If you raise an\u00a0irrational number\u00a0to a rational power, it is possible to get something rational. For instance, raise Sqrt[2] to the power 2 and you’ll get 2.<\/p>\n\n\n\n
But what happens if you raise an irrational number to an irrational power? Can this ever be rational?<\/p>\n\n\n\n
The answer is yes, and we’ll prove it\u00a0without having to find specific numbers that do the trick<\/em>!<\/p>\n\n\n\n Theorem. There exist irrational numbers A and B so that AB<\/sup> is rational.<\/p>\n\n\n\n Proof. We know that Sqrt[2] is irrational. So, if A=Sqrt[2] and B=Sqrt[2] satisfy the conclusion of the theorem, then we are done. If they do not, then Sqrt[2]Sqrt[2]<\/sup> is irrational, so let A be this number. Then, letting B=Sqrt[2], it is easy to verify that AB<\/sup>=2 which is rational and hence would satisfy the conclusion of the theorem. QED.<\/p>\n\n\n\n This proof is\u00a0non-constructive<\/em>\u00a0because it (amazingly) doesn’t actually tell us whether Sqrt[2]Sqrt[2]<\/sup>\u00a0is rational or irrational!<\/p>\n\n\n\n The Math Behind the Fact:<\/strong> But you don’t need Gelfond-Schneider to construct an explicit example, assuming you know\u00a0transcendental\u00a0numbers exist (numbers that are not roots of non-zero polynomials with integer coefficients). Let x be any transcendental and q be any positive rational. Then xlog_x(q)<\/sup>=q so all we have to show is that log_x(q) is irrational. If log_x(q)=a\/b then q=xa\/b<\/sup>, implying that xa<\/sup>-qb<\/sup>=0, contradicting the transcendentality of x.<\/p>\n\n\n\n
Actually, Sqrt[2]Sqrt[2]<\/sup> can be shown to be irrational, using something called the Gelfond-Schneider Theorem (1934), which says that if A and B are roots of polynomials, and A is not 0 or 1 and B is irrational, then AB<\/sup> must be irrational (in fact, transcendental).<\/p>\n\n\n\n