Page 134 - HOW TO PROVE IT: A Structured Approach, Second Edition

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120 Proofs
Solution
Theorem. Suppose B is a set and F is a family of sets. If ∪F ⊆ B then
F ⊆ P (B).
Proof. Suppose ∪F ⊆ B. Let x be an arbitrary element of F. Let y be an
arbitrary element of x. Since y ∈ x and x ∈ F, clearly y ∈∪F. But then since
∪F ⊆ B, y ∈ B. Since y was an arbitrary element of x, we can conclude that
x ⊆ B,so x ∈ P (B). But x was an arbitrary element of F, so this shows that
F ⊆ P (B), as required.

This is probably the most complex proof we’ve done so far. Read it again
and make sure you understand its structure and the purpose of every sentence.
Isn’t it remarkable how much logical complexity has been packed into just a
few lines?
It is not uncommon for a short proof to have such a rich logical structure.
This efficiency of exposition is one of the most attractive features of proofs, but
it also often makes them difficult to read. Although we’ve been concentrating
so far on writing proofs, it is also important to learn how to read proofs written
by other people. To give you some practice with this, we present our last proof
in this section without the scratch work. See if you can follow the structure of
the proof as you read it. We’ll provide a commentary after the proof that should
help you to understand it.
For this proof we need the following definition: For any integers x and y,
we’ll say that x divides y (or y is divisible by x)if ∃k ∈ Z(kx = y). We use the
notation x | y to mean “x divides y.” For example, 4 | 20, since 5 · 4 = 20.

Theorem 3.3.6. For all integers a, b, and c, if a | b and b | c then a | c.
Proof. Let a, b, and c be arbitrary integers and suppose a | b and b | c. Since
a | b, we can choose some integer m such that ma = b. Similarly, since b | c,
we can choose an integer n such that nb = c. Therefore c = nb = nma,so
since nm is an integer, a | c.

Commentary. The theorem says ∀a ∈ Z∀b ∈ Z∀c ∈ Z(a | b ∧ b | c → a | c),
so the most natural way to proceed is to let a, b, and c be arbitrary integers,
assume a | b and b | c, and then prove a | c. The first sentence of the proof indi-
cates that this strategy is being used, so the goal for the rest of the proof must
be to prove that a | c. The fact that this is the goal for the rest of the proof is not
explicitly stated. You are expected to figure this out for yourself by using your
knowledge of proof strategies. You might even want to make a givens and goal
list to help you keep track of what is known and what remains to be proven as