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Function proofs

  1. Oct 2, 2008 #1
    the function f(x) = 1 if x is rational
    f(x) = 0 if x is irrational is not continuous for all real numbers, c



    the function f(x) = x if x is rational
    f(x) = 0 if x is irrationa is continuous at x=0 and not continuous for all other real numbers c

    the function f(x) = 1/q if x is rational and x = p/q in lowest terms
    f(x) = 0 if x is irrational
    is continuous at c if c is irrational and not continuous at c if c is rational

    I'm terrible at proofs, please help!
     
  2. jcsd
  3. Oct 2, 2008 #2

    tiny-tim

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    Hi phyguy321;! :smile:

    (have an epsilon: ε and a delta: δ :smile:)

    Let's start with:
    Can you use an ε and δ method to prove that it is continuous at x = 0?

    Have a go. :smile:
     
  4. Oct 3, 2008 #3
    I've got the proof for when f(x) is continuous at x=0 but I'm not sure how to prove that its discontinuous at every other value using delta epsilon. I understand it has to do with the density property of real numbers and that its full of holes be cause between every real number there exists rational and irrational numbers.
     

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  5. Oct 4, 2008 #4

    tiny-tim

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    Hi phyguy321! :smile:

    Same method … start "if x ≠ 0, then for any ε < x, … " :wink:
     
  6. Oct 4, 2008 #5
    but how do i say if |x-c|<delta then |f(x) - f(not zero)|< epsilon?
     
  7. Oct 5, 2008 #6

    tiny-tim

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    But it's not continuous, so there isn't a δ.

    You should be trying to prove that, no matter how small δ is, the neighbourhood will contain values that further away than ε. :smile:
     
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