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F(x) is a rational number

  1. Mar 13, 2014 #1
    question.png

    Please see attached.
    The question is to determine f(0).

    In my opinion the question wants me to find f(x)..

    If f(x)=1, then

    1. f:[0,1] -> R is continuous (do I need to prove?)

    2. Obviously, f(1) = 1

    3. f(x) is a rational number for all xε[0,1]

    so f(0) = 1

    Am I correct?
     
  2. jcsd
  3. Mar 13, 2014 #2

    jbunniii

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    Editing because I think I misunderstood your question in my first reply.

    Yes, that's reasonable. Technically you only need to find ##f(0)##, but enough information is given that you can find ##f(x)## for all ##x \in [0,1]##.

    I take this to mean you want to set ##f(x) = 1## for all ##x \in [0,1]## and verify that it meets the conditions. And it does meet the conditions:

    Assuming you know that a constant function is continuous, then you don't need to prove it again.

    Yes.

    However, all you have done is to show that the constant function ##f(x) = 1## satisfies the conditions. Can you prove that it is the ONLY function that meets the conditions? How do I know there isn't some other function with ##f(0) = 0## which works?

    Indeed, you can answer this problem without even stating the values of ##f(x)## for ##0 < x < 1##. The key facts you need are that ##f## is continuous and that ##f## takes on only rational values. To do this, suppose that ##f(0) \neq 1## and look for a contradiction. Hint: there's a fundamental theorem about continuous functions that makes this a one-liner.
     
    Last edited: Mar 13, 2014
  4. Mar 13, 2014 #3
    Thanks a lot.
    I think I am done in this question but I really want to know more...

    If I prove f(0)=1 by contradiction
    which theorem can we use?
    I think extreme values theorem , IVT are unrelated..
     
  5. Mar 13, 2014 #4

    HallsofIvy

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    What you need to prove is that if f is continuous and f takes only rational number values for all x in [0, 1], then f is a constant function. Use IVT. If f(a) is NOT equal to f(b) for some a and b in [0,1]) then f takes on all values between f(a) and f(b). That is an interval and any interval contains irrational numbers.
     
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