Heres the proof from Undergraduate Analysis by Serge Lang.
we take the property of well-ordering as an axiom of the natural numbers. It says that "Every non-empty set of natural numbers has a least element."
Suppose that for each positive integer we are given an assertion A(n), and that we can prove the following two properties:
(1) The assertion A(1) is true.
(2) For each positive integer n, if A(n) is true, then A(n + 1) is true.
Then for all positive integers n, the assertion A(n) is true.
proof: Let S be the set of all positive integers n for which the assertion A(n) is false. We wish to prove that S is empty, i.e that there is no element in S. Suppose there is some element in S. By well-ordering, there exists a least element, n_0 in S. By assumption, n_0 != 1, and hence n_0 > 1. Since n_0 is least it follows that n_0 - 1 is not in S, in other words the assertion A(n_0 - 1) is true. But then by property (2), we conclude that A(n_0) is also true because n_0 = (n_0 - 1) + 1. This is a contradiction, which proves what we wanted.
Or we can do:
Suppose that for each positive integer we are given an assertion A(n), and that we can prove the following two properties:
(1) The assertion A(1) is true.
(2) For each positive integer n, if A(n - 1) is true, then A(n) is true.
Then for all positive integers n, the assertion A(n) is true.
proof: Let S be the set of all positive integers n for which the assertion A(n) is false. We wish to prove that S is empty, i.e that there is no element in S. Suppose there is some element in S. By well-ordering, there exists a least element, n_0 in S. By assumption, n_0 != 1, and hence n_0 > 1. Since n_0 is least it follows that n_0 - 1 is not in S, in other words the assertion A(n_0 - 1) is true. But then by property (2), we conclude that A(n_0) is also true. This is a contradiction, which proves what we wanted.
|
|
Bookmarks