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Analysis Proof

  1. Sep 5, 2011 #1
    1. The problem statement, all variables and given/known data
    Prove that if x in R and epsilon > 0 are arbitrary, then there exist r in Q such that |x - r | < epsilon


    2. Relevant equations



    3. The attempt at a solution

    I'm stumped on this one. I tried using the reverse triangle inequality, but I seemingly hit dead ends with it.
     
  2. jcsd
  3. Sep 5, 2011 #2
    Can you use the fact that the rationals are dense in the reals? This means that any real number can be approximated by a rational number.
     
  4. Sep 5, 2011 #3
    I know what you are saying, but how do I write that (this is my first proofs class).

    By the denseness of Q, you can say that epsilon < p < q < ... < r < x

    But I'm not sure how you get the inequality in there. Feel like I'm missing something
     
  5. Sep 5, 2011 #4
    The assumption is that you are given an [itex]x \in \mathbb{R}[/itex] and an [itex]\varepsilon > 0[/itex]. Using the fact that the rationals are dense in the reals, any real number can be approximated by a rational number. In other words, there exists a [itex]r \in \mathbb{Q}[/itex] such that [itex]r[/itex] is arbitrarily close to [itex]x[/itex], or [itex]|x - r| < \varepsilon[/itex].
     
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