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Evaluating magnitudes of non-algebraic numbers

  1. Jun 26, 2009 #1
    Just like it is possible to show that [tex]e^\pi > \pi^e[/tex], is it possible to show that

    [tex]\sqrt{2}^{\sqrt{2}} > \frac{1+\sqrt{5}}{2}[/tex]


    [tex]\sqrt{2}^{\sqrt{2}} < \sqrt{3}[/tex]
  2. jcsd
  3. Jun 26, 2009 #2


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    Science Advisor

    Since all these numbers are well defined, it is possible, although at times it might require a little work.
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