Are my questions invalid or difficult cause I'm not getting answers since many days?

Question 1:
     By definition absolute value gives just no of units and does not indicate any direction neither positive nor negative then why in practice we use +ve direction like $\left|4\right|=+4$ it should be just 4 not +4

Question 2:
     We know that   $\sqrt{{x}^2} = \pm x = \left|x\right| $  

Then why   $\left|4\right|=+4$   and not   $\left|4\right| = \pm 4$
Also     $\sqrt{{4}^2} = +4$   and not   $\sqrt{{4}^2} = \pm 4$

Is this all have do with involvement of a variable only (when variable involved use $\pm$ and when constants involved don't use $\pm$)   ............ or there are some other rules !


The plus does nothing, it's $+4=4$. It's there to draw attention to the fact that the absolute value left a positive value unchanged -- they wanted to stress the difference between $-$ and $+$ case.

Also, $\sqrt{x^2}=\pm x$ is wrong. It's $\sqrt{x^2}=|x|$. The result of square root is always positive by definition. You can't get two values from a function. If you want to solve a quadratic equation, for instance, $y^2=x$, then you have to explicitly put $\pm$ in front of the square root to specify both solutions: $y=\pm \sqrt{x}$.

  • $\begingroup$ Is this true, $$ \left|k\right| < x \implies -k < x < +k $$ $\endgroup$ – Danish ALI Nov 15 '14 at 19:55
  • 1
    $\begingroup$ @DanishALI No, it's the other way around. If $x>0$, then you just bounded $k$ from both sides, so $-x<k<x$ (you can interpret this statement as "the distance of k from the coordinate origin is less than x". However, if $x\leq 0$, then no $k$ satisfies this condition (the absolute value cannot be negative). $\endgroup$ – orion Nov 16 '14 at 9:52
  • $\begingroup$ I have a link to this question: math.stackexchange.com/questions/96065/… here and in this post a reply by David Mitra states that |x + 1| is either (x + 1) or -(x + 1) if we interpret it here we could say |x| is either x or -x so is David Mitra saying wrong ? because |x| can't be -x because absolute value function will always produce positive numbers $\endgroup$ – Danish ALI Nov 16 '14 at 15:49
  • $\begingroup$ @DanishALI You are understanding it wrong. $|x|$ is one of the options $x$ and $-x$: the one that makes the value positive! It's not both and it's not optional what you choose! So, $|-5|=-(-5)$ because this makes it positive, but $|5|=(5)$ because that stays positive. $\endgroup$ – orion Nov 16 '14 at 16:21
  • $\begingroup$ Alright I get it this way |x| will always be positive but x could be any positive number or a negative number but when x goes under || operator it's returned in positive regardless what original direction was. Now have I understood it correctly ? $\endgroup$ – Danish ALI Nov 17 '14 at 20:00

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.