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Math Help - Continuous Functions Proof

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    Continuous Functions Proof

    Define f : R R by f(x) = 5x if x is rational and f(x) = x + 6 if x is irrational. Prove that f is discontinuous at 1 and continuous at 2. Are there any other point besides 2 at which f is continuous?
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    Super Member Gamma's Avatar
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    at x=1 any open neighborhood about it will contain both rational and irrational numbers, but the irrational part is converging to 7 while the rational part is converging to 5 hear.

    however at 2 both parts converge to 10.
    to find the other point consider where
    x^2+6=5x \Rightarrow x^2-5x + 6 =0 \Rightarrow (x-3)(x-2)=0

    3 might be another good place to look.
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  3. #3
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by bearej50 View Post
    Define f : R R by f(x) = 5x if x is rational and f(x) = x + 6 if x is irrational. Prove that f is discontinuous at 1 and continuous at 2. Are there any other point besides 2 at which f is continuous?
    Hi bearej50.

    To prove that f is discontinuous at 1, let \epsilon=1. Then for any \delta>0, the interval (1-\delta,\,1+\delta) contains irrational numbers. If y is one of them, then |f(y)-f(1)|=|y^2+6-5|=|y^2+1|\geq1=\epsilon.

    To prove that f is continuous at 2, let \epsilon>0 be given. Then if \delta=\min\left\{1,\frac\epsilon5\right\}, consider y in the interval (2-\delta,\,2+\delta). If y is rational, we have |f(y)-f(2)|=|5y-10|=5|y-2|<5\delta\leq\epsilon. If y is irrational, then |f(y)-f(2)|=|y^2-4|=|y+2||y-2|<3\delta<\epsilon.

    There is one other point at which f is continuous, and that is found by solving the equation 5x=x^2+6 for x.
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    Or consider  (a_n) = 1+ \frac{\sqrt{2}}{n} . Then  a_n \to 1 . But  f(a_n) \not \to f(1) . Thus  f is discontinuous at  x = 1 . It seems that if both rational parts and the irrational parts are equal at  x then  f is continuous at  x (e.g. in the case of  x = 2 ).
    Last edited by manjohn12; May 5th 2009 at 05:43 PM.
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