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Homework Help: Proof in Real Analysis

  1. Feb 1, 2010 #1
    Prove that abs(x-y) < ε for all ε>0, then x=y.

    I really do not know how to start this... I have tried to do the contra positive which would be If x does not equal y, then there exist a ε>0 such that abs(x-y) >= ε. Can someone help me and lead me to the right direction.
     
  2. jcsd
  3. Feb 1, 2010 #2

    jgens

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    Here's how I might approach the problem. Given any two real numbers [itex]x,y \in \mathbb{R}[/itex] we clearly have that [itex]x-y \in \mathbb{R}[/itex] and similarly that [itex]|x-y| \in \mathbb{R}[/itex]. Now, suppose that [itex]|x-y| > 0[/itex], can you see how this imediately results in a contradiction? What does this contradiction suggest about the value of [itex]|x-y|[/itex]?
     
  4. Feb 1, 2010 #3

    jgens

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    However, if you want to prove the contrapositive, you could do it this way. Pick any two real numbers [itex]x,y[/itex] such that [itex]x-y \neq 0[/itex]. Then clearly [itex]|x-y| > 0[/itex]. Now, what happens if you choose something like [itex]\varepsilon = \frac{|x-y|}{2}[/itex]?
     
  5. Feb 1, 2010 #4

    Hurkyl

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    Quick: without thinking, tell me a positive number that is either equal to or less than 5.
     
  6. Feb 1, 2010 #5
    Ok....I tried the contrapositive....I think that i got it so suppose that abs(x-y)=m>0. Therefore we make ε=m/2 which makes abs(x-y)=m>(m/2)=ε and therefore we have found an ε>o and which makes abs(x-y)>= ε.....is that right?
     
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