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

Question

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Is this a mistake or what? Can anyone help?

Thanks in advance.

Why not? The square root has two values (except for the radicand $0$). But for me, it returns $1$, boringly, as the principal root. — Daniel Fischer, Jul 05 '13 at 19:13
It doesn't even say $\sqrt {1}=-1$. Where are you getting this? — Git Gud, Jul 05 '13 at 19:15
$\sqrt{1}$ 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. — Julien, Jul 05 '13 at 19:21
Related / possible duplicates: Square roots — positive and negative, Why is the even root of a number always positive?, Why is $\sqrt{4} = 2$ and Not $\pm 2$? — Zev Chonoles, Jul 05 '13 at 19:27

score 10 · Accepted Answer · answered Jul 05 '13 at 19:15

10

Did you read the results carefully? It does give $1$ as the answer:

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

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

answered Jul 05 '13 at 19:15

Zev Chonoles

129,973

but why ? mistake or what ? – mnsh Jul 05 '13 at 19:16
1

@hmedan.mnsh: Do you understand that $(-1)\times(-1)=1$? – Zev Chonoles Jul 05 '13 at 19:17
if $\sqrt{1}=-1$ and $\sqrt{1}=1$ so 1=-1 – mnsh Jul 05 '13 at 19:18
sure (−1)×(−1)=1 – mnsh Jul 05 '13 at 19:19
2

@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. – Git Gud Jul 05 '13 at 19:21
@hmedan.mnsh: Note that no one - not Wolfram Alpha, not me, not anyone else - is saying that $\sqrt{1}=-1$, so I don't know where you're getting that from. And clearly your statement that $1=-1$ is false. I recommend reading the Wikipedia article on the square root – Zev Chonoles Jul 05 '13 at 19:21
4

@hmedan.mnsh, you seem to think there is only one square root of $1$. Maybe you are wrong? – Mariano Suárez-Álvarez Jul 05 '13 at 19:21
yes I say that there is only one square root of 1 – mnsh Jul 05 '13 at 19:23
1

@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 $y^2 = x$. – Najib Idrissi Jul 05 '13 at 19:31

score 3 · Answer 2 · edited Apr 13 '17 at 12:21

3

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.

edited Apr 13 '17 at 12:21

Community

1

answered Jul 05 '13 at 19:16

amWhy

209,954

$+1{}{}{}{}{}{}{}{}{}$ – Mikasa Jul 07 '13 at 07:49

score 1 · Answer 3 · answered Jul 05 '13 at 19:15

1

$-1$ 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.

answered Jul 05 '13 at 19:15

Nick Peterson

32,430

score 1 · Answer 4 · answered Jul 05 '13 at 19:18

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)

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

4 Answers4