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Finite Elements in a Set of Rational Numbers Proof

  1. Oct 6, 2009 #1
    1. The problem statement, all variables and given/known data
    This problem is insanely intuitive.

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

    [tex]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} [/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]


    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. jcsd
  3. Oct 6, 2009 #2

    Office_Shredder

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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?
     
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