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Math Help - absolute value proof

  1. #1
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    absolute value proof

    Prove  |x| \leq y \Longleftrightarrow -y \leq x \leq y .

    If  x \geq 0 , then  |x| = x . So  x \leq y . Then  |x| = x \geq -y and so  -y \leq x \leq y . If  x < 0 , then  |x| = -x . So  |x| = -x \leq y and  |x| = -x \geq -y . Thus  -y \leq x \leq y .

    If  -y \leq x \leq y and  x \geq 0 , then  |x| = x \leq y and  |x| = x \geq -y . Similar process if  x < 0 .


    Is this correct?
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  2. #2
    Moo
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    Looks correct to me, though there are some weird things...


    This is of course assuming that y \geq 0.

    ------------------------------------
    Prove  |x| \leq y \Longleftrightarrow -y \leq x \leq y .

    If  x \geq 0 , then  |x| = x . So  x \leq y . Then (<< then ? it's not a consequence of x \leq y but of x \geq 0)  \underbrace{{\color{red}|x| =}}_{\text{useless}} x {\color{red}\geq 0} \geq -y and so  -y \leq x \leq y . If  x < 0 , then  |x| = -x . So {\color{red}\dots}-x \leq y and  {\color{red}\dots} -x {\color{red} \geq 0} \geq -y . Thus  -y \leq x \leq y (<< by multiplying each inequality by -1)

    If  -y \leq x \leq y and  x \geq 0 , then  |x| = x \leq y and  |x| = x \geq -y . Similar process if  x < 0 . (<< this one is good)
    Last edited by Moo; July 4th 2008 at 01:07 AM. Reason: no modification
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