Proof that $\sqrt[p]{n}$ is irrational if $n$ is not a perfect pth power

by Elias   Last Updated September 11, 2019 18:20 PM - source

i'm using Courant's book for self study, i would like to know if my proof that $\sqrt[p]{n}$ is irrational if $n$ is not a perfect pth power. Also would appreciate if someone do know where i can look for the solutions.


If $n$ is not a perfect pth power i can express it in terms of $n^{pm + 1}$. Assuming that $\sqrt[p]{n^{pm+1}}$ is rational i would have:

$$ \sqrt[p]{n^{pm+1}} = \frac{k}{j} \\ n^{pm + 1} = \frac{k^p}{j^p} \\ n^{pm + 1}j^p = k^p \\ n^mn^{\frac{1}{p}}j = k \\ j^p = \frac{k^p}{n^{pm+1}} \\ j = \frac{k}{n^mn^{\frac{1}{p}}} $$

Which is a contradiction because they do have a common factor.



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