Page 305 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Strong Induction 291
The division algorithm can also be extended to negative integers n, and it
can be shown that for every m and n the quotient and remainder q and r are
unique. For more on this, see exercise 13.
Our next example is another important theorem of number theory. We used
this theorem in our proof in the introduction that there are infinitely many
primes. For more on this theorem, see exercise 18.
Theorem 6.4.2. Every integer n > 1 is either prime or a product of primes.

Scratch work
We write the goal in the form ∀n ∈ N[n > 1 → (n is prime ∨ n is a product
of primes)] and then use strong induction. Thus, our inductive hypothesis is
∀k < n[k > 1 → (k is prime ∨ k is a product of primes)], and we must prove
that n > 1 → (n is prime ∨ n is a product of primes). Of course, we start by
assuming n > 1, and according to our strategies for proving disjunctions, a
good way to complete the proof would be to assume that n is not prime and
prove that it must be a product of primes. Because the assumption that n is not
prime means ∃a∃b(n = ab ∧ a < n ∧ b < n), we immediately use existential
instantiation to introduce the new variables a and b into the proof. Applying
the inductive hypothesis to a and b now leads to the desired conclusion.
Proof. We use strong induction. Suppose n > 1, and suppose that for every
integer k,if1 < k < n then k is either prime or a product of primes. Of course,
if n is prime then there is nothing to prove, so suppose n is not prime. Then we
can choose natural numbers a and b such that n = ab, a < n, and b < n. Note
that since a < n = ab, it follows that b > 1, and similarly we must have a > 1.
Thus, by inductive hypothesis, each of a and b is either prime or a product of
primes. But then since n = ab, n is a product of primes.
The method of recursion studied in the last section also has a strong form.
As an example of this, consider the following definition of a sequence of num-
bers, called the Fibonacci numbers after the Italian mathematician Fibonacci
(ca 1174–1250) who first defined them.
F 0 = 0;
F 1 = 1;
for every n ≥ 2, F n = F n−2 + F n−1 .

For example, plugging in n = 2 in the last equation we find that F 2 = F 0 +
F 1 = 0 + 1 = 1. Similarly, F 3 = F 1 + F 2 = 1 + 1 = 2, F 4 = F 2 + F 3 = 1 +
2 = 3, and so on. Note that, starting with F 2 , each Fibonacci number is com-
puted using, not just the previous number in the sequence, but also the one
before that. This is the sense in which the recursion is strong. It shouldn’t be