# Finite Elements in a Set of Rational Numbers Proof

1. Oct 6, 2009

### RPierre

1. The problem statement, all variables and given/known data
This problem is insanely intuitive.

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

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

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

2. Relevant equations
Must be a rigorous proof. Thats about it.

3. 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.

2. Oct 6, 2009

### Office_Shredder

Staff Emeritus
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?