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*To*: cse471-f06@parichaalak.eas.asu.edu*Subject*: In case the (ir)rationality of sqrt(2)^sqrt(2) is bugging you... + Constructive vs. Existential math.*From*: "Subbarao Kambhampati" <rao@asu.edu>*Date*: Tue, 7 Nov 2006 21:30:22 -0700*Domainkey-signature*: a=rsa-sha1; q=dns; c=nofws; s=beta; d=gmail.com; h=received:message-id:date:from:sender:to:subject:mime-version:content-type:x-google-sender-auth; b=t5LAvDX9RscsHlGbaajWqKxX7aEPyDFu2BmlVe6YBDyTZwteniPVmCW0rdbNdGjBaSSDl97+RxGmruq8lGhGj/yQCzPP1yoqSRgL/OeSkeXCB38LECQ3AG7cs2wYhlUoukZDfdsjFKzO8YvtSbSHii0kOjqBKy9NKJYKzIkcjaw=*Sender*: subbarao2z2@gmail.com

..In case you are dying to know whether sqrt(2)^sqrt(2) is rational or irrational, you can be rest assured

that it is irrational (actually transcendental (*)). So a constructive proof for

our theorem is with p=sqrt(2)^sqrt(2) and q=sqrt(2)

see http://www.math.hmc.edu/funfacts/ffiles/10004.3-5.shtml

(which also points out a more general and easy to understand constructive proof. Consider

e^{log_e q} for any transcendental number e and rational number q--which will be q. All you need to show is log_e(q) is irrational and you can show this easily (If log_e(q) = m/n with integers m and n without common factors, then

q = e^{m/n}. This would mean that e is the root of an algebraic equation x^m - q^n = 0. But the definition of trancendental number is that it cannot be the root of any algebraic equation!).

Rao

(*) By the way, transcendental => irrational but not vice versa. In particular, transcendentals are those irrational numbers that cannot be roots of any algebraic equation. Two famous examples of course are e and pi. Notice that proving that a number e *is* transcendental involves showing that e^r for any rational number r cannot be rational (since if it is, then e will be the root of an algebraic equation). Thus, proving transcendentality is not all that easy.

(ps 2:

Check out

http://digitalphysics.org/Publications/Cal79/html/cmath.htm

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