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In case the (ir)rationality of sqrt(2)^sqrt(2) is bugging you... + Constructive vs. Existential math.

..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!).


(*) 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


for a nice discussion on the Constructive vs. Classical mathematics--and how during Hilbert's time there was a pretty big controversy in mathematics--with mathematicians such as Brouer insisted that all math that depended on existential proofs be thrown out.Papa Hilbert had to come to rescue--pretty heady stuff.

You might also look at


which also talks about the "slick" irrational power irrational can be rational proof...