Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For some $x$,

$\sqrt{x^2} = |x|$

However, for $x= -1$.

$\sqrt{(-1)^2} = (-1^2)^{1/2} = (-1)^{2/2} = (-1)^1 = -1$

Isn't this paradoxical?

share|cite|improve this question
$\sqrt{-1^2}\neq\sqrt{(-1)^2}$. – Sanath K. Devalapurkar Feb 27 '14 at 2:01
The trick is that you can't generally say that $(a^p)^q = a^{pq}$ unless $a$ is positive. – Omnomnomnom Feb 27 '14 at 2:01
@SanathDevalapurkar Typo, sorry about that. – dfg Feb 27 '14 at 2:03
@SanathDevalapurkar and Eleven-Eleven: $((-1)^2)^{1/2}$ is, by definition, $\sqrt{(-1)^2}$. However, $(-1^2)^{1/2} \neq (-1)^{2/2}$. This "rule" does not always hold, so long as you define $x^{1/n}$ to be a function on $x$. – Omnomnomnom Feb 27 '14 at 2:08
@Eleven-Eleven: ah, fair enough, I guess I misunderstood your objection. Also, "root of the issue" is a very appropriate phrase here. – Omnomnomnom Feb 27 '14 at 2:16

As I said in the comment, the problem with this series of equalities is that we cannot generally say that $(a^{p})^q = a^{pq}$. So, in this instance, $$ ((-1)^{2})^{1/2} \neq (-1)^{2/2} $$ The equation $(a^{p})^q = a^{pq}$ will hold, however, as long when either $p$ and $q$ are both integers or $a$ is a positive number.

share|cite|improve this answer

To get to the problem here, we start with $\sqrt{(-1)^2}$. Please note that, before dealing with the square root, that $(-1)^2\neq -1^2$. $(-1)^2=(-1)(-1)=1$, while $-1^2=-(1^2)=-1$.

We can rewrite this using rational exponents, so $$\sqrt{(-1)^2}=((-1)^2)^{1/2}$$ I believe here, though we can just remedy the situation using our order of operations. Since we do what is in parentheses before handling exponents, we can go on and say that $$((-1)^2)^{1/2}=(1)^{1/2}=\sqrt{1}=1$$

We should also take into account the fact, though, that, as Omnomnomnomnom pointed out that we can't generally say using rules of exponents that $(a^m)^n=a^{mn}$ for all $a$. Since in this problem $a=-1$, we can't use exponential rules either.

share|cite|improve this answer
Eventually someone has to answer it though, right? – Ryan Sullivant Feb 27 '14 at 2:07
@RyanSullivant These questions, some confusion with square roots, never ever end. – Sawarnik Feb 27 '14 at 7:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.