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How is this statement true? "For any real number $x$ we have $\sqrt{x^2} = |x|$"? Because putting $x=2$ $\sqrt{x^2}$ gives BOTH $2$ and $-2$ But $|x|$ only gives $2$

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    $\begingroup$ Indeed $\sqrt{x^2}=|x|$,can be confuse as you said.... As convention, the square root function is only the positive part, so the statement is true. $\endgroup$ – Brethlosze Sep 18 '17 at 10:21
  • $\begingroup$ Do you mean that by convention, √4 is only 2 and not -2? $\endgroup$ – user167573 Sep 18 '17 at 10:27
  • $\begingroup$ Square root a number always gives a positive answer. $\endgroup$ – abc... Sep 18 '17 at 10:27
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    $\begingroup$ By convention, the square root function always gives a positive number, though both signs are valid. Hence $\sqrt{x}>0$ always. $\endgroup$ – Brethlosze Sep 18 '17 at 10:31
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    $\begingroup$ Your assertion that $\sqrt{4} = \pm 2$ is false. If $x$ is a real number, the notation $\sqrt{x}$ means the principal (nonnegative) square root of $x$. Also, see this related question. $\endgroup$ – N. F. Taussig Sep 18 '17 at 12:09
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Remember that by definition and for $x \in \mathbb{R}$ $$|x| = \begin{cases} x, & \mbox{if } x \ge 0 \\ -x, & \mbox{if } x < 0. \end{cases}$$

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Let a number $n \in \mathbb{R}$, Then
$$\sqrt{n} \geq 0.$$

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    $\begingroup$ What if $n \in \mathbb{C}$? $\endgroup$ – Kevin Sep 18 '17 at 10:39
  • $\begingroup$ I just edited it. $\endgroup$ – abc... Sep 18 '17 at 10:43
  • $\begingroup$ Type \geq in math mode to obtain $\geq$. For $\leq$, type \leq in math mode. Here is a tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Sep 18 '17 at 12:04
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    $\begingroup$ Adding to what @N.F.Taussig said, you can also type \ge or \le to generate the same, respectively. Or, for a more traditional approach, type \geqslant or \leqslant to generate $\geqslant$ or $\leqslant$. $\endgroup$ – Mr Pie May 10 '18 at 2:25

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