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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(2): Possible solution? | Comment 24 of 33 |
(In reply to re: Possible solution? by Steve Herman)

Thanks for the kind words, Steve.

I don't know if "deep thinker" is a fair assessment, especially looking at some of my blunders in previous posts on this site, but thank you.  I still think the solution looks good, though, and I'm glad you concur.

I did briefly address the numbers outside [0,1] when I said "For numbers outside the range 0 to 1, just prepend the integer portion before the decimal point."  So ww(3.5) = 3.1 and ww(3.75) = 3.101 and ww(-3.75) = -3.101.  (Now I'm using your idea of keeping it all in decimal--forget binary and ternary...I agree there's no need!).  This function is already discontinuous for all the integers except 0. So, all we need to do is add 1 for the positive numbers and subtract 1 for the negative numbers (just like you suggested in your comment dated 8/19 at 12:35:47).  I.e.,

`WW(x) = ww(x) + 1 for x>0WW(0) = 0WW(x) = ww(x) - 1 for x<0`

This one's sure, been a lot of fun, hasn't it?   JLo?  I think we're ready to hear your solution.

Ken

Edited on August 23, 2006, 11:32 pm
 Posted by Ken Haley on 2006-08-23 23:17:55

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