## henpen Group Title Prove that the nth root of a prime number is irrational, where n>1. one year ago one year ago

1. satellite73 Group Title

damn lost it

2. satellite73 Group Title

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

3. satellite73 Group Title

want more details?

4. henpen Group Title

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

5. henpen Group Title

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