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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)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(5): Oops... Oops myself... | Comment 19 of 33 |
(In reply to re(4): Oops... Oops myself... by JLo)

JLo, I didn't see where you mentioned "monotonic" at all in your last hint.

Anyway, I assume you mean your revised definition of monotonic is:

x<y implies that f(x) <= f(y)

...instead of f(x) < f(y).  Is that correct?

 Posted by Ken Haley on 2006-08-20 23:54:30

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