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Construct a counterexample

  1. Jan 23, 2006 #1
    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. jcsd
  3. Jan 24, 2006 #2
    Construct a counterexample.
  4. Jan 24, 2006 #3


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    "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?
  5. Jan 24, 2006 #4
    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 [tex]\sqrt 2 < 2[/tex] as the irrational number. Put A = f(a) and B = f(b). You can find a natural number n which satisfies [tex] B - A < \frac 1 n[/tex] Then, at least one of[tex]\sqrt 2 \frac m {2n} [/tex] (where m is an integer) must be between A and B. (notice step of [tex]\sqrt 2 \frac m {2n}[/tex] is smaller than B-A.)

    Then you can prove f(x) isn't continuous : If f(x) is continuous, it must pass such a [tex]\sqrt 2 \frac m {2n}[/tex] 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
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