1
$\begingroup$

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".

$\endgroup$
  • 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
3
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.