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This question already has an answer here:

Why does Wolfram Alpha say that $\sqrt{1}=-1$?

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Is this a mistake or what? Can anyone help?

Thanks in advance.

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marked as duplicate by user67258, Pedro Tamaroff, Zev Chonoles, Ross Millikan, mhd.math Jul 7 '13 at 8:55

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    $\begingroup$ Why not? The square root has two values (except for the radicand $0$). But for me, it returns $1$, boringly, as the principal root. $\endgroup$ – Daniel Fischer Jul 5 '13 at 19:13
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    $\begingroup$ It doesn't even say $\sqrt {1}=-1$. Where are you getting this? $\endgroup$ – Git Gud Jul 5 '13 at 19:15
  • $\begingroup$ put by plot $\sqrt{x}=y$ the plot give me positive y $\endgroup$ – mhd.math Jul 5 '13 at 19:15
  • $\begingroup$ $\sqrt{1}$ can be interpreted as the zero set of the polynomial $x^2-1$. Depending on the ring, this can have all sorts of solutions. But simply $\{\pm 1\}$ in $\mathbb{R}$, $\mathbb{C}$, like in any field. $\endgroup$ – Julien Jul 5 '13 at 19:21
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Did you read the results carefully? It does give $1$ as the answer:

enter image description here

It then helpfully mentions that $-1$ is also a 2nd root of $1$:

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which is perfectly correct, because $(-1)^2=1$.

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  • $\begingroup$ but why ? mistake or what ? $\endgroup$ – mhd.math Jul 5 '13 at 19:16
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    $\begingroup$ @hmedan.mnsh: Do you understand that $(-1)\times(-1)=1$? $\endgroup$ – Zev Chonoles Jul 5 '13 at 19:17
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    $\begingroup$ @hmedan.mnsh You are giving the same meaning to two different terms. The square root of a positive number $x$ is the only positive $a$ such that $a^2=x$. While a square root of a positive number $x$ is any number $a$ such that $a^2=x$. The indefinite and definite articles a and the make all the difference. $\endgroup$ – Git Gud Jul 5 '13 at 19:21
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    $\begingroup$ @hmedan.mnsh, you seem to think there is only one square root of $1$. Maybe you are wrong? $\endgroup$ – Mariano Suárez-Álvarez Jul 5 '13 at 19:21
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    $\begingroup$ @hmedan.mnsh: How do you reconcile that with the fact $(-1)^2=1$? A square root of a number $x$ is a number $y$ such that $y^2 = x$. $\endgroup$ – Najib Idrissi Jul 5 '13 at 19:31
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There are two real values which are roots of $1$: $$(-1)^2 = (1)^2 = 1$$

Exactly one of those values, the non-negative root, is called the principal square root of $1$, and is the real value returned by the the square root function $f$: $$f: \mathbb R^{\geq 0} \to \mathbb R^{\geq 0},\quad f(x) = \sqrt x,\quad\text{where}\;f(1) = \sqrt 1 = 1$$

but $(-1)$ is also, by definition, a real (non-principal) root of $1$.

See also principal square root in Wikipedia.

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  • $\begingroup$ $+1{}{}{}{}{}{}{}{}{}$ $\endgroup$ – mrs Jul 7 '13 at 7:49
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$-1$ is a perfectly valid square root of 1... think about it: $(-1)^2=1$, after all!

It is not what we usually call the "principle branch" of the square root, where you would expect to get 1.

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By writing $\sqrt{x}$ with $x>0$ , you implicitly mean the principal square root which is always positive. This is called the square root of $x$.

There is always another (negative) number whose square is $x$, namely $-\sqrt{x}$, in your case $-1$. The wolfram alpha has made it clear by saying:

1 (real, principle root)
-1 (real root)
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