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 Weird function challenge (Posted on 2006-08-15)
Find a function f:R->R (R the set of real numbers), such that

1. f has a discontinuity in every rational number, but is continous everywhere else, and
2. f is monotonic: x<y → f(x)<f(y)

Note: Textbooks frequently present examples of functions that meet only the first condition; requiring monotonicity makes for a slightly more challenging problem.

 See The Solution Submitted by JLo Rating: 4.3000 (10 votes)

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 re: Nice solution!!! Now, you Want to try this | Comment 27 of 33 |
(In reply to Nice solution!!! Now, you Want to try this by JLo)

JLo,

Is this another hint to the solution you had in mind?  If not, what was your solution?  Like Steve, I'm curious to know...

Plus, I'm not sure I understand what you're asking.  If the function has the required properties, it has jump disontinuities at every rational, and p^-2 (where p is prime) is rational..so that's a redundant requirement, isn't it?  I'm obviously missing something here.

Edited on August 25, 2006, 1:07 am
 Posted by Ken Haley on 2006-08-25 01:02:30

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