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Math Help - Continuity at 0 (Rational/Irrational)

  1. #1
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    Continuity at 0 (Rational/Irrational)

    Prove that the function f defined by f(x)= x if is rational and f(x)=-x if x is irrational is continuous at 0 only .

    Ok so I was thinking doing it this way:

     \exists \{x_n\} \in \mathbb{Q} such that x_n \to c, x_n \neq c , \forall n and c \neq 0
     \exists \{y_n\} \in \mathbb{R} \backslash \mathbb{Q} such that y_n \to c , y_n \neq c , \forall n and c \neq 0

    Then,

    lim f(x_n) = lim (x_n)=c
     <br />
lim f(y_n) = lim -(y_n)=-c<br />

    So then if I say something like there exists a sequence \{x_n\} of rational numbers such that x_n \to 0 and a sequence \{y_n\} of irrational number such that  y_n \to 0. And then show that there limits are the same would that show that it is only continuous at 0???
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  2. #2
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    Does this give a verry strong hint?
    \left| x \right| < \delta  = \varepsilon  \Rightarrow \quad \left| {0 - \left( { - x} \right)} \right| = \left| {0 - \left( x \right)} \right| < \delta  = \varepsilon
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  3. #3
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    Yes, it does give a strong hint...

    Ok so how would I word my answer??? Would something like that be correct?

    Let \epsilon > 0 Choose \delta = \epsilon.
    Then for \left| f(0) - f(x) \right| = \left| {0 - \left( { - x} \right)} \right| = \left| {0 - \left( x \right)} \right|< \epsilon = \delta \quad \forall x \in \mathbb{R}, \left| x - 0 \right| < \delta \quad \delta > 0

    This shows that the function is continous at 0 but it doesn't show that it is only at 0, right???
    So I have to show that the function is discontinuous at every other points? If I showed that there are two sequences converging to different limit would that be enough?
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  4. #4
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    \left( {\forall z \ne 0} \right)\left( {\forall n} \right)\left( {\exists x_n  \in \mathbb{Q}} \right)\left( {\exists y_n  \in \mathbb{R}\backslash \mathbb{Q}} \right)\left[ {\left( {x_n } \right) \to z\,\& \,\left( {y_n } \right) \to z} \right]
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