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This definition is correct:

"let $A,B \in \mathbb{R}$, $B$ is absolute value of $A$, $B \triangleq|A|$, if $B=\begin{cases} A, & \mbox{if }A \geq0 \\ (-A), & \mbox{if }A \leq 0 \end{cases}$"

??

Thanks in advance!

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Can you explain the last but one line? Why do you think so? –  eccstartup Aug 12 '13 at 10:23
    
You probably mean $-(A)$ where you wrote $(-A)$, even though it is the same thing. –  Git Gud Aug 12 '13 at 10:23
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@GarnakOlegovitc Yes. That's why your definition, the way you wrote it, doesn't work. What you looking for is something like: given $A\in \Bbb R$, define the absolute value of $A$ and denote it by $|A|$ as $$|A|=\begin{cases} A, & \mbox{if }A \geq0 \\ (-A), & \mbox{if }A \leq 0 \end{cases}.$$ –  Git Gud Aug 12 '13 at 10:36
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@GarnakOlegovitc Yes, $|A|\in \Bbb R$, but that's by definition. You should think of absolute value as a function which is written in sort of an infix notation: $$|\cdot |\colon \Bbb R\to \Bbb R, A\mapsto \begin{cases} A, & \mbox{if }A \geq0 \\ (-A), & \mbox{if }A \leq 0 \end{cases}$$ –  Git Gud Aug 12 '13 at 10:45
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@GarnakOlegovitc Yes, that's correct and that's an alternative way to define absolute value. Bu the use of $\iff$ is very important. You didn't use it in your question. –  Git Gud Aug 12 '13 at 10:52

1 Answer 1

up vote 2 down vote accepted

Yes, I think it is correct because absolute value just means the distance from the origin.

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