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$$(-1)^ \frac{2}{4} = \sqrt[4]{(-1)^2} = \sqrt[4]{1} = 1$$

$$(-1)^ \frac{2}{4} = (-1)^ \frac{1}{2} = i$$

Came across an interesting Y11 question that made pose this one to my self. I can't for the life of me, find the flaw in the initial reasoning.

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The question is essentially the same as asking what's wrong with:

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

The error stems from being ambiguous about the meaning of $\sqrt{x}$ or, more generally, of $\sqrt[p]{x}$. A (real or) complex number $x\neq 0$ has a set of $p$ complex $p^{th}$ roots. If $x$ is a positive real number, exactly one of those roots is going to be a positive real number, so in some contexts (mostly when one is considering only positive real numbers), one talks about that single positive real root as the $p^{th}$ root of $x$, ignoring the other roots.

But you have to be consistent. You can't say "look, $-2$ is a square root of $4$, but the square root of $4$ is $2$, so $-2=2$".

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The existing answers are not addressing the real issue, and hence are in my opinion misleading. The problem has nothing to do with multiple roots.

When you write "$a^b$", this is a notation that indicates exponentiation. In modern mathematics, this is typically understood as the exponentiation function applied to the two inputs $a$ and $b$. If we use $pow$ to denote this function, then $a^b = pow(a,b)$.

Now the whole idea of exponentiation is repeated multiplication, and that is why we usually define $pow$ such that for every real $x$ and natural $n$ we have:

$pow(x,0) = 1$

$pow(x,n+1) = pow(x,n) · x$.

To extend this to rational and then real exponents is very non-trivial, and one has to prove the properties that the resulting $pow$ function has. One cannot just assume that the properties that were true for natural exponents remains true for real exponents! (See the linked post for a brief sketch of how it is done.)

In particular, after defining $x^a = pow(x,a)$ for reals $x,a$ such that $x>0$, it will still take a lot of hard work to prove the following nice properties:

(1) $pow(x,a+b) = pow(x,a) · pow(x,b)$ for any real $x>0$ and reals $a,b$.

(2) $pow(x,a·b) = pow(pow(x,a),b)$ for any real $x>0$ and reals $a,b$.

I have purposely written the properties in this manner to make it clear that it is unjustifiable to assume that they hold unless you have proven them. Specifically, you have attempted in your first line to use property (2) for negative $x$:

(Wrong) $\color{red}{ pow(-1,2·\frac14) = pow(pow(-1,2),\frac14) }$.

And that is the real mistake; you have moved the factor of "$2$" from the second input of the $pow$ function to the first input, without any justification. It turns out that it is false, and that is why you get nonsense from the first line.

In fact, in this previous post I gave the exact same example to show why one cannot blindly apply previously known facts about some objects to other objects, expecting them to still hold.

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We don't define rational power for negative numbers (accept odd roots, as user21820 noticed). Now, you can use branch cut (as user28120 noticed) to find root of $-1$ but if you do it so, you need to say that $(-1)^{1/2}=e^{ln(-1)\cdot1/2}$, and when you are doing it in this way, paradoxes won't appear.

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    $\begingroup$ You post is simply wrong. Some authors define the cube-root function to be the inverse of the cube function on the whole real line. Same for odd roots. Secondly, complex exponentiation is typically defined via a branch-cut and so there is nothing wrong with saying that the principal square root of $-1$ is $i$, where we take the branch of the natural logarithm corresponding to the argument $(-π,π]$. $\endgroup$
    – user21820
    Jul 23 '18 at 8:33
  • $\begingroup$ Yes, you are right about odd root and branch-cut. I will edit my post. $\endgroup$ Jul 23 '18 at 9:55
  • $\begingroup$ Thanks for editing your post. Unfortunately, it still doesn't answer the question. Indeed taking the principal branch $(-1)^{1/2} = e^{\ln(-1)·1/2} = i$. So that is not the source of the error. The other answer by Anonymous is equally misleading as well... $\endgroup$
    – user21820
    Jul 24 '18 at 6:59

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