Why Read This Book? - Index of

3.4 The Principle of Mathematical Induction 81
define T = N − S = N ∩ S C , then T ⊆ N and n ∈ T . Thus T is a non-empty subset
of the positive integers, which, by the WOP, contains a smallest element a. Now it
is impossible that a = 1 because 1 ∈ S. Thus a>1, so that a − 1 ∈ N. Furthermore,
since a is the smallest element of T , it must be that a − 1 /∈ T . Thus a − 1 ∈ S. But
by I 2, since a − 1 ∈ S, it follows that a − 1 + 1 = a ∈ S. But if a ∈ S, then a/∈ T .
This is a contradiction. Thus there is no smallest element of T , which means that
T is empty. Therefore, N ⊆ S, whence S = N.
What doesTheorem 3.4.6 have to do with proving the claims in Examples 3.4.1–
3.4.5? Think of these examples as statements about the positive integers. They
say effectively that a certain formula or statement is true for all n ≥ 1. To be as
general as possible, address such a statement by P(n). Let’s define S to be the
set of all positive integers n for which the statement P(n) is true. The trick is to
show that S has properties I 1–I 2. Then the PMI will allow us to conclude that
S = N, which is the same as saying P(n) is true for all n. First we show that 1 ∈ S by
showing P(1) is true. This is called establishing the base case. Then we show that
n ∈ S ⇒ n + 1 ∈ S by supposing P(n) is true, and using this to show that P(n + 1)
is true. The assumption that n ∈ S is called the inductive assumption, and the part
of the proof where we show that n ∈ S implies n + 1 ∈ S is called the inductive
step. Having shown that S has properties I 1–I 2, we then conclude S = N; that is,
P(n) holds true for all n. Here is a sample theorem.
Theorem 3.4.7 (Sample). For all positive integers n, P(n).
Proof. We prove by induction on n ≥ 1.
(I 1) P(1).
(I 2) Suppose n ≥ 1 and P(n). Then .... Thus P(n + 1).
Therefore, by induction, P(n) is true for all n ≥ 1.
There will be some point in step I 2 where you use the inductive assumption
P(n) to get over the hump of showing P(n + 1). In the proof of the next theorem,
we rewrite a sum in the following way:
�n+1
�
�n
ak = (a1 + a2 +···+an) + an+1 =
k=1
Theorem 3.4.8 For all n ≥ 1,
n�
k 2 =
k=1
Proof. We prove by induction on n ≥ 1.
n(n + 1)(2n + 1)
6
k=1
ak
�
+ an+1
(3.37)
(3.38)