# $r = \frac{m}{n} = \frac{p}{q} \Rightarrow (b^m)^{1/n} = (b^p)^{1/q}$; help with proof

This is from Baby Rudin chapter 1, exercise 6, and I'm using the unofficial answer key found here: https://minds.wisconsin.edu/handle/1793/67009

There is also this proof here: Prove that $(b^m)^{1/n} = (b^p)^{1/q}$ if $r = m/n = p/q$

which I find unconvincing for reasons I'll explain.

The proof provided by the Wisconsin professor proceeds thus:

Let $k = mq = np$. Because there is only one positive real number c such that $c^{nq} = b^k$, to show that $(b^m)^{1/n}$ and $(b^p)^{1/q}$ have this property, it follows that they are equal. Thus:

$((b^m)^{1/n})^{nq} = (b^m)^q = b^{mq} = b^k$, and similarly for $(b^p)^{1/q}$

My issue is in the computation, specifically the first and second steps; it seems like if I can distribute the exponents like this (which is not a proved property) the proof becomes trivial anyway, because then $(b^m)^{1/n} = b^{m/n} = b^{p/q} = (b^{p})^{1/q}$

In the thread to which I linked above, this issue is brought up, and the OP proposes a proof by induction which includes the following steps

$(x^s)^{k+1} =(x^s)^k(x^s) =(x^{sk})x^s =x^{sk+s} =x^{s(k+1)}$

The part you are missing is that we do know the following facts:

1. If $n,m\in\mathbb N$, then $(x^{n})^m=x^{nm}$ (which can be shown by induction, as indicated)
2. For $n\in\mathbb N$, then $(x^{1/n})^n=x$ (which is by definition of the $n$'th root).

Now we have $$((b^m)^{1/n})^{nq}=(((b^m)^{1/n})^n)^q=(b^m)^q=b^{mq}=b^{np}=(b^p)^n=(((b^p)^{1/q})^q)^n=((b^p)^{1/q})^{np},$$ then by the uniqueness of the positive root and the fact that $np=mq$, $(b^m)^{1/n}=(b^p)^{1/q}$.

EDIT: To see the first point, we have to understand the definition of $x^n$ for $n\in\mathbb N$. In fact, notice that Rudin does not even bother defining this (probably because it's more technical than required for the purpose of his book).

Let's first look at some example. Consider the multiplication operation on the natural numbers. How is it actually defined? We define it by induction as follows: $$n\cdot m=n\cdot (m-1)$$ with $n\cdot 1=n$. Note that this actually gives us an algorithm for multiplying two natural numbers. Now, when we want to define $n^p$, for $p$ a natural number, we must also define this inductively as $$n^p=n^{p-1}\cdot n.$$ Now, with a bit of work, we can extend this to a definition of multiplication on real numbers (Rudin might actually have done this in the appendix, in which he constructs the real numbers), and in the same way define exponentiation similarly as $$x^p=x^{p-1}\cdot x,$$ and I'll leave it to you to check that $x^p\cdot x^q=x^{p+q}$ (which you can show by induction on $q$).

Now, coming back to the inductive step we see that $$(x^s)^{k+1}=(x^s)^k\cdot(x^s)=(x^{sk})x^s=x^{sk+s}=x^{s(k+1)},$$ as required.

• Apologies, but if you look at my question, I bring up an issue with your point 1), which is that the induction proof that I cited seems to assume the conclusion. Can you please explain the induction, since that proof seems to be the crux of the issue?
– BenL
Commented Apr 22, 2017 at 19:27
• I'm sorry I originally missed that, hope my edit helps. Commented Apr 22, 2017 at 22:01

Since $$r=m/n=p/q$$, we have $$nr=m, qr=p$$ Then by above $(x^n)^{1/n}=x$

$$(b^{nr})^{1/n}=(b^m)^{1/n}=b^r$$ $$(b^{qr})^{1/q}=(b^p)^{1/q}=b^r$$ this imply $b^r=(b^m)^{1/n}=(b^p)^{1/q}$