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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)

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 re(3): Oops... Maybe another hint? | Comment 17 of 33 |
(In reply to re(2): Oops... Maybe another hint? by JLo)

I'm not getting any closer.  Here are my answers to your hint questions (and I'm adding the constraint that f_r be monotonic).

1. Given a rational number r, what is the simplest function f_r you can think of, that is discontinuous at r and continuous elsewhere?

f_r(x) = x for x < r
f_r(x) = x+1 for x = r
f_r(x) = x+2 for x > r

2. Now imagine that you have a whole bunch of rational numbers, how could you use the corresponding f_r's to construct a function that is discontinuous at all your rational numbers?

Assume that the "bunch" of rationals are arranged in ascending order and numbered r0, r1... rN.  The function would be:

f_r(x) = x for x<r0
f_r(x) = x+1 for x=r0
f_r(x) = x+2 for r0<x<r1
f_r(x) = x+3 for x=r1
etc.

3. Can that be extended if the "whole bunch" consists of ALL rational numbers?

Um...no.  First of all, you can't put ALL the rational numbers in ascending order--you can't even start.  Second, it's impossible to declare which two unique rational numbers an irrational number is between for purposes of deciding how much to add to x to obtain the function's value.

I must not be thinking of the same "simplest" function that you have in mind, JLo.  Maybe Steve (or someone else) can make more of this hint.  I do suspect that the function will have an infinite definition (finite for any particular x in the domain), but I don't see how it will be defined for the irrationals.

(hmm.. I just thought of something...I'll be back..)

--- (later that day...) ---

Dead end.  I thought by considering the m/n representation of every rational, I could devise a function of m and n that would show promise.  No luck.

Edited on August 20, 2006, 11:50 pm
 Posted by Ken Haley on 2006-08-20 11:53:26

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