Proofs of Existence and Optimality · Non-constructive Proofs of Existence

Lesson 2

Nikolai Chukhin · Alexander S. Kulikov

To give another example, we prove that there exist irrational numbers \(x,y\) such that \(x^{y}\) is rational. To do this, consider two cases.

  • The number \({\sqrt 2}^{\sqrt 2}\) is rational. Then we can take \(x=\sqrt 2\) and \(y=\sqrt 2\).

  • The number \({\sqrt 2}^{\sqrt 2}\) is irrational. Then we can take \(x={\sqrt 2}^{\sqrt 2}\) and \(y =\sqrt 2\): \[x^{y}=\left({\sqrt 2}^{\sqrt 2}\right)^{\sqrt 2}={\sqrt 2}^{2}=2 \ .\]

Thus, we have proven that such \(x,y\) exist, but we have not presented a specific pair of such numbers (although we have significantly narrowed the search space).

For this problem, a constructive proof can also be given: \[\left(\sqrt{2}\right)^{\log_2 9}=3 \ .\] It is easy to show the irrationality of the number \(\log_{2}9\): suppose \(\log_{2}9=a/b\) for positive integers \(a,b\), then \(9^{b}=2^{a}\), but this is impossible, since these numbers have different parities.

Later we will see non-constructive existence proofs for which no constructive analogs are still known.

For the curious 🤓
But what about \(\sqrt{2}^{\sqrt{2}}\)? Is it irrational? Yes, it is even transcendental (it is not a root of any non-zero polynomial with integer coefficients). This follows from the Gelfond–Schneider theorem, which provides a positive answer to Hilbert's seventh problem.