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Intermediate Value Theorem

  1. Jul 21, 2015 #1
    Hi, I'm confused about an explanation in Elementary Calculus, the infinitesimal approach regarding Intermediate Value Theorem.


    The explanation:

    Suppose g is a continuous function on an inverval I, and g(x) # 0 for all x in I

    (i) If g(c) > 0 for at least one c in I, then g(x) > 0 for all X in I
    (ii) if g(c) < 0 for at least one c in I, Then g(x) < 0 for all x in I

    Proof: Let g(c) >0 for some c in I. If g(x1) < 0 for some other point x1 in I, Then by Intermediate value theorem, there is a point x2 between c and x1 such that g(x2) = 0, contrary to hypothesis. Therefore we conclude that g(x) > 0 for all x in I

    The one thing I don't understand is this:
    If g(x) # 0 for all x in I, then how can there be a point x2 such that g(x2)=0 between c and x1 (assuming c and x1 and x2 are within the inverval I) if g(x) is # 0 for all x??
    "g(x) is # 0 for all x" tells me that the equation g(x) can never be zero for any x right?

    And how can we conclude that g(x) > 0 for all x in I, if there exists a point x2 such that g(x)=0?
     
  2. jcsd
  3. Jul 21, 2015 #2

    Orodruin

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    There cannot be such a point. Since assuming that g(x1) < 0 leads to the existence of such a point, the assumption cannot be correct. This is the entire point of a proof by contradiction: Assume A. If A leads to a contradiction, then the negation of A must be true. In your case, A is "there exists a x1 such that g(x1) < 0". Since it leads to a contradiction, there is no x1 for which g(x1) < 0, thus proving that g(x) > 0 for all x.
     
  4. Jul 22, 2015 #3

    HallsofIvy

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    There can't exist such a point. That's exactly what this proof is saying. Where did you get the idea that there could?

    We can't! That is the exact opposite of what this proof is saying.
     
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