Geometry and Proof: Course Summary
... then m disjoint copies of EF can be placed in n disjoint copies of GH. This solution depends on assuming the Axiom of Archimedes: for any two segments AB and CD there is a natural number n such that CD is contained in n disjoint copies of AB (or vice versa). 3) Hilbert provides a third solution. As ...
... then m disjoint copies of EF can be placed in n disjoint copies of GH. This solution depends on assuming the Axiom of Archimedes: for any two segments AB and CD there is a natural number n such that CD is contained in n disjoint copies of AB (or vice versa). 3) Hilbert provides a third solution. As ...