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Math Help - Two questions on functions

  1. #1
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    Two questions on functions

    Okay, here is one:

    Suppose that f:R-->R is continuous for all x and let alpha be in R. Show that if there is a b in R such that f(b)>alpha, then there is a delta>0 so that for all x such that |x-b|<delta, we have f(x)>alpha

    So far, I know that you're going to have to make use of the definition of continuity. THat's about as far as I got though.

    Another:

    Suppose that f:R-->R is continuous for all x and let f(q)=0 for all rationals q. Prove that f(x)=0 for all x
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  2. #2
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    Quote Originally Posted by jmoney90 View Post
    Suppose that f:R-->R is continuous for all x and let alpha be in R. Show that if there is a b in R such that f(b)>alpha, then there is a delta>0 so that for all x such that |x-b|<delta, we have f(x)>alpha
    In the definition of continunity let \varepsilon  = \frac{{f(b) - \alpha }}{2} > 0.

    Now the way this forum works, you should show some effort on your part.
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  3. #3
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    Think a bit about the last one, what happens if f(x)=k>0 where x is irrational?
    What if you pick \varepsilon such that 0<\varepsilon<k?

    Know that every open interval contains irrational and rational numbers.

    Try to visualize it.
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  4. #4
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    Quote Originally Posted by Plato View Post
    In the definition of continunity let \varepsilon  = \frac{{f(b) - \alpha }}{2} > 0.

    Now the way this forum works, you should show some effort on your part.
    I'm still not seeing what I should be doing. I substituted it into the definition and tried tinkering around, but I'm not seeing anything.
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  5. #5
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    Try this:
    Define a function g(x)=f(x)-\alpha. g is continuous, because f(x) is and \alpha is a constant.

    Now g(b)>0, and if g is continuous then there must be an interval I about b such that for every x\in I , g(x)>0.

    Is it easier to see now ?
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  6. #6
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    Quote Originally Posted by hjortur View Post
    Think a bit about the last one, what happens if f(x)=k>0 where x is irrational?
    What if you pick \varepsilon such that 0<\varepsilon<k?

    Know that every open interval contains irrational and rational numbers.

    Try to visualize it.
    I don't see how being rational or irrational has anything to do with this problem!
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  7. #7
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    Suppose that f:R-->R is continuous for all x and let f(q)=0 for all rationals q. Prove that f(x)=0 for all x
    Then please let me know how you solve that without referring to rational or irrational numbers.
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