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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.

 Submitted by JLo Rating: 4.3000 (10 votes) Solution: (Hide) ************************************************ First solution (my original idea): As is generally known, the set of all rational numbers Q is countable, so say Q={r1, r2,...}. Now define f as follows: f(x) = sum 2-i, where we sum over all i such that ri0 an e>0 such that from |x-a| < e follows |f(x)-f(a)|k) is smaller than delta. Now choose epsilon small enough, such that no rational number rj with j<=k is in the epsilon interval around a. Now |f(x)-f(a)| = sum 2-i (sum going over i such that a<=rik) < d Obviously the same approach works not only for the rational numbers, but for any countable infinite set of numbers. One can even prescribe arbitray jumps - we have chosen jumps 2-i - just as long as their sum converges in order to end up with something well-defined. Another choice of jumps that would work is e.g. the set of numbers p-2 with primes p. Finally note that f(x) can be written as the infinite weighted sum of functions fr, where f(x)=0 for x

 Subject Author Date re: Function Steve Herman 2015-12-21 20:21:01 re: Function Steve Herman 2015-12-21 20:21:01 Function Math Man 2015-12-21 19:52:59 Nice work all, especially Ken Steve Herman 2006-08-27 17:57:19 Yet another solution vswitchs 2006-08-25 18:06:37 re(2): Nice solution!!! Now, you Want to try this JLo 2006-08-25 11:31:07 re: Nice solution!!! Now, you Want to try this Ken Haley 2006-08-25 01:02:30 re: Nice solution!!! Now, you Want to try this Steve Herman 2006-08-24 23:00:08 Nice solution!!! Now, you Want to try this JLo 2006-08-24 18:00:36 re(2): Possible solution? Ken Haley 2006-08-23 23:17:55 re: Possible solution? Steve Herman 2006-08-22 10:22:27 Possible solution? Ken Haley 2006-08-22 00:44:13 re(4): Oops... Maybe another hint? Ken Haley 2006-08-22 00:22:12 re(3): Oops... Maybe another hint? Steve Herman 2006-08-21 12:50:07 re(5): Oops... Oops myself... Ken Haley 2006-08-20 23:54:30 re(4): Oops... Oops myself... JLo 2006-08-20 14:18:06 re(3): Oops... Maybe another hint? Ken Haley 2006-08-20 11:53:26 re(2): Oops... Maybe another hint? JLo 2006-08-20 10:42:45 re(2): Oops... Ken Haley 2006-08-20 09:15:10 re: Oops... Steve Herman 2006-08-20 08:44:19 Oops... Ken Haley 2006-08-20 00:39:50 re: Needless to say this is very confusing Steve Herman 2006-08-19 18:34:11 Needless to say this is very confusing Charlie 2006-08-19 16:44:09 re(3): Bigger hint (and proposed problem redefinition) Steve Herman 2006-08-19 12:35:47 re(2): Bigger hint (and proposed problem redefinition) Ken Haley 2006-08-19 10:04:18 re: Bigger hint (and proposed problem redefinition) Steve Herman 2006-08-19 08:24:30 Bigger hint JLo 2006-08-18 18:36:17 re(3): Uncle! (1st condition satisfied) Bractals 2006-08-18 11:25:20 re(2): Uncle! (1st condition satisfied) Steve Herman 2006-08-18 10:32:52 re: Uncle! Don't give up yet! JLo 2006-08-16 17:34:05 re: Uncle! Bractals 2006-08-16 11:42:33 Uncle! Steve Herman 2006-08-16 09:10:37 Awe and quibbles Steve Herman 2006-08-15 10:02:07

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