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Help with proof

  1. Dec 15, 2005 #1
    Hi, im trying to understand the following proof: If f is continuous on [a,b] and f(a) < 0 < f(b), then there is some number x in [a,b] such that f(x) = 0.

    It starts with defining A to be {x: a <= x <= b and f is negative on [a,x]}. It uses the least upper bound property to determine that a least upper bound for A does exist and its between a and b. Call this least upper bound "u". Now suppose f(u) < 0. By a continuity theorem there is a d > 0 st f(x) < 0 for u-d < x < u+d. Now there is some x0 in A which satisfies u-d < x0 < u.
    This is where i dont understand. Intuitively it makes sense, but how do you know that theres any x between u-d and u+d or u-d and u. All i know is that any x0 in A must be smaller or equal to u. Im trying to make as little assumptions as possible and it seems as if this statement assumes the intermediate value theorem, and thats basically what its trying to prove.
  2. jcsd
  3. Dec 15, 2005 #2


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    So you know that, by construction, any x with a≤x<u has f(x)<0. You want to show that f(u)=0. This means eliminating the possibilities that is is either greater or less than 0. It looks like first you want to show it can't be less than 0 by assuming it is and then deriving a contradiction. The way to do this is to start with your (true) claim that, assuming f(u)<0, there exists a d>0 such that f(x)<0 for all u-d<x<u+d. This shows right away that u can't be the least upper bound of A, a contradiction. An analagous argument assuming f(u)>0 leads to the same conclusion.
  4. Dec 15, 2005 #3
    Ya i know the the structure of the proof. Its just this specific statement that i dont understand: " there is some x0 in A which satisfies u-d < x0 < u. " Can you explain to me exactly why thats true.
  5. Dec 15, 2005 #4


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    The proof doesn't require x to be in A, although it can be. All you need is to show that u isn't a least upper bound; that either it isn't an upper bound, or there is an upper bound which is less than it. All that is required is that there is some real x in (u-d,u), which is guaranteed since d is postive.
  6. Dec 15, 2005 #5
    ok as a side question. Here's a theorem that's used to prove this: Suppose f is continuous at a, and f(a) > 0. Then there is a number d > 0 such that f(x) > 0 for all x satisfying [x-a] < d.

    Now when reading this theorem, do i interpret "for all x" as for all x in the domain of f which can be arbitrary, or can i simply pick any x I can think of that satisfies the relation [x-a] < d. And in a sense, the nature of the domain is stated by the theorem.

    Because if its the second case, then ya, i see why the Intermediate Value proof makes sense.
  7. Dec 16, 2005 #6


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    The function is continuous on [a,b], which means [a,b] is part of its domain. So I don't see where the distinction between those two interpretations would come up in the proof. But, in general, d must be chosen so that f is defined at every point with |x-a|<d. If no such d can be chosen, the function isn't continuous at a.
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