Hilbert’s 3rd problem and the Axiom of Choice
I’ve just learned something exciting: contrary to what I thought, the solution of Hilbert’s 3rd Problem ( http://en.wikipedia.org/wiki/Hilbert’s_third_problem ) does not in fact depend on the Axiom of Choice.
I was under the wrong impression that Dehn’s counterexample was something weird and non-constructive, similar to the Banach-Tarski paradox, because it required the AC. Wikipedia too, only mentions Dehn’s proof that relies on the existence of Hamel’s basis in R over Q.
Well, Dehn’s proof is really cool, and I love Hamel’s basis (I wrote about it here https://bbzippo.wordpress.com/2009/11/05/i-think-therefore-vector-basis-exists/ ), but I was always kind of sad when I was thinking that this problem that is so “finitistic” and “linear” could not be solved without the AC.
Apparently, it could 🙂
The other day I got bored and googled “Hilbert’s Third Problem Axiom of Choice” and found this
Now I’m happy, except I don’t like the terms “scissors-congruent” and “equidecomposable”. What’s wrong with “equicomposable”? 🙂