The principle of mathematical induction can be proved if the following axioms are assumed:
- (A1) The set of all natural numbers is well-ordered (that is, every non-empty set of natural numbers has a least element).
- (A2) if m > 0, then there exists n such that m = n + 1
A simplified version is given here. This proof does not use the standard mathematical symbols for there exists and for all to make it more accessible to less mathematically motivated readers.
Suppose
- (1) P(0)
and
- (2) For all n ≥ 0, P(n) ⇒ P(n + 1)
We wish to prove:
- (3) P is true for all integer values of m.
Let S be the set of numbers for which P is false.
Using (1), we see that 0 is not an element of S and is therefore not the minimal element of S.
Using (A2), if m > 0, then m = n + 1. Now if n is in S, then m being bigger cannot be the minimal element of S. But if n is not in S, P(n) and using (2) P(n + 1) and so m is not in S and cannot be the minimal element of S.
Thus, S has no minimal element, and by (A1) must be empty. Thus, P is true for all integer values of m
Converse
Conversely, the axiom can be proved by the principle of mathematical induction. Indeed, the two are equivalent.
Let S be a set of natural numbers. We want to prove that either S has a smallest element or else that S is empty. Let P(n) be the statement that no element of S is smaller than n. P(0) is certainly true, since there is no natural number smaller than 0. Suppose that P(n) is true for some n. If P(n + 1) were false, then S would have an element smaller than n + 1, but it could not be smaller than n, because P(n) was true, and so S would have a minimal element, namely n, and we would be done. So P(n) implies P(n + 1) for all n, or else S has a minimal element. But if P(n) implies P(n + 1) for all n, then by induction we know that P(n) is true for all n, and therefore for all n, no element of S is smaller than n. But this can only be vacuously true, if S has no elements at all, since every natural number is smaller than some other natural number. Thus we are done.