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Math Help - Need help on another proof

  1. #1
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    Need help on another proof

    Hello thanks for everyone who helped me on the previous implication proof, here's another problem I'm stuck on:
    (Prove or disprove)


    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 real x and positive real e, there exists one or more positive real delta so that if |x - y| is smaller than delta then |x^2 - y^2| must be smaller than e, which works for any real y.
    But I'm lost at where to go next, since delta 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!
    Last edited by Selena; February 23rd 2010 at 10:17 AM.
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  2. #2
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    Quote Originally Posted by Selena View Post
    Hello thanks for everyone who helped me on the previous implication proof, here's another problem I'm stuck on:
    (Prove or disprove)


    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 real x and positive real e, there exists one or more positive real delta so that if |x - y| is smaller than delta then |x^2 - y^2| must be smaller than e, which works for any real y.
    But I'm lost at where to go next, since delta 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!
    Perhaps you could assume the antecendent is true, then try and show the consequent?
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  3. #3
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    Thanks I tested out the consequent by plugging in several values for delta, x and y, and I got all three questions. My answer was that i) and iii) works while ii) doesn't, since there's no limit on both x and y.
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  4. #4
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    Quote Originally Posted by Selena View Post
    Hello thanks for everyone who helped me on the previous implication proof, here's another problem I'm stuck on:
    (Prove or disprove)


    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 real x and positive real e, there exists one or more positive real delta so that if |x - y| is smaller than delta then |x^2 - y^2| must be smaller than e, which works for any real y.
    But I'm lost at where to go next, since delta 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!
    .

    (i) tell you to prove that the function f(x) = x^2 is continuous at xεR

    (ii) tell you to prove that the function f(x) = x^2 is uniformly continuous over the Real Nos,and

    (iii) tell you to prove that the function f(x) = x^2 is uniformly continuous at the interval [1,2]

    I will prove (iii) and let you prove the other two.

    So given an ε>0 we must find a δ>0 such that :

    for all ,x,y : if xε[1,2] and yε[1,2] and |x-y|<δ,then |x^2-y^2|<\epsilon

    But if xε[1,2] and yε[1,2] then  2\leq x+y\leq 4 which implies that |x+y|\leq 4 which implies that:

    |x^2-y^2| =|x+y||x-y|\leq 4|x-y|

    Hence if we choose δ = ε/4 we have :

     |x-y|<\delta\Longrightarrow |x-y|<\frac{\epsilon}{4}\Longrightarrow |x+y||x-y|<\frac{\epsilon}{4}.4 \Longrightarrow |x^2-y^2|<\epsilon ,for all x,y belonging to the interval [1,2].

    Hence the function f(x)= x^2 is uniformly continuous in the interval [1,2].

    Notice the said function is uniform at any No belonging to the Real Nos ,but it is not uniformly continuous over the Real Nos
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