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.)
Yet another solution | Comment 29 of 33 |
One might also define a function with an upwards jump of exp(-n) at each rational m/n (m, n relatively prime) and an additional jump at 0.  It has the beautiful property of the w function from previous posts, namely that the larger the jumps, the rarer they are.  It is continuous at the irrationals because the size of the jumps decreases in their vicinity (as n grows), and is finite because the sum of n * exp(n) converges.
  Posted by vswitchs on 2006-08-25 18:06:37
Please log in:
Remember me:
Sign up! | Forgot password

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

Copyright © 2002 - 2019 by Animus Pactum Consulting. All rights reserved. Privacy Information