Finite Elements in a Set of Rational Numbers Proof

RPierre
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Homework Statement


This problem is insanely intuitive.

Define f : (0,1) \rightarrow \Re by

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

Suppose \epsilon &gt; 0. Prove that there are at most a finite number of elements y\in(0,1) such that f(y)\geq\epsilon


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?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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