In the Hilbert problems, you will find the cryptic phrasing "the
equality of the volumes of two tetrahedra of equal
bases and equal altitudes". David Hilbert knew that this is
true; for that matter, Euclid knew that the volume of any pyramid
is 1/3*A*h, where A is the area of its base and h its altitude. Using
calculus, one can easily derive this formula.

So Hilbert was *not* asking for a proof that this is true.
Rather, he wanted a very elementary proof: one which would use only
a few "obvious" axiomatic properties of volume, and show the desired
equality. The existing proofs for the volume of pyramids all use
calculus (or some other limiting principle, such as the archimedean axiom). This requires much stronger motivation to accept. Hilbert wanted the equality of pyramids, *without taking limits*.

Wouldn't it be nice if we could just use the obvious properties of volume? It turns out that we *can* do just that in **R**^{2}, for all polygons.

So what's "obvious" about volumes? Well, (almost) anyone would agree
that (no matter how you define "volume"):

- Volume exists for all polyhedra.
- Congruent bodies have equal volumes.
- Volume is (finitely) additive: if A and B are disjoint
bodies with volumes v(A) and v(B) respectively, then A∪B has
volume, and v(A∪B)=v(A)+v(B).

If we know these 3 facts

(or hold them to be self evident), we can immediately deduce the volume of a

triangle:

C G.____C____________.H
/`-. : /`-. :
/ : `-. -------\ : / : `-. :
/ : `-. -------/ : / : `-. :
/ : `-. :/ : `-. :
/____J___________:. K____J___________:J
B F A B F A
C D C
_. ._________________.
_,-' : : _,-' :
_,-' : : _,-' :
_,-' : : _,-' :
,-' : : ,-' :
c=________________J K=________________J
B A B A

Given a right triangle (see 2nd diagram) ABC, we can take a congruent copy of the triangle DCB, and form a rectangle from the two. Thus, from the facts/axioms of area above, we see (as Euclid shows in his Elements...) that the area of a right triangle must be half the area of the rectangle with equal perpendiculars.

Given *any* triangle (see 1st diagram) ABC, we can drop the perpendicular from C to its foot F on AB; decomposing ABC into right triangles CBF and AFC and forming congruent copies of these, we can apply the previous trick to show that the area of ABC is half the area of the rectangle with width AB and height CF. Euclid also knew how to do this, of course.

Thus, we see that Hilbert's 3rd problem is easy in two dimensions: the areas of 2 triangles with equal bases and heights are equal.

In contrast, Euclid's proofs for the volume of pyramids, and all succeeding proofs, always appealed to some limiting argument. The popular tactic in high school is to divide the pyramid into thin slices. The volume of the pyramid within each slice may be bounded between the volume of the prism drawn from the pyramid's intersection with the top of the slice and the prism drawn from its intersection with the bottom of the slice. Taking *limits*, the formula for pyramidal volume (and the desired equality) follows.

What Hilbert wanted was a similar argument to Euclid's, but in three dimensions (unlikely), or a proof that no such proof was to be had.

Enter the concepts of equicomplementability and equidecomposability: Any 2 polygons of equal area are equicomplementable (congruent polygons can be added to each to get congruent polygons) and equidecomposable (can be divided into congruent polygons). This was known to Hilbert. And we've seen that these 2 concepts are key to formulating an elementary concept of area.
So, since we're only going to be concerned with polyhedra, with finite divisions of them, and with copies of these, the problem becomes to show that all pyramids with equal base and height are equidecomposable and/or equicomplementable (which see for exact formulation), or to find 2 which are not.

Tetrahedra are a good place to start, and it turns out that Hilbert's intuition was correct, at least here: there exist many pairs of pyramids with equal volumes which are neither equidecomposable nor equicomplementable. No amount of geometric cleverness will suffice to define the concept of volume: some calculus will always be required.

I have furnished an outline of Dehn's proof for Hilbert's third problem. The proof itself uses essentially the concepts of first year undergraduate studies in Mathematics, mostly from linear algebra. To follow it, you will probably need some familiarity with vector spaces, and in particular with the idea of the real numbers **R** and subsets of the real numbers as vector spaces over the field of rational numbers **Q**.

I've tried to minimize the amount of number theory required by appealing to the excellent trigonometric number writeup, instead of proving irrationality of some number myself.

However, the ideas are used by Dehn and others in profound and unexpected ways. The very *use* of algebra and number theory to disprove a conjecture in three dimensional geometry is profound.

Read on!