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Need help on a proof

  1. Feb 23, 2010 #1
    1. The problem statement, all variables and given/known data
    Hello thanks for everyone who helped me on the previous implication proof, here's another problem I'm stuck on:
    (Prove or disprove)
    14PMD.jpg


    2. Relevant equations



    3. The attempt at a solution
    I think it has something to do with x^2 - y^2 = (x + y)(x - y), and here's my interpretation of "i)":
    For every natural x and positive natural e, there exists one or more positive natural sigma so that if |x - y| is smaller than sigma then |x^2 - y^2| must be smaller than e, which works for any natural y.
    But I'm lost at where to go next, since sigma could be any number, and with the absolute sign there won't be negative numbers. The same thing with "ii)" and "iii)", where "ii)" simply switched the ordering of the sets and "iii)" limits x and y to 1 and 2.
    Thanks for any help!
     
  2. jcsd
  3. Feb 23, 2010 #2

    Mark44

    Staff: Mentor

    You're not working with the natural numbers, which are {0, 1, 2, 3, ...}. All of the variables are elements of the real numbers, with epsilon and delta being positive reals.

    Also, this Greek letter -- [itex]\delta[/itex] -- is lower case delta. This is lower-case sigma -- [itex]\sigma[/itex].

    It looks to me like i) and ii) say the exact same thing, which is essentially the statement in terms of delta and epsilon that the function f(x) = x^2 is continuous at an arbitrary real number y. In short, these are saying that if x is close to y (within delta), then x^2 will be close to y^2 (within epsilon).
    The third statement limits x and y to the interval [1, 2], but otherwise says the same thing as i) and ii).

    The way these work is that you start off with |x^2 - y^2| < epsilon, and work with the left side of the inequality until you get |x - y| < some expression. You can use the factorization - x^2 - y^2 = (x - y)(x + y), and the fact that the absolute value of a product is the product of the absolute values of its factors. The trick is that when you divide by |x + y|, you need to have some idea of how large or small it will be.
     
  4. Feb 23, 2010 #3
    Thanks! I got it after numerous tests of plugging in numbers.
     
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