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 The Fix Is In (Posted on 2006-11-17)
Prove: Let f be a nondecreasing, but not necessarily continuous, mapping of the closed interval [0,1] into itself. Then f has a fixed point, i.e., there is some x in [0,1] such that f(x)=x.

(Just to avoid any misunderstanding, [0,1] is the set of all real numbers between 0 and 1, with both 0 and 1 also included.)

 See The Solution Submitted by Richard Rating: 4.0000 (2 votes)

 Subject Author Date re: Silly corollary Robby Goetschalckx 2006-11-19 20:47:50 Silly corollary JLo 2006-11-19 05:54:54 Brief proof JLo 2006-11-19 05:51:03 re(5): No Subject Richard 2006-11-18 01:55:42 re(4): No Subject Joel 2006-11-18 00:14:27 re(3): No Subject Richard 2006-11-17 21:35:44 re(2): No Subject Joel 2006-11-17 20:45:20 re: No Subject Richard 2006-11-17 19:58:09 re(2): Solution Bractals 2006-11-17 18:01:24 re: Solution Richard 2006-11-17 16:45:58 Solution Bractals 2006-11-17 14:24:02 No Subject Joel 2006-11-17 13:37:39

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