Page 304 - HOW TO PROVE IT: A Structured Approach, Second Edition

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290 Mathematical Induction
As our first example of the method of strong induction, we prove an important
fact of number theory known as the division algorithm.

Theorem 6.4.1. (Division algorithm) For all natural numbers n and m, if
m > 0, then there are natural numbers q and r such that n = mq + r and
r < m. (The numbers q and r are called the quotient and remainder when n is
divided by m.)
Scratch work

We let m be an arbitrary positive integer and then use strong induction
to prove that ∀n∃q∃r(n = mq + r ∧ r < m). According to the description
of strong induction, this means that we should let n be an arbitrary natu-
ral number, assume that ∀k < n∃q∃r(k = mq + r ∧ r < m), and prove that
∃q∃r(n = mq + r ∧ r < m).
Our goal is an existential statement, so we should try to come up with values
of q and r with the required properties. If n < m then this is easy because we can
just let q = 0 andr = n. But if n ≥ m, then this won’t work, since we must have
r < m, so we must do something different in this case. As usual in induction
proofs, we look to the inductive hypothesis. The inductive hypothesis starts
with ∀k < n, so to apply it we should plug in some natural number smaller
than n for k, but what should we plug in? The reference to division in the
statement of the theorem provides a hint. If we think of division as repeated
subtraction, then dividing n by m involves subtracting m from n repeatedly.
The first step in this process would be to compute n − m, which is a natural
number smaller than n. Perhaps we should plug in n − m for k. It’s not entirely
clear where this will lead, but it’s worth a try. In fact, as you’ll see in the proof,
once we take this step the desired conclusion follows almost immediately.
Notice that we are using the fact that a quotient and remainder exist for some
natural number smaller than n to prove that they exist for n, but this smaller
number is not n − 1, it’s n − m. This is why we’re using strong induction rather
than ordinary induction for this proof.
Proof. We let m be an arbitrary positive integer and then proceed by strong
induction on n.
Suppose n is a natural number, and for every k < n there are natural numbers
q and r such that k = mq + r and r < m.
Case 1. n < m. Let q = 0 and r = n. Then clearly n = mq + r and r < m.
Case 2. n ≥ m. Let k = n − m < n and note that since n ≥ m, k is a nat-
ural number. By inductive hypothesis we can choose q and r such that

k = mq + r and r < m. Then n − m = mq + r ,so n = mq + r + m =

m(q + 1) + r . Thus, if we let q = q + 1 and r = r , then we have n =
mq + r and r < m, as required.