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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.)
re(2): Bigger hint (and proposed problem redefinition) | Comment 9 of 33 |
(In reply to re: Bigger hint (and proposed problem redefinition) by Steve Herman)

Wow. Nice job, Steve...and great problem, JLo.

Can't we just fix the zero problem as follows?  Let's call JLo's example function w (for "weird").  Define the solution function, W (upper-case) as

W(x) = w(x) for x>0
        = -1 for x=0
        = w(x) - 2 for x<0

This gives us the needed discontinuity at 0 and maintains the characteristics of w(x) everywhere else, and the whole function is monotonic if w(x) is.

In any case...I rate this puzzle a 5!

  Posted by Ken Haley on 2006-08-19 10:04:18
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