Page 169 - HOW TO PROVE IT: A Structured Approach, Second Edition

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More Examples of Proofs 155
3.7. More Examples of Proofs

So far, most of our proofs have involved fairly straightforward applications of
the proof techniques we’ve discussed. We end this chapter with a few exam-
ples of somewhat more difficult proofs. These proofs use the techniques of this
chapter, but for various reasons they’re a little harder than most of our earlier
proofs. Some are simply longer, involving the application of more proof strate-
gies. Some require clever choices of which strategies to use. In some cases it’s
clear what strategy to use, but some insight is required to see exactly how to
use it. Our earlier examples, which were intended only to illustrate and clarify
the proof techniques, may have made proof-writing seem mechanical and dull.
We hope that by studying these more difficult examples you will begin to see
that mathematical reasoning can also be surprising and beautiful.
Some proof techniques are particularly difficult to apply. For example, when
you’re proving a goal of the form ∃xP(x), the obvious way to proceed is to try
to find a value of x that makes the statement P(x) true, but sometimes it will
not be obvious how to find that value of x. Using a given of the form ∀xP(x)is
similar. You’ll probably want to plug in a particular value for x, but to complete
the proof you may have to make a clever choice of what to plug in. Proofs that
must be broken down into cases are also sometimes difficult to figure out. It is
sometimes hard to know when to use cases and what cases to use.
We begin by looking again at the proofs from the introduction. Some aspects
of these proofs probably seemed somewhat mysterious when you read them
in the introduction. See if they make more sense to you now that you have a
better understanding of how proofs are constructed. We will present each proof
exactly as it appeared in the introduction and then follow it with a commentary
discussing the proof techniques used.

Theorem 3.7.1. Suppose n is an integer larger than 1 and n is not prime. Then
n
2 − 1 is not prime.
Proof. Since n is not prime, there are positive integers a and b such that
b b 2b
a < n, b < n, and n = ab. Let x = 2 − 1 and y = 1 + 2 + 2 + ··· +
2 (a−1)b . Then

b
b
2b
xy = (2 − 1) · (1 + 2 + 2 +· · · + 2 (a−1)b )
2b
b
2b
b
b
= 2 · (1 + 2 + 2 +· · · + 2 (a−1)b ) − (1 + 2 + 2 +· · · + 2 (a−1)b )
ab
b
2b
3b
b
2b
= (2 + 2 + 2 + ··· + 2 ) − (1 + 2 + 2 +· · · + 2 (a−1)b )
= 2 ab − 1
n
= 2 − 1.