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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.)
Awe and quibbles | Comment 1 of 33
JLo:

AWE: Yes, this sounds like an odd function.  I am aware of a function that is continuous at every rational number, discontinuous elsewhere, but it is not monotonic.  This requires more thought.

QUIBBLE: x<y → f(x)<f(y)  defines strictly increasing monotonic, which is a more restrictive concept than just monotonic.
f monotonic increasing only requires that x<=y → f(x)<=f(y).
For instance, f(x) = 0 is monotonic, but not strictly monotonic.

Edited on August 20, 2006, 6:59 pm
  Posted by Steve Herman on 2006-08-15 10:02:07

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