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Math Help - help with absolute value

  1. #1
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    help with absolute value

    Problem : prove that |x|<|y| <=> x^2 < y^2 by giving a reason for each step :

    My teacher give solution

    step 1 : |x|<|y| => |x||x| =< |x||y| and |x||y| < |y||y|

    step 2 : => |x|^2 < |y|^2

    step 3 : => x^2<y^2

    can anybody justify the first step? i can't understand which properties values to use in order to get to first step. tq

    problem :

    suppose  |x| =< 2 . Use properties of absolute value to show that

    |(2x^2+3x+2)/(x^2+2)| =< 8
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  2. #2
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    Quote Originally Posted by bobey View Post
    Problem : prove that |x|<|y| <=> x^2 < y^2 by giving a reason for each step :

    My teacher give solution

    step 1 : |x|<|y| => |x||x| =< |x||y| and |x||y| < |y||y|

    step 2 : => |x|^2 < |y|^2

    step 3 : => x^2<y^2

    can anybody justify the first step? i can't understand which properties values to use in order to get to first step. tq


    He's using the basic property of inequalities: a<b\Longrightarrow ac<bc \,\,\,for\,\,\,c>0

    problem :

    suppose  |x| =< 2 . Use properties of absolute value to show that

    |(2x^2+3x+2)/(x^2+2)| =< 8

    Remember (definition/properties of abs. val.) that for b>0\,,\,\,|a|<b\Longleftrightarrow -b<a<b , so:

    |(2x^2+3x+2)/(x^2+2)| =< 8\Longleftrightarrow -8\leq \frac{2x^2+3x+2}{x^2+2}\leq 8 ,and now you have an "and" inequality system: both the left AND the right inequalities have to be true, so solve them separately and then take the intersection of the separate solutions. For example:

    -8\leq \frac{2x^2+3x+2}{x^2+2} \Longrightarrow \frac{2x^2+3x+2}{x^2+2}+8\geq 0\Longrightarrow \frac{10x^2+3x+18}{x^2+2}\geq 0 , and since both the numerator and the denominator are ALWAYS positive (why?)

    this inequality is true for all x ;

    \frac{2x^2+3x+2}{x^2+2}\leq 8\Longrightarrow \frac{2x^2+3x+2}{x^2+2}-8\leq 0\Longrightarrow \frac{-6x^2+3x-14}{x^2+2}\leq 0...and now can you see why this is ALWAYS negative?

    Interestingly enough, the given inequality is true for all x, imo, and thus I can't understand why they gave you that |x|<2...

    Tonio
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