Finite Elements in a Set of Rational Numbers Proof

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RPierre
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Homework Statement


This problem is insanely intuitive.

Define [tex]f : (0,1) \rightarrow \Re[/tex] by

[tex]f(x)=\begin{cases} <br /> 1/q&\text{if } x \neq 0 \text{, is rational, and }x = p/q \text{in lowest terms}\\ <br /> 0&\text{otherwise }\end{cases}[/tex]

Suppose [tex]\epsilon > 0[/tex]. Prove that there are at most a finite number of elements [tex]y\in(0,1)[/tex] such that [tex]f(y)\geq\epsilon[/tex]


Homework Equations


Must be a rigorous proof. Thats about it.


The Attempt at a Solution


I have no Idea where to START in solving this. All that I know is over (0,1), f(y) is always going to be 1/q, and therefore y must be rational in the form p/q. Otherwise, I have nothing on this one.
 
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f(y)=1/q only if y=p/q. If y is irrational which it surely can be between 0 and 1, then f(y)=0.

This is pretty straightforward. Answer the question: For what values of q is 1/q >= epsilon? So what does that tell you about the values of y such that f(y) >= epsilon?