1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Gold Member

    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

    User Avatar
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook