Page 142 - HOW TO PROVE IT: A Structured Approach, Second Edition

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128 Proofs
2
We must now find an integer j such that x = 2 j + 1. Plugging in 2k + 1 for
2
2
2
2
2
x we get x = (2k + 1) = 4k + 4k + 1 = 2(2k + 2k) + 1, so j = 2k + 2k
looks like the right choice.
Before giving the final write-up of the proof, we should make a few explana-
tory remarks. The two conditional statements we’ve proven can be thought of
as representing the two directions → and ← of the biconditional symbol ↔ in
the original goal. These two parts of the proof are sometimes labeled with the
symbols → and ←. In each part, we end up proving a statement that asserts the
existence of a number with certain properties. We called this number j in the
scratch work, but note that j was not mentioned explicitly in the statement of
the problem. As in the proof of Theorem 3.3.6, we have chosen not to mention
j explicitly in the final proof either.
Solution
2
Theorem. Suppose x is an integer. Then x is even iff x is even.
Proof. (→) Suppose x is even. Then for some integer k, x = 2k. Therefore,
2
2
2
2
2
x = 4k = 2(2k ), so since 2k is an integer, x is even. Thus, if x is even
2
then x is even.
(←) Suppose x is odd. Then x = 2k + 1 for some integer k. Therefore,
2
2
2
2
2
x = (2k + 1) = 4k + 4k + 1 = 2(2k + 2k) + 1, so since 2k + 2k is an
2
2
integer, x is odd. Thus, if x is even then x is even.
Using the proof techniques we’ve developed, we can now verify some of the
equivalences that we were only able to justify on intuitive grounds in Chapter
2. As an example of this, let’s prove that the formulas ∀x¬P(x) and ¬∃xP(x)
are equivalent. To say that these formulas are equivalent means that they will
always have the same truth value. In other words, no matter what statement
P(x) stands for, the statement ∀x¬P(x) ↔¬∃xP(x) will be true. We can
prove this using our technique for proving biconditional statements.
Example 3.4.3. Prove that ∀x¬P(x) ↔¬∃xP(x).
Scratch work
(→) We must prove ∀x¬P(x) →¬∃xP(x), so we assume ∀x¬P(x) and try
to prove ¬∃xP(x). Our goal is now a negated statement, and reexpressing it
would require the use of the very equivalence that we are trying to prove! We
therefore fall back on our only other strategy for dealing with negative goals,
proof by contradiction. We now have the following situation:
Givens Goal
∀x¬P(x) Contradiction
∃xP(x)