proof of Dehn’s theorem
We define the Dehn’s invariant^{}, which is a number given to any polyhedron which does not change under scissorequivalence.
Choose an additive function $f:\mathbb{R}\to \mathbb{R}$ such that $f(\pi )=f(0)=0$ and define for any polyhedron $P$ the number (Dehn’s invariant)
$$D(P)=\sum _{e\in \{\text{edges of}P\}}f({\theta}_{e})\mathrm{\ell}(e)$$ 
where ${\theta}_{e}$ is the angle between the two faces of $P$ joining in $e$, and $\mathrm{\ell}(e)$ is the length of the edge $e$.
We want to prove that if we decompose $P$ into smaller polyhedra ${P}_{1},\mathrm{\dots},{P}_{N}$ as in the definition of scissorequivalence, we have
$$D(P)=\sum _{k=1}^{N}D({P}_{k})$$  (1) 
which means that if $P$ is scissor equivalent^{} to $Q$ then $D(P)=D(Q)$.
Let ${P}_{1},\mathrm{\dots},{P}_{N}$ be such a decomposition of $P$. Given any edge $e$ of a piece ${P}_{k}$ the following cases arise:

1.
$e$ is contained in the interior of $P$. Since an entire neighbourhood of $e$ is contained in $P$ the angles of the pieces which have $e$ as an edge (or part of an edge) must have sum $2\pi $. So in the right hand side of (1) the edge $e$ gives a contribution of $f(2\pi )\mathrm{\ell}(e)$ (recall that $f$ is additive).

2.
$e$ is contained in a facet of $P$. The same argument as before is valid, only we find that the total contribution is $f(\pi )\mathrm{\ell}(e)$.

3.
$e$ is contained in an edge ${e}^{\prime}$ of $P$. In this case the total contribution given by $e$ to the right hand side of (1) is given by $f({\theta}_{{e}^{\prime}})\mathrm{\ell}(e)$.
Since we have choosen $f$ so that $f(\pi )=0$ and hence also $f(2\pi )=0$ (since $f$ is additive) we conclude that the equivalence (1) is valid.
Now we are able to prove Dehn’s Theorem. Choose $T$ to be a regular tetrahedron^{} with edges of length $1$. Then $D(T)=6f(\theta )$ where $\theta $ is the angle between two faces of $T$. We know that $\theta /\pi $ is irrational, hence there exists an additive function $f$ such that $f(\theta )=1$ while $f(\pi /2)=0$ (as there exist additive functions which are not linear).
So if $P$ is any parallelepiped^{} we find that $D(P)=0$ (since each angle between facets of $P$ is $\pi /2$ and $f(\pi /2)=0$) while $D(T)=6$. This means that $P$ and $T$ cannot be scissorequivalent.
Title  proof of Dehn’s theorem 

Canonical name  ProofOfDehnsTheorem 
Date of creation  20130322 16:18:07 
Last modified on  20130322 16:18:07 
Owner  paolini (1187) 
Last modified by  paolini (1187) 
Numerical id  7 
Author  paolini (1187) 
Entry type  Proof 
Classification  msc 51M04 
Defines  Dehn invariant^{} 
Defines  Dehn’s invariant 