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My main problem is that when you write $(-1)^1=(-1)^{\frac{2}{2}}= \sqrt{(-1)^2}=\sqrt{1}=1$. As I understand the problem is in the first equal $(-1)^1 \neq (-1)^{\frac{2}{2}}$. So $1 \neq \frac{2}{2}$ in exponents. And I raise a question, so in general for $\frac{a}{b}$ to be equal to $\frac{c}{d}$ in exponents: if $a$ is even then $c$ must be even, and if $a$ is odd then $c$ must be odd. And is similar in the case of $b$ and $d$. Am I wrong with this conclusions? I am not a mathematician, just a die-hard lover of them.

sebastian t
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    I would question $(-1)^{\frac22} \overset{?}= \left[(-1)^2\right]^{\frac12}$. – peterwhy Sep 28 '23 at 02:03
  • $(\sqrt{-1})^2=-1\neq 1= \sqrt{(-1)^2}.$ The rules for complex numbers are different from those for real numbers. Here is an article about it: https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/ – Marius S.L. Sep 28 '23 at 02:07
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    Everyone is being very coy about it, so let me be explicit: $1 = 2 \cdot \frac{1}{2}$ is always true, but $a^{bc}=(a^b)^c$ is not reliably true when $a$ is not positive. If you search the site you can find lots of previous answers that get into the details. (I'm not good at searching the site, or I'd provide a link.) – JonathanZ Sep 28 '23 at 02:18
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    Cf. this question – J. W. Tanner Sep 28 '23 at 02:36

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The issue is that the square root is not the same as raising something to the $1/2$ power. Think of how you write $\pm \sqrt{b^2-4ac}$ when you write the quadratic formula. This is because $\sqrt {1}$ is $1$ but $1^{1 \over 2}$ can be $2$ different values.

It does not really depend on being even or odd. Say you took ${-1 \over 2}+{\sqrt3 \over 2}i$ and raised it to the ${3\over 3}$ since $({-1 \over 2}+{\sqrt3 \over 2}i)^3$ is 1 the answer would be $\root{3}\of{1}=1$. Basically the issue is when you replace a fractional exponent with a radical which can only be done when the base or the number inside is the principal value such as we how pick the positive number for square roots (generally if we are doing exponents in the complex plane the principal branch of the logarithm is the value between $\pi$ and $-\pi$ which results in picking the value closest to the positive x-axis).

TheJack
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    I disagree with this answer. While fractional powers are occasionally used to denote multi-valued functions, it's pretty much unheard-of that, without explicit comment, anyone would regard $1^{1/2}$ as anything other than $1$. The question posed here avoids complex number entirely, and rests on the fact that it is not in general the case that $a^{bc} = (a^b)^c$ even when all quantities are real. – Damian Pavlyshyn Sep 28 '23 at 03:24
  • @DamianPavlyshyn Maybe it does but if $\ln$ was single-valued then since $\exp$ is single-valued and multiplication is single-valued you would never have this issue. It's only an issue because you can have multiples of $2\pi i$ in the exponent which don't stay 1. In other words the reason $x^2$ is single-valued is because $2 n \pi i$ stays $2 n \pi i$ with some integer n after getting multiplied by 2. The same does not happen with anything other than integers. This whole thing arises because $e^{2 \pi i}$ is 1 so it doesn't make sense to avoid complex numbers. – TheJack Sep 28 '23 at 16:51
  • @DamianPavlyshyn Your other point is a question of definitions. In most areas of math I've dealt with it's more useful to say the issue arises when we exchange it for a $\sqrt{}$ but maybe in other areas it is not so useful. This doesn't really matter because the question was really asking when it happens and when it doesn't so if you allow complex numbers it happens with any non-integer and if you don't allow them it happens if the denominator is even and the numerator isn't. – TheJack Sep 28 '23 at 16:53