Rules on exponents

Question

What's wrong with this ? $$-2=(-2)^{1}=(-2)^{\frac22}=(-2)^{2\times\frac12}=((-2)^2)^{\frac12} = 4^{\frac12} = 2$$ Note that Wolfram Alpha gives : $-2=(-2)^{2/2} = (-2)^{2\times\frac12}$, and on the other hand $((-2)^2)^{\frac12}=2$. So, based on Wolfram Alpha, I would say that the rule $a^{bc}=(a^b)^c$ doesn't hold when $a<0$.

But at the same time, Wolfram Alpha still gives $$-2=(-2)^{\frac12\times2}=((-2)^{\frac12})^2$$ so it looks like the rule $a^{bc}=(a^b)^c$ holds after all.

I mention that I don't mind working in $\mathbb C$ (so we can have $(-2)^{\frac12}=i\sqrt2$). My question is :

What are the conditions on $a, b, c$ for $a^{bc}=(a^b)^c$ to be true ?

Answer 1

You have a domain error. You should not take a rational exponent of a negative number if that exponent has an even denominator.

Answer 2

Instead of $-2$ lets consider an arbitrary real number $x$, then following your same proof we get $$x = (x^2)^{1/2}=\sqrt{x^2}$$ however it is well known that in fact $\sqrt{x^2}=|x|$ so what went wrong? The rule that $$x^{a/b}=(x^{a})^{1/b}$$ actually breaks down in the very special case that both $x$ is negative and both $a$ and $b$ are even. In general, this property doesn't necessarily hold when the base $x$ is a negative number. For your problem in particular, consider the equation $x^2 = 4$, which has solutions of $2$ or $−2$. The principal value of $4^{1/2}$ is $2$, but $−2$ is also a valid "square root" in the complex numbers, so if the definition of exponentiation of real numbers is extended to allow negative results then the result is no longer well-behaved. In conclusion, your problem arises due to the fact that taking the $b^{th}$ root is not unique so you cannot "undo" this operation in a well defined manner if you do not restrict yourself to a specific domain. In general, your result does not hold when the base of the exponent is negative.

If you would like to learn more about the non-uniqueness of $x^{1/b}$ see https://en.wikipedia.org/wiki/Root_of_unity. It turns out that when $b$ is a positive integer, there are exactly $b$ unique complex numbers $x^{1/b}$ that satisfy $$(x^{1/b})^b=x$$