Definition or proof of $a^{\frac mn}=\sqrt[n]{a^m}$?

Question

Just a curiosity in this book of an high school with the title The powers with exponent rational

the $$\large \color{red}{a^{\frac mn}=\sqrt[n]{a^m}}, \quad a\geq 0$$ is given like a definition. I don't remember if there was a proof. But is there one or should we give it as a definition?

Answer 1

Let $r$ be a positive rational number, represented as the quotient of two positive integers, $r=m/n$. We will show that two definitions of $a^r$, where $a\ge0$ are equivalent: $a^r=\sqrt[n]{a^m}$, and $a^r=(\sqrt[n]{a})^m$. Furthermore, we shall show that either definition is independent of the representation of $r$ as the quotient of positive integers.

---

PART $(1)$:

Let $y\ge 0$ be defined

$$y=\sqrt[n]{a^m}=(a^m)^{1/n}\tag1$$

By definition of the $n$'th root, we have from $(1)$

$$y^n=a^m\tag2$$

---

Now, let $y=z^m$ for some number $z\ge 0$. Then, substituting into $(2)$ reveals

$$(z^m)^n=a^m\tag3$$

Using $(z^m)^n=(z^n)^m$ in $(3)$, we find that $z^n=a$, which implies that

$$y=(a^{1/n})^m\tag4$$

Equating $(1)$ and $(4)$ yields

$$(a^m)^{1/n}=(a^{1/n})^m$$

---

So, the two definitions of $a^r$, $(a^m)^{1/n}$ and $(a^{1/n})^m$ that are equivalent. That is to say, that

$$\bbox[5px,border:2px solid #C0A000]{\sqrt[n]{a^m}=(\sqrt[n]{a})^m}\tag5$$

---

---

PART $(2)$:

Now, we shall show that the definitions of $a^r=a^{m/n}$ in $(5)$ are independent of the representation of $r$.

Suppose $\ell$ is an arbitrary integer. Then, $a^r=a^{m/n}=a^{(\ell m)/(\ell n)}$. We wish to show that

$$a^r=\sqrt[\ell n]{a^{\ell m}}=\sqrt[ n]{a^{ m}}$$

We let $w$ be defined as

$$w=(a^{\ell m})^{1/(\ell n)}\tag6$$

so that $a^{\ell m}=w^{\ell n}$, which implies $a^m=w^n$, from which we have

$$w=(a^m)^{1/n}$$

And similarly, we wish to show that

$$(\sqrt[\ell n]{a})^{\ell m}=(\sqrt[n]{a})^{ m}$$

Let $v= (\sqrt[\ell n]{a})^{\ell m}$. But from $(5)$ we have

$$v=\sqrt[\ell n]{a^{\ell m}}$$

which we just showed was equal to $\sqrt[n]{a^m}$. Therefore, we have proven the independence of the representation of $r$ as the quotient of positive integers.

Answer 2

$a^{\frac mn}$ can be defined as $\sqrt[n]{a^m}$, in which case there is no point of talking about a proof of $a^{\frac mn} = \sqrt[n]{a^m}$. What you should prove is that this is well defined, i.e. if $\frac mn = \frac{m'}{n'}$ then $\sqrt[n]{a^m} = \sqrt[n']{a^{m'}}$. Indeed \begin{align*} &&mn' &= m'n\\ \Rightarrow &&a^{mn'} &= a^{m'n}\\ \Rightarrow &&(a^m)^{n'} &= (a^{m'})^n\\ \Rightarrow &&((a^m)^{n'})^{1/nn'} &= ((a^{m'})^n)^{1/nn'}\\ \stackrel{(*)}{\Rightarrow} &&(((a^m)^{n'})^{1/n'})^{1/n} &= (((a^{m'})^n)^{1/n})^{1/n'}\\ \Rightarrow &&(a^m)^{1/n} &= (a^{m'})^{1/n'}. \end{align*} $(*)$ Here we make use of the property $b^{1/\kappa\lambda}=(b^{1/\kappa})^{1/\lambda}$ for $b \geq 0$ and $\kappa,\lambda \in \mathbb{Z}_{>0}$. This follows from the fact that both LHS and RHS are non-negative solutions of the equation $x^{\kappa\lambda}=b$.

You could also talk about the motivation behind the definition: We would like $(a^x)^y = a^{x\cdot y}$ to hold for rational exponents, as it holds when $x,y \in \mathbb{N}$. So we would like to define $a^{\frac mn}$ in a way so that $(a^{\frac mn})^n = a^m$.