perplexus.info :: Calculus : Weird function challenge

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: Function

| Comment 32 of 33 |

(In reply to Function by Math Man)

Sorry, Math Man.

That very interesting function does not meet either condition.

It fails the 1st condition because it is continuous at 0.

It fails the 2nd condition in too many ways to count.

Posted by Steve Herman on 2015-12-21 20:21:01

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