# Evaluating magnitudes of non-algebraic numbers

1. Jun 26, 2009

### espen180

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

$$\sqrt{2}^{\sqrt{2}} < \sqrt{3}$$

2. Jun 26, 2009

### mathman

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