Evaluating magnitudes of non-algebraic numbers

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espen180
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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]

or

[tex]\sqrt{2}^{\sqrt{2}} < \sqrt{3}[/tex]
 
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espen180 said:
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]

or

[tex]\sqrt{2}^{\sqrt{2}} < \sqrt{3}[/tex]

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