I'm working through Rudin's "Principles of Mathematical Analysis" on my own, so I don't want the full answer. I'm only looking for a hint on this problem.

The problem states that $b > 1$ and $m, n, p, q$ are integers, $n > 0$, $q > 0$, and $r = m/n = p/q$ and asks me to prove that $$ (b^m)^{1/n} = (b^p)^{1/q} $$

"hence it makes sense to define $b^r = (b^m)^{1/n}$" (to quote the book).

Previously in the chapter I proved the theorem that for every $x>0, x \in \mathbb{R}$ and $n>0, n \in \mathbb{Z}$, there is one and only one real $y$ such that $y^n = x$. However, the book hasn't established that $(x^s)^t = x^{st}$ yet, so I don't believe I can use that (which would make the solution trivial anyway).

My first thought was to try something like this, in which I start by simply reversing that theorem a few times like this:

\begin{align} (b^m)^{1/n} &= y\\ b^m &= y^n \\ b &= (y^n)^{1/m} \end{align}

but I don't see where that gets me. Plus, this doesn't seem to be the right path because I'm not given that $m > 0$, so the exponent in the last step may not even be defined. I can't use the fact that $b^r = (b^m)^{1/n}$ at this stage because that would be proving something using what I'm trying to prove.

This problem is the first step in the exercise in Rudin that walks me through proving the product rule for rational, and later real, exponents, which is detailed really nicely in this answer.

Am I just missing the first step, or am I on the wrong track to begin with?

Second attempt

I think I've figured it out now, thanks to the hints. First, I need to prove that $(x^s)^t = x^{st}$ where $x \in \mathbb{R}, s,t \in \mathbb{z}$, and $t > 0$. Using the hint of induction, I first proved that this is true for $t = 1$, which is trivial, since $(x^s)^1 = x^s = x^{s \cdot 1}$.

Now, I assume that this is true for $t = k$, i.e. $(x^s)^k = x^{sk}$, and prove that it's true for $t = k + 1$.

\begin{align} (x^s)^t &= (x^s)^{k+1} \\ &= (x^s)^k (x^s) \\ &= (x^{sk}) x^s \\ &= x^{sk+s} \\ &= x^{s(k+1)} \\ &= x^{st} \end{align}

With that proved, I can move on to the next part of the proof.

Let $\alpha$ and $\beta$ be reals such that

$$ \alpha^n=b^m \qquad \beta^q=b^p $$

We know these are unique by the theorem proved earlier in Rudin. Therefore:

\begin{align} \alpha^n &= b^m \\ (\alpha^n)^q &= (b^m)^q \\ \alpha^{nq} &= b^{mq} \end{align}

and I can prove the same way that

$$ \beta^{nq} = b^{np} = b^{mq} $$

Applying the uniqueness of real roots theorem again means that $\alpha = \beta$. So,

\begin{align} \alpha^n &= b^m \\ \alpha &= (b^m)^{1/n} \end{align}


\begin{align} \beta^q &= b^p \\ \alpha^q &= b^p \\ \alpha &= (b^p)^{1/q} \end{align}

Therefore $(b^m)^{1/n} = (b^p)^{1/q}$.

Is that all correct?

  • 1
    $\begingroup$ But surely you can prove and use $(x^s)^t = x^{s t}$ when $s$ and $t$ are integers? $\endgroup$ – user66081 Jan 14 '14 at 0:48
  • $\begingroup$ @user66081 I think I proved it using induction (which was given as a hint in another answer). Does that look correct? $\endgroup$ – M T Jan 14 '14 at 23:16
  • $\begingroup$ In the proof for $(x^s)^t = x^{st}$, are you allowed to use the identity $b^{n+m}=b^nb^m$ where $n,m \in \mathbb{N}$? $\endgroup$ – Björn Lindqvist Jun 19 '18 at 23:20

The result you're looking to obtain is: $$ \frac{m}{n} = \frac{p}{q} \implies (b^m)^{1\over{n}} = (b^p)^{1\over{q}} $$

This is equivalent to: $$ (b^m)^{1\over{n}} \not = (b^p)^{1\over{q}} \implies \frac{m}{n} \not = \frac{p}{q} $$

If you first prove $(x^s)^t = x^{st}$ for integers s and t as suggested by another user, it should be fairly simple to proceed in this direction (noting that the function $f(x) = x^l$ is injective for non-negative x and an integer root is non-negative).

Edit: Since you have this part done there's no need for me to be coy. My method goes as follows.

Let $f(x) = x^{nq}$ be a function on the non-negative reals.

$$ (b^m)^{1\over{n}} \not = (b^p)^{1\over{q}} \implies f((b^m)^{1\over{n}}) \not = f((b^p)^{1\over{q}}) $$

$ f((b^m)^{1\over{n}}) = (((b^m)^{1\over{n}})^n)^q = (b^m)^q = b^{mq} $

$ f((b^p)^{1\over{q}}) = (((b^p)^{1\over{q}})^q)^n = (b^p)^n = b^{pn} $

Hence $ b^{mq} \not = b^{pn} $. This implies $mq \not = pn $, from which follows $ \frac{m}{n} \not = \frac{p}{q} $. Q.E.D.

  • $\begingroup$ It's easy to prove that $(x^s)^t = x^{st}$ when $s > 0, t > 0$ and $s,t \in \mathbb{Z}$ (I think the proof is just repeated multiplication), but since Rudin hasn't defined what negative exponents mean yet, is it ok to take that definition as given? I mean, can I take it as given that $x^s = \frac{1}{x^{-s}}$ when $s < 0$ and proceed with proving $(x^s)^t = x^{s t}$ when $s, t \in \mathbb{Z}$ as before? Other than that I think I have that part of the proof finished. $\endgroup$ – M T Jan 14 '14 at 21:22
  • $\begingroup$ @Michael The fact that exponents in the expressions key to the question can be negative tells me it's not meaningful to try and prove the result without a definition for a negative exponent. I would simply take it as given that they're defined as you've written above. Alternately, you can take one step backwards and assume we wish to preserve the validity of $x^s x^t = x^{s+t}$, then assert a definition of negative exponents from that. $\endgroup$ – G. H. Faust Jan 14 '14 at 21:46
  • $\begingroup$ I feel like I must have some background missing here, because this stuff seems super difficult, and I don't think I ever proved stuff like $x^0 = 1$ (which seems related to all this, and I'm not even sure how I would go about doing that). That's tangential to the question at hand, though. $\endgroup$ – M T Jan 14 '14 at 21:56
  • $\begingroup$ @Michael It's probably just that you're trying to prove things which aren't necessarily set in stone. Exponentials are just arbitrary functions, we could have defined some of their behaviour differently and kept the fundamental part (repeated multiplication) as is without issue. Their behaviour as currently defined is a result of wanting them to have nice properties, from which solid definitions follow. (Though $x^0 = 1$ would follow without the identity/property I mentioned, possibly even without extending powers to all the positive reals.) $\endgroup$ – G. H. Faust Jan 14 '14 at 22:10

After the proof that for every positive number $y$ and for every positive integer $n$ there exists a unique positive real number $x$ such that $x^n=y$, Rudin proves that $(ab)^{1/n}=a^{1/n}b^{1/n}$ for positive reals $a,b$ and positive integer $n$. In the proof, he defines some numbers $\alpha$ and $\beta$ suche that $\alpha^n=a$ and $\beta^n=b$ and proceed with his proof. In a similar fashion, for this problem we can define $\alpha$ and $\beta$ to be the unique reals such that $$ \alpha^n=b^m \qquad \beta^q=b^p $$ (Note that we can always assume that $n>0$ and $q>0$). And then show that $\alpha=\beta$ using the uniqueness of the roots. Also I think that it is more convenient to read $m/n=p/q$ as $mq=pn$.


I think that a simpler way is first to show that, for every $b>1$ and integers $m,n>0$, we have $$(b^{1/n})^m=(b^m)^{1/n}\qquad (*)$$ or, equivalently, that $((b^{1/n})^m)^n=b^m$. You should be able to do this with the property $$a^{pq}=(a^p)^q \qquad \forall a>1,\ p,q>0\text{ integers,}\qquad(**)$$ which should already have been shown (by induction, probably). With this, you can "commute roots and powers", so you can, for example, put all power to the right and all roots to the left and use property $(**)$ to simplify things.

Now, given $b>1$ and $p,q,m,n>0$ integers such that $p/q=m/n$, you want to show that $(b^m)^{1/n}=(b^p)^{1/q}$, or, equivalently, $((b^m)^{1/n})^q=b^p$. You should do this easily using $(*)$ and $(**)$ (and, of course, the equality $pn=qm$).

  • $\begingroup$ This really makes me feel that I somehow don't have the background for analysis yet, since (**) isn't proved anywhere in Rudin's book, and I don't remember proving it before. On the surface it doesn't look so hard though. I guess it's just frustrating because when I see so many different ways of proving things, I have trouble approaching new proofs and really seeing where to start. $\endgroup$ – M T Jan 14 '14 at 22:49
  • $\begingroup$ (Don't get me wrong though; this is still a really helpful answer!) $\endgroup$ – M T Jan 14 '14 at 22:49
  • $\begingroup$ But don't get frustrated. Try to prove $(**)$ then. This simply says that "if we write $(a^p)$ $q$ times, we get $pq$ copies of $a$, or, equivalenty, that $\underbrace{\underbrace{(a\cdots a)}_{p\text{ times}}\cdots\underbrace{(a\cdots a)}_{p\text{ times}}}_{q\text{ times}}=\underbrace{a\cdots a}_{pq\text{ times}}$. This is (intuitively) implied by "multiplication being recursive addition", and that you're just counting the $a$'s. You should really focus, at this point, on "what does the definitions mean" (for example, the definition of exponatiation as recursive multiplication). $\endgroup$ – Luiz Cordeiro Jan 16 '14 at 11:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.