Proof for Hilbert's third problem (idea)

See Hilbert's third problem for what we want to prove, motivation, discussion and importance. Due to the moderate length of the proof, I give an outline of the ideas and principal constructs used. More details are in the linked nodes.

Max Dehn proved a counterexample to Hilbert's third problem; in fact, his method gives a means to construct very many counterexamples, as well as a more general framework for studying how polyhedra can fit together in three dimensions. The proof shows how to construct pairs of tetrahedra of equal volume which are neither equidecomposable nor equicomplementable. As such, the equality of their volumes cannot be justified without the use of calculus.

The proof uses some linear algebra in an ingenious way. It may help to "recall" that the real numbers R are a vector space over the field of rational numbers Q. A basis for this vector space is often called a Hamel basis, to emphasise that only finite linear combinations are considered here (unlike, e.g., the bases used in Hilbert spaces and other Banach spaces). The existence of such a Hamel basis requires the axiom of choice; luckily, Dehn doesn't really require such a basis. The proof can proceed without relying on the axiom of choice. It is completely constructive: the tetrahedra may be specified explicitly.

Outline of proof:

I recommend you first read just the outline, then follow the links to learn about what happens in each step.

1. Define Dehn invariants. A Dehn invariant D_f is a function mapping polyhedra in R³ to numbers. See next steps for the essential property of Dehn invariants, which provides motivation for the strange definition. Dehn invariants are preserved by all congruent transformations.
2. Show that Dehn invariants are "additive". If you cut up a polyhedron into smaller polyhedra, the Dehn invariant of their "sum" is the sum of the Dehn invariants of the smaller polyhedra. It immediately follows that for any Dehn invariant D_f, if A and B are either equidecomposable or equicomplementable then D_f(A)=D_f(B). This should show the relevance of Dehn invariants to Hilbert's third problem...
3. "Note" (or, if you're like me, get some hints) that there are zero and nonzero Dehn invariants of some polyhedra (the problem is actually showing that any polyhedra with nonzero invariants). In particular, find tetrahedra R,T of equal volume and some D_f with D_f(S)≠D_f(T).
4. It follows that R and T are neither equidecomposable nor equicomplementable!

   Since there are two tetrahedra R and T such that all Dehn invariants D(R) of R are zero, and one Dehn invariant D_f for which D_f(T)≠0. These R and T have equal bases (indeed, they have congruent bases -- the isoceles right triangle) and equal heights above these bases (1).