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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(3): Bigger hint (and proposed problem redefinition) | Comment 10 of 33 |
(In reply to re(2): Bigger hint (and proposed problem redefinition) by Ken Haley)

Ken:

You've done it!  Congratulations all around.

A formulation I like a little better (using Ken's) notation is:

W(x) = w(x) + sign(x)

where sign(x) = 1 if x > 0
-1 if x < 0
0 if x = 0

This has the pleasing quality that W(x) = -W(-x).

But I'm just gilding the lily, now.

Great problem and great hints, JLO!
Great contribution, Ken!

I'm rating this a 5 also!

And I withdraw my earlier suggestion about redefining the problem to only consider positive real numbers.

 Posted by Steve Herman on 2006-08-19 12:35:47

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