Can you please help me with the proof of this? I kwon this is for the properties of the absolute value, $|\cdot|$, but i can't figurate how get the "or".

$|x|>a \implies x>a \; \;\text{or} \;\; x<-a$

How you get that conclusion and not

$|x|>a \implies x>a \; \;\text{and} \;\; x<-a$

What is the best form to write the proof ? and get the "or".

  • 1
    $\begingroup$ $|x|>a\implies x>a\text{ and }x<-a$ is of course wrong ! $\endgroup$ – Surb Apr 11 '18 at 6:00
  • $\begingroup$ $|x| =$ either $x$ or $-x$ so if $|x|> a$ then either $x > a$ or $-x > a$. $\endgroup$ – fleablood Apr 11 '18 at 7:47

The definition of the absolute value is $$|x|:=\begin{cases}x&x\ge 0\\ or\\ -x&x<0\end{cases}.$$

Therefore $$|x|>a\implies \begin{cases}x>a&x\ge 0\\ or\\ -x>a&x<0\end{cases}\implies x>a\text{ or }x<-a.$$


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