From the pythagorean theorem we can construct the lengths $\sqrt2$ and $\sqrt3$
Is it possible to construct the lengths $\sqrt2^{\sqrt3 }$ and $\sqrt3^{\sqrt2 }$ in order to show ( geometrically ) that:
$\sqrt2^{\sqrt3 }$ $<$ $\sqrt3^{\sqrt2 }$
thank you in advance!
It is sufficient to demonstrate that the following are true:
$$1 < \sqrt 2 < e$$
$$1 < \sqrt 3 < e$$
$$ \sqrt 2 < \sqrt 3$$
These can be demonstrated geometrically as long as you are happy that $e>2$.
Having established the above geometrically, we can continue using calculus and algebra.
Consider $f(x)=\frac {\ln x}x$
$f'(x)=\frac 1{x^2}-\frac {\ln x}{x^2}$
$f'(x)=\frac {1-\ln x}{x^2}$
For $1<x<e$, we know that $0<\ln x <1 $
This means that $1-\ln x>0$ so $f'(x)>0$ - the function is increasing.
Since the function is increasing, $a<b \Rightarrow f(a)<f(b)$
$\sqrt 2<\sqrt 3 \Rightarrow f(\sqrt 2)<f(\sqrt 3)$
$\frac {\ln \left(\sqrt 2 \right)}{\sqrt 2}<\frac {\ln \left(\sqrt 3 \right)}{\sqrt 3}$
${\sqrt 3}{\ln \left(\sqrt 2 \right)}<{\sqrt 2}{\ln \left(\sqrt 3 \right)}$
${\ln \left({\sqrt 2}^{\sqrt 3} \right)}<{\ln \left({\sqrt 3}^{\sqrt 2} \right)}$
${{\sqrt 2}^{\sqrt 3}}<{{\sqrt 3}^{\sqrt 2}}$ as required.