Is $(-1)^{1/3} = -1$ , but $(-1)^{2/6} = 1$. Why aren't these the same?

Question

So if you try to solve $(-1)^{1/3}$ you can do

$(-1)^{1/3} = \sqrt[3]{-1} = -1$ (cubic root of $-1$)

But what if I write $1/3$ as $2/6$?

$$(-1)^{1/3} = (-1)^{2/6}$$

So $(-1)^{2/6} = \sqrt[6]{(-1)^2}$ (sixth root of $(-1)^2$)

So we have $\sqrt[6]{(-1)^2}= \sqrt[6]{1} = + 1$

Then we have: $+1 = -1$

How is that possible?

Answer 1

Exponentiation $a^b$ is not well-defined for all combinations of real numbers $a$ and $b$. Some of the cases where it has good definitions are

• When $a$ is arbitrary and $b$ is a nonnegative integer.

• When $a\ne 0$ and $b$ is any integer.

• When $a\ge 0$ and $b$ is a positive rational number.

• When $a\ge 0$ and $b$ is a positive real number.

• When $a$ is $e$ and $b$ is an arbitrary complex number.

These gives rise to quite different definitions which fortunately happen to agreee on the result where the domains overlap.

Your case $(-1)^{1/3}$ is not in any of these domains, and there is no generally accepted well-behaved definition that extends the domain to $(-1,1/3)$. Your own calculation shows that such a definition cannot exist while also respecting the usual power rules.

... Well ... you could posit an ad-hoc definition that gives meaning to $a^b$ in the case that

• $a$ is an arbitrary real, and $b=1/k$ for some positive odd integer $k$.

But that still would not enable your calculation with $b=2/6$.