Question

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

Answer 1

damn lost it

Answer 2

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

Answer 3

want more details?

Answer 4

Sorry for rebumping, the connection was so bad that it failed to notify me of your answer.

Answer 5

Much more concise than other answers on sqrt2 proof- thanks!