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Sep 8, 2014 at 14:41 comment added Włodzimierz Holsztyński @Keiwan--nice! (clean :-)
Aug 31, 2010 at 12:27 comment added Tom Goodwillie I guess I was in a hurry, too, when I wrote "question" for "answer".
Aug 31, 2010 at 12:17 history edited Keivan Karai CC BY-SA 2.5
added 2 characters in body
Aug 31, 2010 at 12:06 vote accept CommunityBot moved from User.Id=6976 by developer User.Id=69903
Aug 31, 2010 at 12:05 vote accept CommunityBot moved from User.Id=6976 by developer User.Id=69903
Aug 31, 2010 at 12:05
Aug 31, 2010 at 12:05 vote accept CommunityBot moved from User.Id=6976 by developer User.Id=69903
Aug 31, 2010 at 12:05
Aug 31, 2010 at 11:42 comment added Keivan Karai Sorry Tom, I think we started to edit almost at the same time!
Aug 31, 2010 at 11:41 comment added Keivan Karai Sorry, I was in a run, so the proof was not clear. I just rewrote it.
Aug 31, 2010 at 11:40 history edited Keivan Karai CC BY-SA 2.5
added 390 characters in body
Aug 31, 2010 at 11:36 comment added Tom Goodwillie I have just edited the question for clarity. I had seen this argument before, but I had forgotten it. Note that it actually implies that the dual of the product is the sum.
Aug 31, 2010 at 11:32 history edited Tom Goodwillie CC BY-SA 2.5
added 50 characters in body; added 3 characters in body
Aug 31, 2010 at 10:53 comment added Stefan Geschke Why can you assume that $x$ is of the form $(a_0,2a_1,4a_2,\dots)$?
Aug 31, 2010 at 10:47 history answered Keivan Karai CC BY-SA 2.5