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Question from Rudin chapter 1:

Let $b \in \mathbb{R}$ and $ b > 0$ . Let $m,n,p,q \in \mathbb{Z}$ with $ n,q>0$ . Assume $r = m/n = p/q$. Prove that $(b^m)^{1/n} = (b^p)^{1/q}$.

The attempt: I think I'm supposed to use the theorem that says "for every real x>0 and every integer n>0, there exists a unique real y such that $y^n=x $."

Since $b^m>0$, I know there is an $x \in \mathbb{R}$ such that $x^n= b^m$ and since $b^p>0$ I know there is a y in R such that $y^q = b^p$.

But I don't know what to do next? I mean clearly $(b^m)^{1/n} = (b^p)^{1/q}$ but how do you show it?

Henry T. Horton
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Since $b > 0$, also $b^m > 0$. By definition, $(b^m)^{(1/n)}$ is the unique $\alpha > 0$ such that $\alpha^n = b^m$. Let $\beta = (\alpha^q)^n > 0$. We have $$\beta = {\underbrace{(\alpha^q)}_{>0}}^n = \alpha^{qn} = (\alpha^n)^q = (b^m)^q = b^{mq} \overset{m/n=p/q}{=} b^{pn} = {\underbrace{(b^p)}_{>0}}^n$$ Since the number $x > 0$ with $x^n = \beta$ is unique, we get $x = \alpha^q = b^p$. This shows $\alpha = (b^p)^{(1/q)}$. Now from the definition of $\alpha$, the claim follows.

azimut
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  • How do we know that $(b^m)^(1/n) > 0 $? –  Feb 12 '13 at 00:21
  • $(b^m)^{(1/n)}$ is defined as the unique rational number $\alpha$ such that $\alpha^n = b^m$. So $\alpha = (b^m)^{(1/n)} > 0$ by definition. Note that for applying the definition, $b^m > 0$ was needed (which we have since $b > 0$). – azimut Feb 12 '13 at 00:32