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More Examples 267
18. Suppose a is a real number and a < 0. Prove that for all n ∈ N,if n is
n
n
even then a > 0, and if n is odd then a < 0.
∗ 19. Suppose a and b are real numbers and 0 < a < b.
n
n
(a) Prove that for all n ≥ 1, 0 < a < b . (Notice that this generalizes
Theorem 3.1.2.)
√ √
(b) Prove that for all n ≥ 2, 0 < n a < n b.
n
n
(c) Prove that for all n ≥ 1, ab + ba < a n+1 + b n+1 .
(d) Prove that for all n ≥ 2,
n n n
a + b a + b
< .
2 2
6.2. More Examples

We introduced mathematical induction in the last section as a method for prov-
ing that all natural numbers have some property. However, the applications of
mathematical induction extend far beyond the study of the natural numbers. In
this section we’ll look at some examples of proofs by mathematical induction
that illustrate the wide range of uses of induction.

Example 6.2.1. Suppose R is a partial order on a set A. Prove that every finite,
nonempty set B ⊆ A has an R-minimal element.

Scratch work

You might think at first that mathematical induction is not appropriate for this
proof, because the goal doesn’t seem to have the form ∀n ∈ NP(n). In fact,
the goal doesn’t explicitly mention natural numbers at all! But we can see that
natural numbers enter into the problem when we recognize that to say that B
is finite and nonempty means that it has n elements, for some n ∈ N, n ≥ 1.
(We’ll give a more careful definition of the number of elements in a finite set
in Chapter 7. For the moment, an intuitive understanding of this concept will
suffice.) Thus, the goal means ∀n ≥ 1∀B ⊆ A(B has n elements → B has a
minimal element). We can now use induction to prove this statement.
In the base case we will have n = 1, so we must prove that if B has one
element, then it has a minimal element. It is easy to check that in this case the
one element of B must be minimal.
For the induction step we let n ≥ 1 be arbitrary, assume that ∀B ⊆ A (B has
n elements → B has a minimal element), and try to prove that ∀B ⊆ A(B has
n + 1 elements → B has a minimal element). Guided by the form of the goal,
we let B be an arbitrary subset of A, assume that B has n + 1 elements, and try
to prove that B has a minimal element.