We define the Dehn’s invariant, which is a number given to any polyhedron which does not change under scissor-equivalence.

Choose an additive function $f\colon\mathbb{R}\to\mathbb{R}$ such that $f(\pi)=f(0)=0$ and define for any polyhedron $P$ the number (Dehn’s invariant)

where $\theta_{e}$ is the angle between the two faces of $P$ joining in $e$, and $\ell(e)$ is the length of the edge $e$.

We want to prove that if we decompose $P$ into smaller polyhedra $P_{1},\ldots,P_{N}$ as in the definition of scissor-equivalence, we have

which means that if $P$ is scissor equivalent to $Q$ then $D(P)=D(Q)$.

Let $P_{1},\ldots,P_{N}$ be such a decomposition of $P$. Given any edge $e$ of a piece $P_{k}$ the following cases arise:

1. 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 $f(2\pi)\ell(e)$ (recall that $f$ is additive).

2. 2.

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

3. 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 $f(\theta_{{e^{{\prime}}}})\ell(e)$.

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 scissor-equivalent.