Mathematical induction

The principle of mathematical induction states that

In order to prove a statement about an integer n, if we can 1. Prove the statement when n=b, for some fixed integer b,

and

2. Show that the truth of the statement for n=k−1 implies the truth of the statement for n=kwhenever k > b, then we can conclude the statement is true for all integersn≥b. As an example, let us give yet another proof that a set with n elements has 2n subsets. This proof uses essentially the the same bijections we used in proving the Pascal Equation. The statement we wish to prove is the statement that “A set of size nhas 2nsubsets.”

Our statement is true when n= 0, because a set of size 0 is the empty set and the empty set has 1 = 20 subsets. (This step of our proof is called a base step.)

Now suppose that k >0 and every set with k−1 elements has 2k−1 subsets. Suppose S = {a1, a2, . . . ak} is a set with k ele-

ments. We partition the subsets of S into two blocks. Block B1 consists of the subsets that do not contain an and blockB2 con- sists of the subsets that do containan. Each set inB1 is a subset of{a1, a2, . . . ak−1}, and each subset of{a1, a2, . . . ak−1}is inB1. Thus B1 is the set of all subsets of {a1, a2, . . . ak−1}. Therefore by our assumption in the first sentence of this paragraph, the size of B1 is 2k−1. Consider the function from B2 to B1 which takes a subset of S including ak and removes ak from it. This

function is defined on B2, because every set in B2 contains ak.

This function is onto, because if T is a set inB1, thenT∪ {ak}

is a set in B2 which the function sends to T. This function is one-to-one because ifV andW are two different sets inB2, then removing ak from them gives two different sets in B1. Thus we have a bijection between B1 and B2, so B1 and B2 have the same size. Therefore by the sum principle the size of B1∪B2 is 2k−1+ 2k−1 = 2k. Therefore S has 2k subsets. This shows that if a set of sizek−1 has 2k−1 _{subsets, then a set of sizek has 2}k

subsets. Therefore by the principle of mathematical induction, a set of size n has 2n subsets for every nonnegative integer n.

The first sentence of the last paragraph is called theinductive hypothesis. In an inductive proof we always make an inductive hypothesis as part of proving that the truth of our statement whenn=k−1 implies the truth of our statement whenn=k. The last paragraph itself is called theinductive step of our proof. In an inductive step we derive the statement for n = k from the statement for n = k−1, thus proving that the truth of our statement whenn=k−1 implies the truth of our statement when n=k. The last sentence in the last paragraph is called the inductive conclusion. All inductive proofs should have a base step, an inductive hypothesis, an inductive step, and an inductive conclusion.

There are a couple details worth noticing. First, in this problem, our base step was the case n = 0, or in other words, we had b = 0. However, in other proofs,bcould be any integer, positive, negative, or 0. Second, our proof that the truth of our statement forn=k−1 implies the truth of our statement forn=krequired thatkbe at least 1, so that there would be an

SOME EXAMPLES OF MATHEMATICAL INDUCTION 35 element ak we could take away in order to describe our bijection. However,

condition (2) of the principle of mathematical induction only requires that we be able to prove the implication for k >0, so we were allowed to assume k >0.

Strong Mathematical Induction

One way of looking at the principle of mathematical induction is that it tells us that if we know the “first” case of a theorem and we can derive each other case of the theorem from a smaller case, then the theorem is true in all cases. However, the particular way in which we stated the theorem is rather restrictive in that it requires us to derive each case from the immediately preceding case. This restriction is not necessary, and removing it leads us to a more general statement of the principal of mathematical induction which people often call the strong principle of mathematical induction. It states:

In order to prove a statement about an integer nif we can 1. Prove our statement whenn=b, and

2. Prove that the statements we get with n = b, n = b+ 1, . . .n=k−1 imply the statement withn=k,

then our statement is true for all integers n≥b.

You will find some explicit examples of the use of the strong principle of mathematical induction in Appendix B and will find some uses for it in this chapter.