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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): Uncle! (1st condition satisfied) | Comment 5 of 33 |
(In reply to re: Uncle! by Bractals)


The function is continuous at irrationals if you can always pick an epsilon so that all values within the epsilon neighborhood are arbitrarily close to the value of the function at the irrational.

My text gives as an example:

F(x) = 0 if x irrational,
       = 1/n where x is a rational expressible as m/n,
           where m and n are  relatively prime.

For instance, this is continuous at Pi.
I can always find a small neighborhood of Pi such that all rationals in the neighborhood evaluate to something that is arbitrarily close to zero.

However, this is discontinuous at 2.  F(2) = 1/2, but every neighborhood of 2, no matter how small, contains irrationals (whose function has a value of 0).

Edited on August 18, 2006, 10:34 am
  Posted by Steve Herman on 2006-08-18 10:32:52

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