Homework Help: Construct a counterexample

1. Jan 23, 2006

loli12

I was asked to show whether this is true: f(x) is defined for all x in [a,b] with f(b) > f(a) [values given]. the values of f at any x in (a,b) is rational. So, is f(x) continous?

I think this is not continuous as this seems like the question is trying to use intermediate value property to imply continuity. But I can;t think of a more proper way to proof the answer. Or am I wrong on this? Please give me some idea!

2. Jan 24, 2006

Muzza

Construct a counterexample.

3. Jan 24, 2006

HallsofIvy

"this seems like the question is trying to use intermediate value property to imply continuity." I would have said it the other way around! If the function were continuous, then it would have the intermediate value property. Whatever f(a) and f(b) are, since f(b)> f(a) they are not the same. Does there exist an irrational number between them? What does that tell you?

4. Jan 24, 2006

maverick6664

Right. It's not continuous. You can prove there is (are) at least an irrational number between f(a) and f(b).

Let's use $$\sqrt 2 < 2$$ as the irrational number. Put A = f(a) and B = f(b). You can find a natural number n which satisfies $$B - A < \frac 1 n$$ Then, at least one of$$\sqrt 2 \frac m {2n}$$ (where m is an integer) must be between A and B. (notice step of $$\sqrt 2 \frac m {2n}$$ is smaller than B-A.)

Then you can prove f(x) isn't continuous : If f(x) is continuous, it must pass such a $$\sqrt 2 \frac m {2n}$$ which is evidently an irrational number.

BTW you can easily prove the opposite theory: If f(x) takes only irrational numbers between [a,b], then f(x) cannot be continuous.

Last edited: Jan 24, 2006