Tiny doubt with square root

Question

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 $x=1$.

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

score 2 · Accepted Answer · answered Mar 13 '21 at 18:41

2

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.

answered Mar 13 '21 at 18:41

RobertTheTutor

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So my reasoning is not flawed, right? – Eisenstein Mar 13 '21 at 18:43
You wrote $\sqrt{1} = \pm 1$, which is incorrect. – RobertTheTutor Mar 13 '21 at 18:44
Tell me this- $-1^2=+1$? right? – Eisenstein Mar 13 '21 at 18:45
$(-1)^2 = 1$. However $-1^2 = -1$, because of order of operations, which startled me when I first learned it, as I always thought of "-1" as one thing, rather than "subtracting $1$". What you wrote says to square $1$, then add a minus sign. – RobertTheTutor Mar 13 '21 at 18:47

score 1 · Answer 2 · answered Mar 13 '21 at 18:43

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 $x\in(-\infty,0]$.

Tiny doubt with square root

2 Answers2