Page 157 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Proofs Involving Disjunctions 143
In fact, this rule is the one we used in our first example of deductive reasoning
in Chapter 1!
Once again, we end this section with a proof for you to read without the
benefit of a preliminary scratch work analysis.

Theorem 3.5.5. Suppose m and n are integers. If mn is even, then either m is
even or n is even.
Proof. Suppose mn is even. Then we can choose an integer k such that mn =
2k.If m is even then there is nothing more to prove, so suppose m is odd. Then
m = 2 j + 1 for some integer j. Substituting this into the equation mn = 2k,we
get (2 j + 1)n = 2k,so2 jn + n = 2k, and therefore n = 2k − 2 jn = 2(k −
jn). Since k − jn is an integer, it follows that n is even.

Commentary. The overall form of the proof is the following:
Suppose mn is even.
If m is even, then clearly either m is even or n is even. Now suppose
m is not even. Then m is odd.
[Proof that n is even goes here.]
Therefore either m is even or n is even.
Therefore if mn is even then either m is even or n is even.

The assumptions that mn is even and m is odd lead, by existential instanti-
ation, to the equations mn = 2k and m = 2 j + 1. Although the proof doesn’t
say so explicitly, you are expected to work out for yourself that in order to prove
that n is even it suffices to find an integer c such that n = 2c. Straightforward
algebra leads to the equation n = 2(k − jn), so the choice c = k − jn works.

Exercises

d 1. Suppose A, B, and C are sets. Prove that A ∩ (B ∪ C) ⊆ (A ∩ B) ∪ C.
p ∗
p d2. Suppose A, B, and C are sets. Prove that (A ∪ B) \ C ⊆ A ∪ (B \ C).
p d3. Suppose A and B are sets. Prove that A \ (A \ B) = A ∩ B.
d 4. Suppose A ∩ C ⊆ B ∩ C and A ∪ C ⊆ B ∪ C. Prove that A ⊆ B.
p ∗
p d5. Recall from Section 1.4 that the symmetric difference of two sets A and B
is the set A B = (A \ B) ∪ (B \ A) = (A ∪ B) \ (A ∩ B). Prove that
if A B ⊆ A then B ⊆ A.
p
d6. Suppose A, B, and C are sets. Prove that A ∪ C ⊆ B ∪ C iff A \ C ⊆
B \ C.
d 7. Prove that for any sets A and B, P (A) ∪ P (B) ⊆ P (A ∪ B).
p ∗