 About 5,960 results  books.google.com 
 books.google.com Remark 22.1.4 This example suggests the following question: If we assume
Archimedes' axiom (A) in addition, are the notions of equidecomposable and
equal content equivalent? We will see that the answer is yes in any Hilbert plane
with (P) ... 

 books.google.com We shall now begin to examine the problems of equidecomposability and
equicomplementabilityforsolid bodies (polyhedra). Two polyhedra are said to be
equidecomposable if one of them can be decomposed into a finite number of
parts in ... 

 books.google.com Miklós Laczkovich  2001  Preview We show that any two rectangles of the same area are equidecomposable using
translations alone. It is clear that a rectangle of size a x b is equidecomposable to
a rectangle of size (a/2) x (26) using two pieces and translations. Repeating this ... 

 books.google.com 9c = 4>c if c C (Joe u sup a2„ \ a2n+i , neti = ip~1cifcC supa2n+i \ a2n+2 neti By
394Be, 9 G G*, so a and b are Grequidecomposable. (c) This is easy to prove
directly from the results in 394B, but also follows at once from (bi); any transitive
... 

 books.google.com any two countably infinite sets of X are countably Gequidecomposable. In
particular, any countably infinite subset of X is countably Gparadoxical. Exercise
2.2.1. Show that finite Gequidecomposability and countable G
equidecomposability ... 

 books.google.com Gardner's Theorem 8.1. Indeed. suppose that D and Q are equidecomposable
using the translations t,t3, t3. Replacing D by n (D) we may assume that t] is the
identity. Then t1, t2, t3 generate a discrete group. which contradicts Theorem 8.1. 

 books.google.com 7s a regular tetrahedron in R3 equidecomposable to a cube using Lebesgue
measurable pieces ? We may add the following question posed by the author in
Section 10.2 of [23]. Problem 6.3. Is every polygon equidecomposable to a
square ... 

 books.google.com Then A − A 1is equidecomposable witha subset of A (it is asubset ofA ), and A is
equidecomposable with asubset of A − A 1 , namely with A 2. Thus PropositionG.
2 implies thatA and A − A 1 are equidecomposable, andsoA1 and A−A 1form the
... 

 books.google.com A bounded set is Jordan measurable if and only if its boundary has Jordan
measure zero if and only if its characteristic function is Riemann integrable (see
Appendix B). Define equidecomposability in 3t„ using the join operation (A v B e
3tv if A ... 

 