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  \section{Problem 5}
  
  Theorem: There exist irrational numbers a,b such that $a^b$ is rational.
  
  Proof:
  We know $\sqrt{2}$ is irrational if $a=\sqrt{2}$ and $b=\sqrt{2}$ then $a^b=(\sqrt{2})^\sqrt{2}$, then there are two cases, $a^b$ is either rational or irrational.
  
  If $(\sqrt{2})^\sqrt{2}$ is rational then there exist irrational numbers a,b such that $a^b$ is rational.
  
  If $(\sqrt{2})^\sqrt{2}$ is irrational then let $a=(\sqrt{2})^\sqrt{2}$ and $b=\sqrt{2}$ then $a^b=((\sqrt{2})^\sqrt{2})^\sqrt{2}$ $=(\sqrt{2})^2=2$ and $a^b$. Since we know that 2 is rational, this shows that there exist irrational numbers a,b such that $a^b$ is rational.
  
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  page revision: 1, last edited: 25 Jan 2018 23:08