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I have a small doubt. Is $\sqrt{x^2}= |x|$ true? For, put $x=-1\implies x^2=1$. Now, $\sqrt 1=\pm 1$, whereas $|1|=+1$ only. $x\in\mathbb{R}$

P.S. This doubt came @Daniel's answer from this post. To quote-

The problem appears when you take square roots, since it is not still true that $\sqrt{(-1)^2}=-1$, in fact, $\sqrt{x^2}= |x|$ so in this case $\sqrt{(-1)^2}=|-1|=1$, which is not equal to $x-2$ when $x=1$.

Eisenstein
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  • What is $x$? An element of $\mathbb{R}$, or? – orlp Mar 13 '21 at 18:40
  • Also $\sqrt{1}$ is defined as just $1$ - the square root on real numbers always provides the positive solution to the quadratic equation by definition. – orlp Mar 13 '21 at 18:41
  • $\sqrt{1}=\pm 1$ is not quite correct, $\sqrt{1}=1$, because $\sqrt{.}:\mathbb{R}^{\geq 0}\rightarrow \mathbb{R}^{\geq 0}$ is defined as the inverse to the restriction $^2:\mathbb{R}^{\geq 0}\rightarrow \mathbb{R}^{\geq 0}$ – Peter Melech Mar 13 '21 at 18:44
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    Also: [Why is $\sqrt{x^2}= |x|$ rather than $\pm x$?](https://math.stackexchange.com/q/1340738/42969) – Martin R Mar 13 '21 at 18:50

2 Answers2

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The definition of the square root symbol is to indicate the positive square root only, which is why we write $\pm \sqrt{x}$ when we want to indicate both answers. So $\sqrt{1} = 1$ only.

RobertTheTutor
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I don't quite understand your reasoning after the question, therefore I'll state my point without addressing it. The standard notation for the real variable (which is the one used in user Daniel's comment) is that, given $\alpha\in\Bbb R$, $\sqrt\alpha$ is the non-negative real number $u$ such that $u^2=\alpha$, if such a number exist. Therefore the domain of $\sqrt\bullet$ is $[0,\infty)$ and its range is $[0,\infty)$ as well. It follows that $\sqrt{x^2}=x$ if $x\in[0,\infty)$ and $\sqrt{x^2}=-x$ if $x\in(-\infty,0]$.