Non-constructive proofs First consider the theorem that there are an infinitude of
prime numbers.
Euclid's
proof is constructive. But a common way of simplifying Euclid's proof postulates that, contrary to the assertion in the theorem, there are only a finite number of them, in which case there is a largest one, denoted
n. Then consider the number
n! + 1 (1 + the product of the first
n numbers). Either this number is prime, or all of its prime factors are greater than
n. Without establishing a specific prime number, this proves that one exists that is greater than
n, contrary to the original postulate. Now consider the theorem "there exist
irrational numbers a and b such that a^b is
rational." This theorem can be proven by using both a constructive proof, and a non-constructive proof. The following 1953 proof by Dov Jarden has been widely used as an example of a non-constructive proof since at least 1970:
CURIOSA 339. A Simple Proof That a Power of an Irrational Number to an Irrational Exponent May Be Rational. \sqrt{2}^{\sqrt{2}} is either rational or irrational. If it is rational, our statement is proved. If it is irrational, (\sqrt{2}^{\sqrt{2}})^{\sqrt{2}} = 2 proves our statement. Dov Jarden Jerusalem In a bit more detail: • Recall that \sqrt{2}
is irrational, and 2 is rational. Consider the number q = \sqrt{2}^{\sqrt2}. Either it is rational or it is irrational. • If q is rational, then the theorem is true, with a and b both being \sqrt{2}. • If q is irrational, then the theorem is true, with a being \sqrt{2}^{\sqrt2} and b being \sqrt{2}, since :\left (\sqrt{2}^{\sqrt2}\right )^{\sqrt2} = \sqrt{2}^{(\sqrt{2} \cdot \sqrt{2})} = \sqrt{2}^2 = 2. At its core, this proof is non-constructive because it relies on the statement "Either
q is rational or it is irrational"—an instance of the
law of excluded middle, which is not valid within a constructive proof. The non-constructive proof does not construct an example
a and
b; it merely gives a number of possibilities (in this case, two mutually exclusive possibilities) and shows that one of them—but does not show
which one—must yield the desired example. As it turns out, \sqrt{2}^{\sqrt2} is irrational because of the
Gelfond–Schneider theorem, but this fact is irrelevant to the correctness of the non-constructive proof.
Constructive proofs A
constructive proof of the theorem that a power of an irrational number to an irrational exponent may be rational gives an actual example, such as: :a = \sqrt{2}\, , \quad b = \log_2 9\, , \quad a^b = 3\, . The
square root of 2 is irrational, and 3 is rational. \log_2 9 is also irrational: if it were equal to m \over n, then, by the properties of
logarithms, 9
n would be equal to 2
m, but the former is odd, and the latter is even. A more substantial example is the
graph minor theorem. A consequence of this theorem is that a
graph can be drawn on the
torus if, and only if, none of its
minors belong to a certain finite set of "
forbidden minors". However, the proof of the existence of this finite set is not constructive, and the forbidden minors are not actually specified. They are still unknown. ==Brouwerian counterexamples==