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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 mathematician Aug 25 '11 at 17:03

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  • $\begingroup$ 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 $\endgroup$ – dato datuashvili Aug 25 '11 at 6:49
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    $\begingroup$ 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 $|x|$. $\endgroup$ – anon Aug 25 '11 at 6:49
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    $\begingroup$ $\sqrt{x^2}$ 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 $(-2)^2$? $\endgroup$ – André Nicolas Aug 25 '11 at 8:12
  • $\begingroup$ 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 $-x$. $\endgroup$ – GEdgar Aug 25 '11 at 12:11

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."


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.


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