# Square root of a number squared is equal to the absolute value of that number [duplicate]

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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.

## marked as duplicate by Jonas Meyer, JavaMan, Hans Lundmark, Asaf Karagila♦, J. M. is a poor mathematicianAug 25 '11 at 17:03

• 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
• $\sqrt{x^2}$ 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
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.