Evaluating magnitudes of non-algebraic numbers

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SUMMARY

This discussion evaluates the magnitudes of non-algebraic numbers, specifically comparing the expressions \(\sqrt{2}^{\sqrt{2}}\), \(\frac{1+\sqrt{5}}{2}\), and \(\sqrt{3}\). It establishes that \(\sqrt{2}^{\sqrt{2}} > \frac{1+\sqrt{5}}{2}\) and explores the possibility of \(\sqrt{2}^{\sqrt{2}} < \sqrt{3}\). The discussion emphasizes that while these comparisons are well-defined, they may require rigorous mathematical proof.

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  • Understanding of non-algebraic numbers
  • Familiarity with exponential functions
  • Basic knowledge of inequalities in mathematics
  • Ability to perform mathematical proofs
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  • Research methods for proving inequalities involving non-algebraic numbers
  • Study the properties of the golden ratio, \(\frac{1+\sqrt{5}}{2}\)
  • Explore the implications of transcendental numbers in mathematical comparisons
  • Learn about the significance of \(\sqrt{2}^{\sqrt{2}}\) in number theory
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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.
 

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