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?

  • $\begingroup$ $\sqrt{-1^2}\neq\sqrt{(-1)^2}$. $\endgroup$ – user122283 Feb 27 '14 at 2:01
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    $\begingroup$ The trick is that you can't generally say that $(a^p)^q = a^{pq}$ unless $a$ is positive. $\endgroup$ – Omnomnomnom Feb 27 '14 at 2:01
  • $\begingroup$ @SanathDevalapurkar Typo, sorry about that. $\endgroup$ – dfg Feb 27 '14 at 2:03
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    $\begingroup$ @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 $x$. $\endgroup$ – Omnomnomnom Feb 27 '14 at 2:08
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    $\begingroup$ @Eleven-Eleven: ah, fair enough, I guess I misunderstood your objection. Also, "root of the issue" is a very appropriate phrase here. $\endgroup$ – 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.


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.

  • $\begingroup$ Eventually someone has to answer it though, right? $\endgroup$ – Ryan Sullivant Feb 27 '14 at 2:07
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    $\begingroup$ @RyanSullivant These questions, some confusion with square roots, never ever end. $\endgroup$ – Sawarnik Feb 27 '14 at 7:24

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