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