# 4.1.6: Mathematical induction revisited

We end this section by returning to the topic of mathematical induction. First, I will give proofs of the two forms of the principle of mathematical induction. These proofs were omitted from the previous chapter, but only for the lack of a bit of set notation. In fact, the principle of mathematical induction is valid only because it follows from one of the basic axioms that define the natural numbers, namely the fact that any non-empty set of natural numbers has a smallest element. Given this axiom, we can use it to prove the following two theorems:

Theorem 4.3.

Let P be a one-place predicate whose domain of discourse includes the natural numbers. Suppose that $$P(0) \wedge(\forall k \in \mathbb{N}(P(k) \rightarrow P(k+1))) .$$ Then $$\forall n \in \mathbb{N}, P(n)$$

Proof. Suppose that both $$P(0)$$ and $$(\forall k \in \mathbb{N}(P(k) \rightarrow P(k+1)))$$ are true, but that $$(\forall n \in$$$$\mathrm{N}, P(n) )$$ is false. We show that this assumption leads to contradiction.

Since the statement $$\forall n \in \mathbb{N}, P(n)$$ is false, its negation, $$\neg \forall n \in \mathbb{N}, P(n) ),$$ is true. The negation is equivalent to $$\exists n \in \mathbb{N}, \neg P(n) .$$ Let $$X=\{n \in \mathbb{N} | \neg P(n)\} .$$ since $$\exists n \in$$ $$\mathbb{N}, \neg P(n)$$ is true, we know that $$X$$ is not empty. Since $$X$$ is a non-empty set of natural numbers, it has a smallest element. Let x be the smallest element of X. That is, x is the smallest natural number such that P(x) is false. Since we know that P(0) is true, x cannot be 0. Let y = x − 1. Since x is a natural number. Since y < x, we know, by the definition of x, that P(y) is true. We also know that ∀k ∈ N (P(k) → P(k + 1))is true. In particular, taking k = y, we know that P(y) → P(y + 1). Since P(y) and P(y) → P(y+1), we deduce by modus ponens that P(y+1) is true. But y+1 = x, so we have deduced that P(x) is true. This contradicts the fact that P(x) is false. This contradiction proves the theorem.

Theorem 4.4.

Let P be a one-place predicate whose domain of discourse includes the natural numbers. Suppose that P(0) is true and that

$$(P(0) \wedge P(1) \wedge \cdots \wedge P(k)) \rightarrow P(k+1)$$

is true for each natural number k ≥ 0. Then it is true that ∀n ∈ N, P(n).

Proof. Suppose that P is a predicate that satisfies the hypotheses of the theorem, and suppose that the statement ∀n ∈ N, P(n) is false. We show that this assumption leads to a contradiction.

Let X = {n ∈ N | ¬P(n)}. Because of the assumption that ∀n ∈ N, P(n) is false, X is non-empty. It follows that X has a smallest element. Let x be the smallest element of X. The assumption that P(0) is true means that 0 ̸∈ X, so we must have x > 0. Since x is the smallest natural number for which P(x) is false, we know that P(0), P(1), ..., and P(x − 1) are all true. From this and the fact that (P(0)∧P(1)∧···∧P(x−1)) → P(x),we deduce that P(x) is true. But this contradicts the fact that P(x) is false. This contradiction proves the theorem.