 A slick classical proof goes as follows. Either \(\sqrt{2}^{\sqrt{2}}\) is rational, in which case we take \(a = b = \sqrt{2}\); or else \(\sqrt{2}^{\sqrt{2}}\) is irrational, in which case we take \(a = \sqrt{2}^{\sqrt{2}}\) and \(b = \sqrt{2}\) (see Dummett 1977 [2000], 6). But as it stands, this proof does not enable us to pinpoint which of the two choices of the pair \((a,b)\) has the required property. In order to determine the correct choice of \((a,b)\), we would need to decide whether \(\sqrt{2}^{\sqrt{2}}\) is rational or irrational, which is precisely to employ our initial interpretation of disjunction with \(P\) the statement “\(\sqrt{2}^{\sqrt{2}}\) is rational”.