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

Question

I have a doubt on how to solve Exercise B11 of Section 3.4 of Advanced Calculus of Watson Fulks. It only says: prove that $\sqrt[n]{a^m}=(\sqrt[n]{a})^m$. Clearly it's not always true if $a<0$, so I guess I most assume $a\geq0$. And given that the previous sections are about continuity, strictly increasing functions and inverses I guess that I most use those topics. The only thing I have done is to define the function $f(x)=\sqrt[n]{x^m}$. It's strictly increasing so it must have an increasing inverse $\varphi(y)=\sqrt[m]{y^n}$ but from this point I don't know how to continue.

Answer 1

I assume $n$ and $m$ are positive integers? Just talk it out. For each positive real number $x$, there is a unique positive real number, called $\sqrt[n]{x}$, whose $n$th power is $x$.

So $\sqrt[n]{a^m}$ is the unique positive real number whose $n$th power is $a^m$. Since $\sqrt[n]{a} \space ^m$ is a positive real number whose $n$th power is $a^m$, they have to be equal.

Answer 2

This depends on you definitions but if you have proven that

Proposition: for any $b > 0$ and any $n\in \mathbb N$ there exists a $c\in \mathbb R$ $c > 0$ so that $c^n = b$ and such a $c> 0$ is unique. We call that $c$ then $n$-th root of $b$ and we write it either as $\sqrt[n]{b}$ or as $b^{\frac 1n}$.

Then this result is basic.

You know $(b^n)^m = b^{nm}$ so

So $[(\sqrt[n]{a})^m]^n = (\sqrt a)^{mn} = [(\sqrt[n]{a}^n)]^m= a^m$. So $\sqrt[n]{a}^m$ is a positive number that when raised to the $n$th power is $a^m$. So $(\sqrt[n]{a})^m = (\sqrt[n] {a^m})$.

But the more important part is proving that proposition.

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And $(b^m)^n = b^{mn}$ because.....

If $b^m= \underbrace{b\times b \times .... \times b}_{m}$ then $(b^m)^n=(b^n)^m = b^{mn} = \underbrace{\underbrace{b\times b \times .... \times b}_{m}\times \underbrace{b\times b \times .... \times b}_{m} \times .... \times \underbrace{b\times b \times .... \times b}_{m}}_n = \underbrace{b\times b \times .... \times b}_{mn}=\underbrace{\underbrace{b\times b \times .... \times b}_{n}\times \underbrace{b\times b \times .... \times b}_{n} \times .... \times \underbrace{b\times b \times .... \times b}_{n}}_m$

Answer 3

First, for any $a>0$ and $n,m\in\mathbb{N}$, we know

$$(a^n)^m=a^{nm}=a^{mn}=(a^m)^n$$

This is easily seen by a simple counting argument:

$$(a^n)^m=\underbrace{(\underbrace{a\cdot a\cdot a\cdot...\cdot a}_{n\text{ times}})\cdot (\underbrace{a\cdot a\cdot a\cdot...\cdot a}_{n\text{ times}})\cdot...\cdot (\underbrace{a\cdot a\cdot a\cdot...\cdot a}_{n\text{ times}})}_{m\text{ times}}$$

$$= \underbrace{a\cdot a\cdot a\cdot ...\cdot a}_{nm\text{ times}}=a^{nm}$$

By the same argument we get $(a^m)^n=a^{mn}$. Of course, since multiplication is commutative this implies

$$a^{nm}=a^{mn}$$

Before continuing, note that for all $x>0$, we know $(\sqrt[n]{x})^n=\sqrt[n]{x^n}=x$. Now, define the function

$$\phi(x)=\sqrt[m]{x^n}$$

We will show that this is the inverse function for both of your functions. Since they are strictly increasing functions on the non-negative reals, this will imply they are the same function.

Now, let us consider your two functions. We have

$$\phi(f(x))=\sqrt[m]{f(x)^n}=\sqrt[m]{(\sqrt[n]{x^m})^n}=\sqrt[m]{x^m}=x$$

If we call your other function $g(x)$, then

$$\phi(g(x))=\sqrt[m]{g(x)^n}=\sqrt[m]{((\sqrt[n]{x})^m)^n}=\sqrt[m]{(\sqrt[n]{x})^{mn}}=\sqrt[m]{(\sqrt[n]{x})^{nm}}=\sqrt[m]{((\sqrt[n]{x})^n)^m}=\sqrt[m]{x^m}=x$$

Thus, $\phi(x)$ is the inverse function for both $f(x)$ and $g(x)$ and we are done.