Existence of Rational Powers of Irrational Numbers/Proof 2

From ProofWiki
Jump to navigation Jump to search

Theorem

There exist irrational numbers $a$ and $b$ such that $a^b$ is rational.


Proof

Given that $2$ is rational and $\sqrt 2$ is irrational, consider the number $q = \sqrt 2^{\sqrt 2}$.

We consider the two cases.

$(1): \quad$ If $q$ is rational then $a = \sqrt 2$ and $b = \sqrt 2$ are the desired irrational numbers.
$(2): \quad$ If $q$ is irrational then $q^{\sqrt 2} = \paren {\sqrt 2^{\sqrt 2} }^{\sqrt 2} = \sqrt 2^{\paren {\sqrt 2} \paren {\sqrt 2} } = \sqrt 2^2 = 2$ is rational, so $a = q = \sqrt 2^{\sqrt 2}$ and $b = \sqrt 2$ are the desired irrational numbers.

$\blacksquare$


Law of the Excluded Middle

This theorem depends on the Law of the Excluded Middle.

This is one of the logical axioms that was determined by Aristotle, and forms part of the backbone of classical (Aristotelian) logic.

However, the intuitionist school rejects the Law of the Excluded Middle as a valid logical axiom.

This in turn invalidates this theorem from an intuitionistic perspective.


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Existence_of_Rational_Powers_of_Irrational_Numbers/Proof_2&oldid=732827"