# Proof about m/nth root of a prime.

1. Jul 17, 2012

### cragar

Lets take a prime number and raise it to m/n where m and n are coprime. x,y are coprime
and I want to show that this is irrational.
Proof: lets assume for the sake of contradiction that
$P^{\frac{m}{n}}=\frac{x}{y}$
P is prime and m,n,x,y are integers.
no we take both sides to the nth power and then multiply the y term over.
$y^n P^m=x^n$
now we factor y and x into their prime factorization.
${p^a.....{P_{t}}^b}^n P^m={{P_{q}}^c........}^n$
okay so if the left side is equal to the right side.
on the left side we know that P has at least m factors, but if P is also contained
in y then it has m+n factors. but if x and y are coprime then x has no common factors with y so y cant have factors of P in it. so now if x contains multiples of P , in order to have the same amount of factors it must be true that ne=m where e is the number of factors of P in x.
but this would imply that m and n are not coprime therefore this is a contradiction and our original number is irrational.

2. Jul 17, 2012

### chiro

Hey cragar.

The thing I see is that when you get Pm = xn/yn, then it means that (x/y) have to be a prime since primes have no factors other than 1 or itself. But the only way for this to happen is if x and y are not co-prime and also relevant powers of that prime.

Thus you have a contradiction and you have proved the result that there exists no rational form given x and y are co-prime. Also because we are dealing with a prime, we know that the RHS of the above must be an integer and also that it has a particular decomposition since primes only have factors of 1 and itself.

Edit: Correction should have been / instead of *

Last edited: Jul 17, 2012
3. Jul 17, 2012

### cragar

how did you get $p^m=x^ny^n$

4. Jul 17, 2012

### chiro

Sorry, I've changed it: should have been / instead of *

5. Jul 18, 2012

### cragar

okay I see, ya that is also a contradiction too and right off the bat.