What happens when a is negative in $ a^{\frac{m}{n}} $.

Question

What happens when a is negative in $ a^{\frac{m}{n}} $.

This question may seem silly but I got confused by it. If you take the example $2^{\frac{4}{3}}$

Method 1: $\sqrt[3]{2^4} \approx 2.51984$

Method 2: ${(\sqrt[3]{2})^4} \approx 2.51984$

But if you take -2 then

Method 1: $\sqrt[3]{-2^4} \approx 2.51984$

Method 2: $(\sqrt[3]{-2})^4 \approx$

Is this wrong or correct?

Answer 1

Every real number has three cube roots, one real and two complex. This means that rational exponentiation leads to several answers over the complex numbers. To simplify things, we often restrict ourselves to the real roots; in the cases where there are two real roots (like square roots, fourth roots, any roots with an even index) we use the principal (positive) root, so ${4}^\frac{1}{2} = 2$, rather than $-2$.

In your example, the (approximate) three cube roots of $2$ are $1.25992, -0.62996 + 1.09112\ i,$ and $-0.62996 - 1.09112\ i$. Similarly, the (approximate) cube roots of $-2$ are $-1.25992, 0.62996 + 1.09112\ i,$ and $0.62996 - 1.09112\ i$.

You seem to be using the real roots of $2$ and $-2$ in your first three examples, but then are switching to a complex root in the fourth example, hence the incongruency. If you're restricting yourself to the real roots, then all four examples should have the approximate answer $2.51984$; if you're working over the complex numbers, however, then every example would have three possible (approximate) answers, namely $2.51984, -1.25992 + 2.18224\ i$ and $-1.25992 - 2.18224\ i$.

Answer 2

Your method 2 yields the wrong answer,$(\sqrt[3]{-2})^4 \in \mathbb{R}$ Since the cuberoot of negative numbers is well defined. You mist have kept -1 isolated in one of them.

Answer 3

If you apply the first method to $-2$, then you should be computing$$\sqrt[3]{(-2)^4}=\sqrt[3]{16}\simeq2.51984,$$whereas if you apply the second method to it, then you should be computing$$\sqrt[3]{-2}^4\simeq(-1.25992)^4\simeq2.51984.$$So, there is no contradiction.

Answer 4

I suppose that you used Wolfram Alpha and here, as with many other software, we face the problem of the cube root.

If I type $(-2)^{\frac 13}$, the result effectively comes as $0.629961 +1.09112 i$ and then the result when you raise this number to the fourth power.

Now, if instead, I type cube root of -2, I get $-1.2599\cdots$

For odd roots of negative numbers, the computational engine of Wolfram|Alpha always select chosen the principal root, which is complex valued. This is why, similar to Sqrt[x], Mathematica also has CubeRoot[x].

But what you did is perfectly correct.