Why does $\sqrt[\leftroot{-2}\uproot{2}b]{x^a} = x^{a/b}$ and what does it mean?

Question

I started to wonder the expression $\sqrt[\leftroot{-2}\uproot{2}b]{x^a} = x^{a/b}$, which was introduced in primary/high school and I don't remember any teacher actually proving this equation. What does the expression $x^{a/b}$ actually mean?

I can understand the other interpretation $\sqrt[\leftroot{-2}\uproot{2}b]{x^a} $: First raise $x$ to the $a$th power and then take the $b$th-root of it. But what does $x^{a/b}$ mean? For example $2^4$ is perfectly clear to me, but what does $2^{3/2} = 2^{1.5}$ mean? For me the expression $x^y$ means: Multiply $x$, $y$ times by itself so multiplying $2$, $1.5$ times by itself is a bit confusing :)

Is it even generally always true that:

$$\sqrt[\leftroot{-2}\uproot{2}b]{x^a} = x^{a/b},$$

if $x\in\mathbb{C}$ and $a,b\in\mathbb{R}$?

And if yes then why? Is it just a definition or agreement on notation? Or is there a actual physical/intuitive reason for this?

Thank you for any help =)

Answer 1

Throughout, $x$ will be a parameter restricted to real values greater than or equal to $1$. When $y$ is not a positive integer, the definition of $x^y$ is more subtle than "$x$ multiplied by itself $y$ times".

• One first defines $x^{1/b}$, for a positive integer $b$, to be the unique positive solution $w$ to the equation $w^b=x$. The existence of this $w$ is from the intermediate value theorem, using the continuity of raising a number to the $b$th power. The notation $\sqrt[b]x$ is just a synonym for $x^{1/b}$.
• Then for any positive rational number $a/b$, one defines $x^{a/b}$ to be $(x^a)^{1/b}$ (equivalently, $\sqrt[b]{x^a}$). In other words, it's the unique positive solution $w$ to $w^b = x^a$.
• Finally, for any positive real number $y$, one defines $x^y = \sup \{ x^{a/b} \colon \frac ab<y \}$.

From these definitions, various things can be proved: the rules of exponents $x^{ab}=(x^a)^b$ and $x^{a+b}=x^ax^b$, the fact that $x^y$ is for fixed $x$ a continuous increasing function of $y$, and so on.

One can extend to negative exponents by declaring $x^{-y} = 1/x^y$, and one can extend to real numbers $0<\xi<1$ by declaring $\xi^y = (1/\xi)^{-y}$. Again one checks that all the results of exponents, and different ways of parsing an expression like $\xi^{a-b}$ or $\xi^{-y}$, are consistent.

So it's not intuitive at all, really. As evidence, notice that we can't do any of this if the base is a negative number - the expression $(-2)^y$ makes sense only if $y$ is an integer. (At least, until you get into complex numbers, when the story gets crazier still....)

Answer 2

The $b$-th root $a=\sqrt[b]{x}$ is defined as a positive solution of the equation $x=a^b$. If you start from the fact $(a^b)^c=a^{bc}$, which you already know for whole number exponents, you generalize it to work for rational and even real exponents. We can prove your equality by simply taking your equation to the $b$-th power:

$$\sqrt[b]{x^a}=x^{a/b}$$ $$(\sqrt[b]{x^a})^b=(x^{a/b})^b$$ On the left, the root is removed because we introduced the root exactly as the inverse of $b$-th power. On the right, use the product rule: $$x^a=x^{(a/b)b}=x^a$$

However, there is one problem here: the equation $x=a^b$ has more than one solution. It has $b$ solutions for $a$ (it is a polynomial of degree $b$). For positive arguments, we always take the positive solution, and everything is fine. But for negative arguments, or if you want to work in complex numbers, this doesn't work anymore. Remember the absolute value:

$$|x|=\sqrt{x^2}$$

In this case you see that negative $x$ don't respect $\sqrt{x^2}=(x^2)^{1/2}\neq x$. Integer powers larger than one are irreversible (for instance, squaring kills the minus and you can't guess what it was), while the roots have multiple solutions and one must be chosen as the standard solution. For real powers, things get even worse.

In short: for positive numbers, you can assume $\sqrt[n]{x}=x^{1/n}$ and $(x^a)^b=x^{ab}$. In any other case, be really careful and avoid using roots and powers if you can. Treat them as solutions of polynomials instead - in that case, you are always aware that there is more than one solution.