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Variation of Epsilon Delta Proof

  1. Jul 19, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove that if

    ##\left |x-x_{0} \right | < \frac{\varepsilon }{2}## and ##\left |y-y_{0} \right | < \frac{\varepsilon }{2}##

    then

    ##|(x+y)-(x_0+y_0)| < \varepsilon ## and ##|(x-y)-(x_0-y_0)| < \varepsilon ##


    2. Relevant equations
    Postulate and proof with real numbers as well as inequalities.

    3. The attempt at a solution

    First we understand that the first two statements imply the existence of ##\varepsilon ## such that

    ##\frac{\varepsilon }{2} - |x-x_0| > 0##

    ##\frac{\varepsilon }{2} - |y-y_0| > 0##

    Hence we can add the two statements.

    ##\varepsilon - |x-x_0| - |y-y_0| > 0##

    ##|x-x_0| + |y-y_0| < \varepsilon ##

    Considering Triangle Inequality

    ##|a+b| \leq |a|+|b|##

    ##|x-x_0+y-y_0| \leq |x-x_0| + |y-y_0| < \varepsilon##

    Rearranging

    ##|(x+y)-(x_0+y_0)| \leq |x-x_0| + |y-y_0| < \varepsilon##

    Considering the inequality we can therefore say

    ##|(x+y)-(x_0+y_0)| < \varepsilon##

    The similar reasoning can be done to get

    ##|(x-y)-(x_0-y_0)| < \varepsilon##

    The proof is done as required. ##\blacksquare##

    I have not read the chapter about limit yet, since this is taken from Spivak's Calculus problem in prologue chapter. So I might miss some important fact or theorem.

    I also hope you guys can give me advice on how to make the above proof smoother or up to the rigorous standard, and additionally to write with LaTex properly. I've spend almost 1 hour just to transfer what I've done in paper to computer. I know I can use my drawing stylus, but I decided to try Latex in spite of that.

    Thank You
     
  2. jcsd
  3. Jul 19, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    I think that is fine.

    I would probably use the opposite direction: reduce the formula you want to show to a true statement. That direction is fine, too.

    They don't "imply" the existence. You just have to show (your long formula) for all x where ##|x-x_0|<\epsilon## is true. Therefore, in your proof, you can assume that this formula is satisfied - if it is not, you don't have to show anything.
     
  4. Jul 19, 2013 #3
    Ok thanks for your remark. Is it okay though, to prove a statement from the "then" part back to the "if" part? Is this what you refer by opposite direction?

    Do you have any good reference to learn Latex by the way?
     
  5. Jul 19, 2013 #4

    Zondrina

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    Homework Helper

    You can find lots of good latex resources with a simple Google search.

    Yes you can also start from 'then' in this case and work backwards if you would like to.

    ##|(x + y) - (x_0 + y_0)| = |(x - x_0) + (y - y_0)| ≤ |x - x_0| + |y - y_0|##
     
  6. Jul 19, 2013 #5
    In terms of "smooth-ness" you could probably start the proof just from the Triangle Inequality. The first bit seems a little unnecessary. As for LaTeX, I like this site: http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:About [Broken] for quick stuff. The best way to learn LaTeX though is just to start working and google as you need instead of trying to learn it then go.
     
    Last edited by a moderator: May 6, 2017
  7. Jul 19, 2013 #6
    Ok I'll check the link out. Thanks for your help guys.
     
  8. Jul 19, 2013 #7

    LCKurtz

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    Science Advisor
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    Gold Member

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