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Math Help - Discontinous (proof required)

  1. #1
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    Discontinous (proof required)

    Given:

    f(x)= X^2 ---------for all rational x
    = -X --------for all irrational x

    Prove that it is continous at x=0 and discontinous at every point.


    I've proved for continous part (with epsilon-delata language). But couldn't do the discontinous part. Tried contradiction, but am getting nothing out of it. If anyone can do it rigorously, it would be highly appreciated.
    Thanks
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  2. #2
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    Quote Originally Posted by concern View Post
    Given:

    f(x)= X^2 ---------for all rational x
    = -X --------for all irrational x

    Prove that it is continous at x=0 and discontinous at every point.


    I've proved for continous part (with epsilon-delata language). But couldn't do the discontinous part. Tried contradiction, but am getting nothing out of it. If anyone can do it rigorously, it would be highly appreciated.
    Thanks
    Note that whatever \delta>0 you choose. The interval about some point
    ]c-\delta,c+\delta[, will always include both rational and irrational numbers.

    Now if you pick \varepsilon<c^2+c then there is no \delta>0 such that when
    0<\vert c - x \vert <\delta , \vert f(x)-f(c)\vert<\varepsilon

    Can you finish this ?
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  3. #3
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    still not that clear. I understand the 1st paragraph, but second paragraph is elusive. sorry, i'm not that good at it, so am needing help. you have said there is no delta>0. but that is the thing we need to show rigorously, isn't it which will solve the problem completely. so can you help in showing rigorously that such delta doesn't exist,can you clarify a bit more? that would be great. thanks for your help.

    Quote Originally Posted by hjortur View Post
    Note that whatever \delta>0 you choose. The interval about some point
    ]c-\delta,c+\delta[, will always include both rational and irrational numbers.

    Now if you pick \varepsilon<c^2+c then there is no \delta>0 such that when
    0<\vert c - x \vert <\delta , \vert f(x)-f(c)\vert<\varepsilon

    Can you finish this ?
    Last edited by concern; November 9th 2009 at 12:27 PM.
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  4. #4
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    The first step is "knowing" what you are going to proof. If you are not sure wether something is true or not
    you can't prove it.

    So if the function is going to be continuous then there must excist a delta for every epsilon.
    You know that no matter what delta you choose there will be rational and irrational numbers there.

    So if you have a rational number c and irrational one b. Then f(c)=c^2 and f(b)=-b.
    If delta is very small then c is very close to b so they are almost the same number,
    so what happens if you pick epsilon that is smaller then the difference between f(c) and f(b) ?

    That is the basic outline.

    I'll show you some steps.

    Assume that the function is continous at c\ne0, then for every \varepsilon>0
    there excists a \delta>0 such that \vert f(x)-f(c)\vert<\varepsilon if 0<\vert x - c \vert<\delta.

    Now let 0<\varepsilon_1<\lvert\frac{c^2+c}{2}\rvert then you know that there excists a \delta_1>0 corresponding to that epsilon.

    Now you can pick a rational number a and irrational number b such that 0<\vert a-c\vert<\delta_1 and 0<\vert b-c\vert<\delta_1,
    and their absolute value is grater than c.

    Then \vert f(a)-f(b)\vert=\vert a^2+b\vert>\vert c^2+c\vert>\varepsilon_1

    a clear contradiction that \vert f(x)-f(c)\vert<\varepsilon_1 if 0<\vert x - c \vert<\delta_1

    Now this is probably not the most elegant solution, but you get the idea?
    Now it is your turn to write something like this, or not like this.
    You don't have to use proof by contradiction.
    Just think about the problem, if epsilon is small enough then there can be no delta .

    Hope that helps.
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  5. #5
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    thanks a lot. this has made it much clear and now i already get it.
    wonderful
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