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Thread: absolute value proof

  1. #1
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    absolute value proof

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

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

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


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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 $\displaystyle y \geq 0.$

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

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

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