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Math Help - Analysis Absolute Value Proofs

  1. #1
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    Analysis Absolute Value Proofs

    Ok, so I'm having some issues working on these proofs:

    a) |xy| = |x| * |y|
    b) If |x-y| < c, then |x| < |y| + c
    and
    c) If |x-y| < E for all E > 0, then x = y.

    Any help would be greatly appreciated!
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by auwesty View Post
    Ok, so I'm having some issues working on these proofs:

    a) |xy| = |x| * |y|
    b) If |x-y| < c, then |x| < |y| + c
    and
    c) If |x-y| < E for all E > 0, then x = y.

    Any help would be greatly appreciated!
    a.) |xy|=\sqrt{(xy)^2}=\sqrt{x^2y^2}=\sqrt{x^2}\sqrt{y  ^2}=|x|\,|y|

    b.) |x|=|y+x-y|\leq|y|+|x-y|<|y|+c

    c.) Assume x\neq y. Then by the properties of a metric, |x-y|=d>0. Thus, taking E=\frac{d}{2} gives us an E>0 such that |x-y|>E, which is a contradiction. So x=y. \square

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  3. #3
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    Thanks!

    That's all we have to do? All the examples we did in class had 2 - 4 cases we needed to prove (based on the definition of absolute value and the problem in question) in order to make this work.
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by auwesty View Post
    Thanks!

    That's all we have to do? All the examples we did in class had 2 - 4 cases we needed to prove (based on the definition of absolute value and the problem in question) in order to make this work.
    Yes. You could prove (a) by cases (assume x is positive and y is negative, etc.) but it's unnecessary if you know that |x|=\sqrt{x^2}.

    For the other two proofs, all you need to know is that |x-y| is a metric, and then you can use properties of metrics to complete the proof. No cases required!
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  5. #5
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    ahhh ok, we haven't learned metrics yet that's why i was a little confused
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  6. #6
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    Quote Originally Posted by auwesty View Post
    b) If |x-y| < c, then |x| < |y| + c

    c) If |x-y| < E for all E > 0, then x = y.
    b) \left| x \right| = \left| {x - y + y} \right| \leqslant \left| {x - y} \right| + \left| y \right| < c + \left| y \right|

    c) Suppose that x\ne y then that means |x-y|>0.
    So let E=|x-y|>0 then you have |x-y|<|x-y|.
    That is a contradiction to x\ne y.
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