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Math Help - Continuous Functions, Rational number/ Irrational number.

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    Continuous Functions, Rational number/ Irrational number.

    Let  f:[a,b] -> R be continuous on [a,b] and suppose that f(x) = 0 for each rational number x in [a,b] .Prove that f(x)=0 \forall x \in [a,b]

    Ok, Do I need to do a proof by contradiction to prove that f(x)=0 \forall x \in [a,b]. What it comes down to is proving that f(x) = 0 for each irrational number x \in [a,b], right???
    Otherwise I don't really see how to do it, I mean even if I do by contradiction the way I thought about doing it was maybe say something like Assume that f(x) \neq 0 \forall x \in [a,b]
    I am not sure... any suggestion?
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    Quote Originally Posted by ynn6871 View Post
    Let  f:[a,b] -> R be continuous on [a,b] and suppose that f(x) = 0 for each rational number x in [a,b] .Prove that f(x)=0 \forall x \in [a,b]

    Ok, Do I need to do a proof by contradiction to prove that f(x)=0 \forall x \in [a,b]. What it comes down to is proving that f(x) = 0 for each irrational number x \in [a,b], right???
    Otherwise I don't really see how to do it, I mean even if I do by contradiction the way I thought about doing it was maybe say something like Assume that f(x) \neq 0 \forall x \in [a,b]
    I am not sure... any suggestion?
    If x is an irrational number then there is a sequence \{ x_n\} of rational numbers so that x_n \to x. Then by continuity f(\lim x_n) = \lim f(x_n) \implies f(x) = 0.
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    So from the problem we know the behavior of the function when x is rational so that is why we just have to look at the function when x is irrational, correct?

    If is an irrational number then there is a sequence of rational numbers so that
    I don't understand that part, how come you if you have x as an irrational number you have to have a sequence of rational numbers so that ??? Can you explain a little?


    Then by continuity .
    I think I get the second part better but lel me check if I understand it well, so since we know that the sequence converges to x then the \lim x_n = x and \lim f(x_n) = f(\lim x_n)= f(x) = 0 is that correct??
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    Quote Originally Posted by ynn6871 View Post
    So from the problem we know the behavior of the function when x is rational so that is why we just have to look at the function when x is irrational, correct?
    I don't understand that part, how come you if you have x as an irrational number you have to have a sequence of rational numbers so that ??? Can you explain a little?
    Do you understand the theorem Between any two real numbers the is a rational number?
    No matter how one approaches this problem, that theorem is used.
    If z is irrational, then \left( {\forall n \in \mathbb{Z}^ +  } \right)\left( {\exists r_n  \in \mathbb{Q}} \right)\left[ {z < r_n  < z + \frac{1}{n}} \right].
    It should be clear to you that \left( {r_n } \right) \to z which means f\left( {r_n } \right) \to f(z) = 0
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    Ok, I understand what you mean. I think I am starting to understand better but I am still a little confused about one little part of your explanation (sorry for asking so many question).

    So you said
    If z is irrational, then .
    That part I follow and understand it no problem now.

    The only thing that I am not sure I understand is
    It should be clear to you that which means
    because the problem states that for each rational number x in , however in your thing you said that f(z) = 0 z being an irrational number which is the opposity of what is stated in the problem isn't it??? Or am I missing something? Or is it because since r_n is rational then f(r_n)=0
    according to the problem?? So since , then f(r_n)=f(z) = 0 more or less???

    I am sorry to ask that many question, I am juste tying to be sure I understand it well and that I don't get this answer for the wrong reason.
    Thanks for all your help.
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    Look at the given: \left( {\forall x \in \mathbb{Q}} \right)\left[ {f(x) = 0} \right].
    If a number is not rational then it is irrational.
    Thus we need to show \left( {\forall x \in \mathbb{R}\backslash \mathbb{Q}} \right)\left[ {f(x) = 0} \right]. That is, if z is not rational f(z)=0.

    Now of any w, by the given f(w)=0 if w is rational and is w is irrational we have proved that f(w)=0.
    So f is zero for all numbers.
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