The result of a negative number raised to equivalent fractional powers

Question

When a negative number is raised to a fractional power, will two equivalent fractions as powers give the result the same sign (i.e., both negative or both positive)? For example, I want to raise -5 to a power of $\frac{1}{3}$. $$(-5)^\frac{1}{3}$$ obviously yields a negative number, but what if I used an equivalent fraction, say, $\frac{2}{6}$? $$(-5)^\frac{2}{6}$$ seems to yield a positive number since it is equal to $\sqrt[6]{(-5)^{2}}$, with the inner square removing the negative sign. What is the reason that both expressions should be or should not be the same (sign)? $$(-5)^\frac{1}{3} = (-5)^\frac{2}{6}?$$

Answer 1

Hypothetically, for the sake of discussion, we should suppose that $(-5)^\frac13$ and $(-5)^\frac16$ are well-defined expressions. Formally, we should work with the algebraic closure of the field of rational numbers $\mathbb{Q}$ for this. IF this is the case, then it is reasonable to state that $$(-5)^\frac16\cdot(-5)^\frac16=(-5)^\frac13$$ because $$(-5)^\frac16\cdot(-5)^\frac16=(-5)^{\frac16+\frac16}=(-5)^\frac26=(-5)^\frac13.$$ On the other hand, you are claiming that $$(-5)^\frac26 = [(-5)^2]^\frac16=25^\frac16\neq(-5)^\frac16\cdot(-5)^\frac16.$$ Your line of reasoning is faulty, and this is probably because you are thinking that $x^{p\cdot{q}}=(x^p)^q=(x^q)^p$, which is not true in general if $p,q$ are not integers. Why is that? Because unlike with exponents $n$ that are integers, $$(x\cdot{y})^p=x^p\cdot{y^p}$$ is not true in general for $p\in{\mathbb{Q}}$. To realize that, here is a counterexample with $x=y=-1,p=\frac12$: $[(-1)\cdot{(-1)}]^\frac12=1^\frac12\neq(-1)^\frac12\cdot{(-1)^\frac12}=-1$. Ultimately, this is not so much a problem with negative numbers, but a problem inherent to the very idea of exponentiation: there is just no satisfactory way to extend exponentiation to accommodate for non-integral powers, even if we limit ourselves to positive rational numbers.

The problem is that exponentiation, like multiplication, is motivated by iterated multiplication. Essentially, by definition, $$x^n:=\prod_{m\in{\mathbb{N}},m\lt{n}}x,$$ much in the same way that, for natural numbers, we motivate multiplication by iterated addition, that is, $$x\cdot{n}:=\sum_{m\in{\mathbb{N}},m\lt{n}}x.$$ However, when we extend the operation of multiplication from the natural numbers to a ring, such as the integers or the rational numbers, we need to use a more fundamental property of multiplication in order to define multiplication in these structures. It turns out that we can do this rigorously: we define multiplication $\cdot$ as being a binary operation that distributes over addition $+$. This manages to replicate the idea of iterated addition in the special case of the natural numbers, while also giving a well-defined product when both quantities are not natural numbers. With exponentiation, we would like to do the same thing: we would like to generalize exponentiation to abstract structures that are not reliant on the natural numbers or integers. But unlike with multiplication, we actually have no satisfactory and consistent way of doing this. One problem is that $f:f(x)=x\cdot{2}$, for example, is an invertible function in a field. However, $f:f(x)=x^2$ is not invertible in most structures we are interested in. As such, $x^\frac12$ is not uniquely well-defined, unless you are willing to make some sacrifices/concessions, but even with those sacrifices, the definition is far from satisfactory. So in many cases, it is actually preferrable to just leave it undefined, depending on the particular structure we are working in and the applications.

In general, we say that $x^\frac{1}{m}$ is defined as the unique solution to the equation $y^m=x$ with the largest real part, but this definition is only applicable within the complex numbers, for example. With this definition, it makes sense to say that $x^\frac12=\sqrt{x}$, but then you definitely lose $(x^p)^\frac12=(x^\frac12)^p$ in general.