All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math > Calculus
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: Nice solution!!! Now, you Want to try this | Comment 27 of 33 |
(In reply to Nice solution!!! Now, you Want to try this by JLo)

JLo,

Is this another hint to the solution you had in mind?  If not, what was your solution?  Like Steve, I'm curious to know...

Plus, I'm not sure I understand what you're asking.  If the function has the required properties, it has jump disontinuities at every rational, and p^-2 (where p is prime) is rational..so that's a redundant requirement, isn't it?  I'm obviously missing something here.

Edited on August 25, 2006, 1:07 am
  Posted by Ken Haley on 2006-08-25 01:02:30

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (1)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information