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Math Help - absolute value proof

  1. #1
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    absolute value proof

    does anybody know how to solve the following..

    prove that, for real numbers x and y, |x| < |y| iff x^2 < y^2.
    Is the result true if the absolute value signs are removed from the statement?
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  2. #2
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    Quote Originally Posted by mathhelp1 View Post
    prove that, for real numbers x and y, |x| < |y| iff x^2 < y^2.
    If x^2<y^2 then \sqrt{x^2}<\sqrt{y^2}.
    How do we write \sqrt{a^2}~?

    If |x|<|y| then |x|^2<|x||y|~\&~|x||y|<|y|^2
    Recall that |a|^2=a^2.
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  3. #3
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    Quote Originally Posted by mathhelp1 View Post
    does anybody know how to solve the following..

    prove that, for real numbers x and y, |x| < |y| iff x^2 < y^2.
    Is the result true if the absolute value signs are removed from the statement?
    First, I think you are misunderstanding what "Linear and Abstract Algebra" means. This probably belongs in the "Pre-University Algebra" section.

    I presume you know that an "iff" (if and only if) statement has to be proved both ways. So you must prove both "if |x|< |y| then x^2< y^2" and "if x^2< y^2 then |x|< |y|".

    Now to prove the first, look at |x|< |y|. As is almost always the case with absolute value, break this into parts. If x\ge 0, |x|= x while if x< 0 |x|= -x. Since each of x and y can be "non-negative" or "negative", there will be four cases.

    1) Suppose x and y are both non-negative. Then |x|< |y| means 0\le x< y. Now what can you say?
    2) Suppose x and y are both negative. Then |x|< |y| means 0< -x< -y< 0. Now what can you say about their squares?
    3) Suppose x is non-negative and y is negative. Then |x|< |y| means 0\le x< -y. Now what can you say about their squares?
    4) Suppose x is negative and y is non-negative. Then |x|< |y| means 0< -x< y. Now what can you say about their squares?

    I rather suspect that you can do the other way by looking at the same four cases.
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