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The proof I got for 1.6 a. in Baby Rudin is different from those I've seen elsewhere. I wonder if I made a mistake?

1.6: Fix $>1$. Prove that if $,,,$ are integers, $>0,>0$ and $=\frac{}{}=\frac{}{}$, then $(^)^{\frac{1}{}}=(^)^{\frac{1}{}}$

My attempt:

Since $=\frac{}{}=\frac{}{}$, then $mq = pn$

Therefore, $b^{mq} = b^{pn}$

So, $({b^m})^q = ({b^p})^n$

Therefore, $\bigl(({b^m})^q\bigr)^{\frac{1}{qn}} = \bigl(({b^p})^n\bigr)^{\frac{1}{qn}}$

So, $({b^m})^{\frac{1}{n}} = ({b^p})^{\frac{1}{q}}$

Bill Dubuque
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lightnesscaster
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    You use the fact that $(B^q)^{1/qn}=B^{1/n}$ in the last step, but the well posedness need to be proved by this exercise. – Liufeng Yang Feb 06 '21 at 13:17
  • @LiufengYang Well I'm using the fact that $(b^p)^n)^{1/(qn)}$ = $(b^p)^{1/q}$, which doesn't seem to be an equivalent property. – lightnesscaster Feb 06 '21 at 13:23
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    Yes, the fact is weaker than the exercise and can be checked by definition easily, as $((B^q)^{1/qn})^{qn}=B^q=(B^{1/n})^{qn}\Rightarrow (B^q)^{1/qn}=B^{1/n}$. Combining with this, your proof might be more complete. – Liufeng Yang Feb 06 '21 at 13:38
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    What is the definition of $b^{1/k}$ in Rudin? – Justin Young Feb 06 '21 at 15:00
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    @JustinYoung - Rudin does the most basic definition: $b^{1/k}$ is the unique positive real number such that $(b^{1/k})^k = b$. – Paul Sinclair Feb 07 '21 at 00:34

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