Page 316 - HOW TO PROVE IT: A Structured Approach, Second Edition

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302 Mathematical Induction
transitive. We let x, y, and z be arbitrary, assume (x, y) ∈ S and (y, z) ∈ S, and
prove that (x, z) ∈ S. According to the definition of S, the statement (x, y) ∈ S
n
+
means ∃n ∈ Z ((x, y) ∈ R ), so we immediately introduce the variable n to
n
standforapositiveintegersuchthat(x, y) ∈ R .Theassumptionthat(y, z) ∈ S
is handled similarly. The goal (x, z) ∈ S is also an existential statement, so to
k
prove it we must find a positive integer k such that (x, z) ∈ R . We use Lemma
6.5.1 to show that k = m + n works.
Finally, for the third goal we use the natural strategy of letting T be arbitrary,
assumingthat R ⊆ T ⊆ A × A andT istransitive,andthenprovingthat S ⊆ T .
Once again, if you don’t see why the conclusion S ⊆ T follows from ∀n ∈
n
+
Z (R ⊆ T ), as claimed in the proof, you should work out the details of the
proof yourself. This last goal is proven by induction, as you might expect based
n
on the recursive nature of the definition of R . For the induction step, we let n
n
be an arbitrary positive integer, assume that R ⊆ T , and prove that R n+1 ⊆ T .
To prove that R n+1 ⊆ T we take an arbitrary element of R n+1 and prove that it
must be an element of T. Writing out the recursive definition of R n+1 gives us
a way to use the inductive hypothesis, which, as usual, is the key to completing
the induction step.

We end this chapter by returning once again to one of the proofs in the intro-
duction. Recall that in our first proof in the introduction we used the formula

b
b
2b
(2 − 1) · (1 + 2 + 2 +· · · + 2 (a−1)b ) = 2 ab − 1.
We discussed this proof again in Section 3.7 and promised to give a more
careful proof of this formula after we had discussed mathematical induction.
We are ready now to give this more careful proof. Of course, we can also state
the formula more precisely now, using summation notation.
Theorem 6.5.3. For all positive integers a and b,
a−1

b kb ab
(2 − 1) · 2 = 2 − 1.
k=0
Proof. We let b be an arbitrary positive integer and then proceed by induction
on a.
Base case: When a = 1wehave

a−1 0
b kb b kb
(2 − 1) · 2 = (2 − 1) · 2
k=0 k=0
b
= (2 − 1) · 1
= 2 ab − 1.