Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicate:
Significance of $\displaystyle\sqrt[n]{a^n} $?

The square root of a number squared is equal to the absolute value of that number. Why is $\sqrt{x^2} = |x|$? Why not just $x$? Please give me a reason and also help me prove it.

share|improve this question

marked as duplicate by Jonas Meyer, JavaMan, Hans Lundmark, Asaf Karagila, J. M. Aug 25 '11 at 17:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
i have studied in school that aritmetic root of any number is such non negative number whose square is given number so it means that we take absolute sign to ensure square is not negative –  dato datuashvili Aug 25 '11 at 6:49
2  
It's basically because of convention. Since $a^2=(-a)^2$ for any $a$, there will always be two square roots of any positive number. We define the function $f(x)=\sqrt{x}$ for positive $x$ to be the positive square root. So if $x$ is negative, then $\sqrt{x^2}$ is $x$ flipped across $0$ to the positive side, or $|x|$. –  anon Aug 25 '11 at 6:49
1  
$\sqrt{x^2}$ couldn't possibly be plain $x$ in general. If the square root function is given the number $4$, how can it know whether the $4$ came from $2^2$ or $(-2)^2$? –  André Nicolas Aug 25 '11 at 8:12
    
In your question you should say "real number" and not just "number". It is false for complex numbers. But at least $\sqrt{x^2}$ is either $x$ or $-x$. –  GEdgar Aug 25 '11 at 12:11
add comment

2 Answers

You are referring to the principal square root. From Wikipedia: "Although the principal square root of a positive number is only one of its two square roots, the designation 'the square root' is often used to refer to the principal square root."

share|improve this answer
add comment

By definition the square root of a nonnegative real number $y$ is the unique real number $z$ for which $z\geq 0$ and $z^2=y$. Consequently, to prove that $\sqrt{x^2}=|x|$ it is enough to show that $|x|\geq 0$ and $|x|^2=x^2$ but these two facts are very easy to verify.

share|improve this answer
add comment

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