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Weird function challenge (Posted on 2006-08-15) Difficulty: 4 of 5
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)

Comments: ( Back to comment list | You must be logged in to post comments.)
Nice work all, especially Ken | Comment 30 of 33 |
Nice problem, JLo. 

I accept JLo's original solution (and vswitch's variation), and agree that they work.  They are not my favorite sort of function, however. I have a preference for functions that can be calculated.  I like a lot that Ken's function has a known, exact, readily calculated value for rationals.  As for irrationals, I intuitively feel that Ken's function  can be calculated more readily and converges more quickly.

But hey, it's just a preference.  JLo and vswitch seem to have a preference for functions where the size of the discontinuities are known and interesting.

Pick your posion ...  
  Posted by Steve Herman on 2006-08-27 17:57:19
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