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Prove that the nth root of a prime number is irrational, where n>1.

Mathematics
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damn lost it
site is not working for me idea is start by contradiction \[p=(\frac{a}{b})^n\] so \[b^np=a^n\] then note that all the primes in \(a^n\) how powers that are as multiple of \(n\) whereas on the left the power of \(p\) is not
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Sorry for rebumping, the connection was so bad that it failed to notify me of your answer.
Much more concise than other answers on sqrt2 proof- thanks!

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