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Math Help - continous proof problem

  1. #1
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    continous proof problem

    Hi Everyone,

    I'm new to the forums, but this place sounds awesome. I have a group assignment due tomorrow and we've gotten all of the problems except this last one. Let me know if anyone has any input on it -- it would be greatly appreciated

    f(x) = 0 if x is rational
    x if x is irrational

    With epsilon = .1, find a delta that makes plausible that f is continuous at 0.

    Is this even possible? Because as you get closer and closer to x (0) there will always be at least one rational and at least one irrational number in the interval? Thanks so much for your help!!!

    Jason
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  2. #2
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    If \varepsilon  = 0.1 then choose \delta  = 0.1 thus we have \left| {x - 0} \right| < \delta \quad  \Rightarrow \quad \left| {f(x) - 0} \right| < \varepsilon .
    This is because f(x) = \left\{ {\begin{array}{lr}<br />
   0 & {x \in Q}  \\<br />
   x & {x \in \Re \backslash Q}  \\<br />
\end{array}} \right. and in either case \left| {f(x)} \right| < \varepsilon
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  3. #3
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    Thanks so much!!!! I understand it now too
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