13 Nov
2011
13 Nov
'11
11 a.m.
I wrote to Gene (inadvertently not sending it to all math-fun) about his solution to the tiling problem: << Yes, that's essentially what I had in mind -- nice solving. and then I added: << Now for Part B: Is there a way to solve this puzzle (for 2 sets) so that one of the sets is a subgroup of R ?
That is: Can R be partitioned into two congruent subsets that are each dense, one of which is a subgroup of R ? --Dan Even though kleptomaniacs can't help themselves, they do.