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Math Help - Popcorn-kind function (I suppose)

  1. #1
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    Popcorn-kind function (I suppose)

    f(x)=\begin{cases}   \frac{q}{q+1}\mbox{ if }x=\frac{p}{q}\mbox{ is a rational number}\\   0\mbox{ if }x\mbox{ is irrational or }0  \end{cases}
    It is assumed here that \gcd(p,q) = 1 and q > 0 so that the function is well-defined and nonnegative.

    I have to show that there is no such closed interval on which f(x) has maximum although it is bounded.
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  2. #2
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    Quote Originally Posted by james_bond View Post
    f(x)=\begin{cases}   \frac{q}{q+1}\mbox{ if }x=\frac{p}{q}\mbox{ is a rational number}\\   0\mbox{ if }x\mbox{ is irrational or }0  \end{cases}
    It is assumed here that \gcd(p,q) = 1 and q > 0 so that the function is well-defined and nonnegative.

    I have to show that there is no such closed interval on which f(x) has maximum although it is bounded.
    You need to show that within any interval, you can find p/q with both p and q arbitrarily large so that q/(q+1) is arbitrarily close to 1: but never equal to 1.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    You need to show that within any interval, you can find p/q with both p and q arbitrarily large so that q/(q+1) is arbitrarily close to 1: but never equal to 1.
    Thx I understand but how can I do that? Little confused :S
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  4. #4
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    Quote Originally Posted by james_bond View Post
    Thx I understand but how can I do that? Little confused :S
    Let a and b be numbers with a< b. If a< p/q< b, for integers p and q (at least q positive), then aq< p< bq. How large does (b-a)q have to be? How large does q have to be for that to be possible? Do you see that for arbitrarily large q it is always possible?
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