Page 297 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Recursion 283
n
Base case: When n = 4wehave n! = 24 > 16 = 2 .
n
Induction step: Let n ≥ 4 be arbitrary and suppose that n! > 2 . Then
(n + 1)! = (n + 1) · n!
> (n + 1) · 2 n (by inductive hypothesis)
n
> 2 · 2 = 2 n+1 .

Example 6.3.2. Prove that for every real number a and all natural numbers
n
m
m and n, a m+n = a · a .
Scratch work

There are three universal quantifiers here, and we’ll treat the first two differ-
ently from the third. We let a and m be arbitrary and then use mathematical
m
n
induction to prove that ∀n ∈ N(a m+n = a · a ). The key algebraic fact in the
n
induction step will be the formula a n+1 = a · a from the recursive definition
of exponentiation.
Solution
Theorem. For every real number a and all natural numbers m and n, a m+n =
m
n
a · a .
Proof. Let a be an arbitrary real number and m an arbitrary natural number.
We now proceed by induction on n.
m
m
m
0
Base case: When n = 0, we have a m+n = a m+0 = a = a · 1 = a · a =
n
m
a · a .
n
m
Induction step. Suppose a m+n = a · a . Then
a m+(n+1) = a (m+n)+1
= a m+n · a (by definition of exponentiation)
n
m
= a · a · a (by inductive hypothesis)
m
= a · a n+1 (by definition of exponentiation).
Example 6.3.3. A sequence of numbers a 0 , a 1 , a 2 ,... is defined recursively
as follows:

a 0 = 0;
for every n ∈ N, a n+1 = 2a n + 1.

Find a formula for a n and prove that your formula is correct.