Proof Involving Rational and Irrational Numbers
Date: 09/13/2004 at 10:23:43 From: John Subject: Proving existence of irrational a, b, where a^b is rational The question I'm stuck on is: Prove the existence of 2 irrational numbers, a and b, where a^b is rational. Thanks for any help!
Date: 09/13/2004 at 10:50:35 From: Doctor Luis Subject: Re: Proving existence of irrational a, b, where a^b is rational Hi John, Good question! This has been considered before, and there's a very interesting (if not famous) proof of this result. The proof is interesting in the sense that it is non-constructive, i.e. we never actually find specific irrational values a,b that are both decidedly irrational. Nonetheless, we can still prove the theorem. We start by noting that sqrt(2) is irrational and that 2 is rational. Next, we define the number x = sqrt(2)^sqrt(2). By logic (law of excluded middle), we can confidently say that our number x is either rational or it is irrational. So, we consider each of the two possibilities: 1) Say x is rational. Then we are done! We have found two values a = sqrt(2), b = sqrt(2) such that a^b is rational. The theorem is proven, at least for this possibility. 2) Say x is irrational. Then we are done, too! Take a = x = sqrt(2)^sqrt(2), and b = sqrt(2). Then a^b = 2 is rational. The theorem is proven for the second possibility. Therefore the theorem is true because we have exhausted the two possibilities. The interesting part is that it doesn't matter whether x is rational or irrational! Either way, the theorem is proven. Incidentally, you can actually PROVE that x = sqrt(2)^sqrt(2) is irrational by using a result known as Gelfond's theorem. You can read more about it if you follow these URLs: Gelfond's Theorem http://mathworld.wolfram.com/GelfondsTheorem.html Gelfond-Schneider Constant: 2^sqrt(2) http://mathworld.wolfram.com/Gelfond-SchneiderConstant.html However, our little proof is quite elegant, albeit indirect. Let us know if you have any other thought-provoking questions. - Doctor Luis, The Math Forum http://mathforum.org/dr.math/
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