Well-ordering principle

In mathematics, the well-ordering principle states that every non-empty subset of positive integers contains a least element. In other words, the set of positive integers is well-ordered by its "natural" or "magnitude" order in which $$x$$ precedes $$y$$ if and only if $$y$$ is either $$x$$ or the sum of $$x$$ and some positive integer (other orderings include the ordering $$2, 4, 6, ...$$; and $$1, 3, 5, ...$$).

The phrase "well-ordering principle" is sometimes taken to be synonymous with the "well-ordering theorem". On other occasions it is understood to be the proposition that the set of integers $$\{\ldots, -2, -1, 0, 1, 2, 3, \ldots \}$$ contains a well-ordered subset, called the natural numbers, in which every nonempty subset contains a least element.

Properties
Depending on the framework in which the natural numbers are introduced, this (second-order) property of the set of natural numbers is either an axiom or a provable theorem. For example:
 * In Peano arithmetic, second-order arithmetic and related systems, and indeed in most (not necessarily formal) mathematical treatments of the well-ordering principle, the principle is derived from the principle of mathematical induction, which is itself taken as basic.
 * Considering the natural numbers as a subset of the real numbers, and assuming that we know already that the real numbers are complete (again, either as an axiom or a theorem about the real number system), i.e., every bounded (from below) set has an infimum, then also every set $$A$$ of natural numbers has an infimum, say $$a^*$$. We can now find an integer $$n^*$$ such that $$a^*$$ lies in the half-open interval $$(n^*-1,n^*]$$, and can then show that we must have $$a^* = n^*$$, and $$n^*$$ in $$A$$.
 * In axiomatic set theory, the natural numbers are defined as the smallest inductive set (i.e., set containing 0 and closed under the successor operation). One can (even without invoking the regularity axiom) show that the set of all natural numbers $$n$$ such that "$$\{0,\ldots,n\}$$ is well-ordered" is inductive, and must therefore contain all natural numbers; from this property one can conclude that the set of all natural numbers is also well-ordered.

In the second sense, this phrase is used when that proposition is relied on for the purpose of justifying proofs that take the following form: to prove that every natural number belongs to a specified set $$S$$, assume the contrary, which implies that the set of counterexamples is non-empty and thus contains a smallest counterexample. Then show that for any counterexample there is a still smaller counterexample, producing a contradiction. This mode of argument is the contrapositive of proof by complete induction. It is known light-heartedly as the "minimal criminal" method and is similar in its nature to Fermat's method of "infinite descent".

Garrett Birkhoff and Saunders Mac Lane wrote in A Survey of Modern Algebra that this property, like the least upper bound axiom for real numbers, is non-algebraic; i.e., it cannot be deduced from the algebraic properties of the integers (which form an ordered integral domain).

Example applications
The well-ordering principle can be used in the following proofs.

Prime factorization
Theorem: Every integer greater than one can be factored as a product of primes. This theorem constitutes part of the Prime Factorization Theorem.

Proof (by well-ordering principle). Let $$C$$ be the set of all integers greater than one that cannot be factored as a product of primes. We show that $$C$$ is empty.

Assume for the sake of contradiction that $$C$$ is not empty. Then, by the well-ordering principle, there is a least element $$n \in C$$; $$n$$ cannot be prime since a prime number itself is considered a length-one product of primes. By the definition of non-prime numbers, $$n$$ has factors $$a, b$$, where $$a, b$$ are integers greater than one and less than $$n$$. Since $$a, b < n$$, they are not in $$C$$ as $$n$$ is the smallest element of $$C$$. So, $$a, b$$ can be factored as products of primes, where $$a = p_1p_2...p_k$$ and $$b = q_1q_2...q_l$$, meaning that $$n = p_1p_2...p_k \cdot q_1q_2...q_l$$, a product of primes. This contradicts the assumption that $$n \in C$$, so the assumption that $$C$$ is nonempty must be false.

Integer summation
Theorem: $$1 + 2 + 3 + ... + n = \frac{n(n + 1)}{2}$$ for all positive integers $$n$$.

Proof. Suppose for the sake of contradiction that the above theorem is false. Then, there exists a non-empty set of positive integers $$C = \{n \in \mathbb N \mid 1 + 2 + 3 + ... + n \neq \frac{n(n + 1)}{2}\}$$. By the well-ordering principle, $$C$$ has a minimum element $$c$$ such that when $$n = c$$, the equation is false, but true for all positive integers less than $$c$$. The equation is true for $$n = 1$$, so $$c > 1$$; $$c - 1$$ is a positive integer less than $$c$$, so the equation holds for $$c - 1$$ as it is not in $$C$$. Therefore, $$\begin{align} 1 + 2 + 3 + ... + (c - 1) &= \frac{(c - 1)c}{2} \\ 1 + 2 + 3 + ... + (c - 1) + c &= \frac{(c - 1)c}{2} + c\\ &= \frac{c^2 - c}{2} + \frac{2c}{2}\\ &= \frac{c^2 + c}{2}\\ &= \frac{c(c + 1)}{2} \end{align}$$ which shows that the equation holds for $$c$$, a contradiction. So, the equation must hold for all positive integers.