espen180
- 831
- 2
Just like it is possible to show that e^\pi > \pi^e, is it possible to show that
\sqrt{2}^{\sqrt{2}} > \frac{1+\sqrt{5}}{2}
or
\sqrt{2}^{\sqrt{2}} < \sqrt{3}
\sqrt{2}^{\sqrt{2}} > \frac{1+\sqrt{5}}{2}
or
\sqrt{2}^{\sqrt{2}} < \sqrt{3}