Page 303 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Strong Induction 289
enough to make the proof work, and we need to assume that all smaller natural
numbers have the property. This is the idea behind a variant of mathematical
induction sometimes called strong induction:
To prove a goal of the form ∀n ∈ NP(n):
Prove that ∀n[(∀k < nP(k)) → P(n)], where both n and k range over the
natural numbers in this statement. Of course, the most direct way to prove this
is to let n be an arbitrary natural number, assume that ∀k < nP(k), and then
prove P(n).
Note that no base case is necessary in a proof by strong induction. All that is
needed is a modified form of the induction step in which we prove that if every
natural number smaller than n has the property P, then n has the property P.
In a proof by strong induction, we refer to the assumption that every natural
number smaller than n has the property P as the inductive hypothesis.
Toseewhystronginductionworks,itmighthelpifwefirstreviewbrieflywhy
ordinary induction works. Recall that a proof by ordinary induction enables us
to go through all the natural numbers in order and see that each of them has
some property P. The base case gets the process started, and the induction step
shows that the process can always be continued from one number to the next.
But note that in this process, by the time we check that some natural number
n has the property P, we’ve already checked that all smaller numbers have the
property. In other words, we already know that ∀k < nP(k). The idea behind
strong induction is that we should be allowed to use this information in our
proof of P(n).
Let’s work out the details of this idea more carefully. Suppose that we’ve
followed the strong induction proof strategy and proven the statement
∀n[(∀k < nP(k)) → P(n)]. Then, plugging in 0 for n, we can conclude that
(∀k < 0P(k)) → P(0). But because there are no natural numbers smaller than
0, the statement ∀k < 0P(k) is vacuously true. Therefore, by modus ponens,
P(0) is true. (This explains why the base case doesn’t have to be checked sepa-
rately in a proof by strong induction; the base case P(0) actually follows from
the modified form of the induction step used in strong induction.) Similarly,
plugging in 1 for n we can conclude that (∀k < 1P(k)) → P(1). The only
natural number smaller than 1 is 0, and we’ve just shown that P(0) is true, so
the statement ∀k < 1P(k) is true. Therefore, by modus ponens, P(1) is also
true. Now plug in 2 for n to get the statement (∀k < 2P(k)) → P(2). Since
P(0) and P(1) are both true, the statement ∀k < 2P(k) is true, and therefore
by modus ponens, P(2) is true. Continuing in this way we can show that P(n)
is true for every natural number n, as required. For an alternative justification
of the method of strong induction, see exercise 1.