Popcorn-kind function (I suppose)

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

I have to show that there is no such closed interval on which $\displaystyle f(x)$ has maximum although it is bounded.

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Originally Posted by james_bond

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

I have to show that there is no such closed interval on which $\displaystyle f(x)$ has maximum although it is bounded.

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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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Originally Posted by HallsofIvy

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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Thx I understand but how can I do that? Little confused :S

Quote:

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Originally Posted by james_bond

Thx I understand but how can I do that? Little confused :S

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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?