewriting A5 via algebraic manipu-lations to reach the desired contradiction that both p and q are even integers.

Those steps and their justifications are provided in the following table.

Statement Reason

A6: 2q²= p². Multiply both sides of A5 by q². A7: p²is even. From A6 because p² is 2 times

some integer; namely, q².

A8: p is even. From Proposition 12.

A9: p = 2k, for some integer k. Definition of an even integer.

A10: 2q²= (2k)²= 4k². Substitute p = 2k from A9 in A6.

A11: q²= 2k². Divide A10 through by 2.

A12: q²is even. From A11 because q² is 2 times some integer; namely, k².

A13: q is even. From Proposition 12.

9.3 ADDITIONAL USES FOR THE CONTRADICTION METHOD 105

So, both p and q are even (see A8 and A13), and this contradicts A3, thus completing the proof.

Proof of Proposition 13. Assume, to the contrary, that r is a rational number of the form p/q (where p and q are integers with q 6= 0) and that r²= 2. Furthermore, it can be assumed that p and q have no common divisor for, if they did, this number could be canceled from both the numerator p and the denominator q. Because r² = 2 and r = p/q, it follows that 2 = p²/q², or equivalently, 2q²= p². Noting that 2q² is even, p², and hence p, are even.

Thus, there is an integer k such that p = 2k. On substituting this value for p, one obtains 2q²= p²= (2k)² = 4k², or equivalently, q²= 2k². From this it then follows that q², and hence q, are even. Thus it has been shown that both p and q are even and have the common divisor 2. This contradiction establishes the claim.

This proof, discovered in ancient times by a follower of Pythagoras, epito-mizes the use of contradiction. Try to prove the statement some other way.

9.3 ADDITIONAL USES FOR THE CONTRADICTION METHOD