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Recently I had a thought that got me thinking.

There is the identity $\left (a^b\right )^c=a^{bc}$.

If I now take this identity I can do this:$$(-1)=(-1)^{2\cdot 0.5}=\left ((-1)^2\right )^{0.5}=1^{0.5}=1.$$Isn't an identity supposed to be true for all values?

Interestingly if I take the part $(-1)^{0.5\cdot 2}$ I can rearange it to both:$$\left ((-1)^2\right )^{0.5}=1^{0.5}=1$$and$$\left ((-1)^{0.5}\right )^2=i^2=-1.$$So in the latter example the identity does hold true.

But isn't exponentiation supposed to be commutative? Why do I get two different solutions?

Edit: As a user has pointed out in Why $\sqrt{-1 \cdot {-1}} \neq \sqrt{-1}^2$? The identity a^0.5b^0.5=(ab)^0.5 only holds true for postive numbers.

Is that also the case for the identity I used? Are all identities covering exponents only true for positive numbers?

Also what I meant with commutative was: (a^b)^c=(a^c)^b?

MJD
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tempdev nova
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    As you've just discovered, exponentiation is not commutative. Consider $2^3≠3^2$. – MJD Jun 18 '22 at 12:28
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    It is not associative either: Consider $(2^2)^3 ≠ 2^{(2^3)}$. – MJD Jun 18 '22 at 12:31
  • "Isn't an identity supposed to be true for all values?" Hehe, yes, for all values where both sides are defined. I don't think of $\left (a^b\right )^c=a^{bc}$ as an identity because, even when both sides are defined, it is inapplicable (as you've suggested) when there's a non-positive base and non-integral exponent. – ryang Jun 18 '22 at 12:53
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    Wait so $\left (a^b\right )^c=a^{bc}$. isn't an identity after all? – tempdev nova Jun 18 '22 at 12:57
  • Maybe call it a conditional equality? Usually an identity doesn't look wrong when it's written with three bars like this: $\left (a^b\right )^c\equiv a^{bc}.$ Semantics! – ryang Jun 18 '22 at 13:00
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    Interesting, I never got that taught in school. – tempdev nova Jun 18 '22 at 13:01
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    Whether a formula is, or isn't, an identity depends on the setting. $x^2\ge0$ is identically true in the setting of the real numbers, but not in the setting of the complex numbers. Similar remarks apply to $\log(xy)=\log x + \log y$. $x+x=0$ is an identity in any field of characteristic two, but not in the reals. – Gerry Myerson Jun 18 '22 at 13:24
  • x+x=0 is an identity in any field of characteristic two, but not in the reals. Wi don't understand this sentence. Isn't x+x=2x? – tempdev nova Jun 18 '22 at 14:50
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    Thanks for the clarification @GerryMyerson. So, for positive $a,b,c,$ we can indeed assert $\left (a^b\right )^c\equiv a^{bc},$ that is, that $\left (a^b\right )^c=a^{bc}$ is an identity. – ryang Jun 18 '22 at 15:21
  • $x+x=2x$, but in any field of characteristic two, $2=0$, so $x+x=0$. But maybe you are not familiar with the concept of a field of characteristic two. Are you familiar with congruences modulo two? By the way, tempdev, if you want to be sure I see a comment intended for me, you have to put @Gerry in it. – Gerry Myerson Jun 19 '22 at 00:41
  • @GerryMyerson Are you familiar with congruences modulo two? I am afraid that my knowledge about maths does not exceed what is tought in high school, so I think that what you are talking about probably far exceeds my knowledge. But thanks for your answer anyways! – tempdev nova Jun 19 '22 at 12:31

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